Create all numbers from 1-100 using 1,3,3,6

Create all the numbers from $1$ to $100$ using the numbers $1$,$3$,$3$, and $6$.

You can only use each number once, except for the $3$, of which you have two.

You can use addition ($x+y$), subtraction ($x-y$), division ($frac{x}{y}$), multiplication ($xtimes y$), exponentiation ($x^y$) and roots ($sqrt[leftroot{-2}uproot{2}x]{y}$).

You can combine numbers like $1$ and $3$ to $13$ etc.

You must use all numbers.
EDIT: no factorials, in squareroots 2 is hidden, no combining results of operations, you can use parentheses and start with negative numbers, no rounding and no decimal points. Good Luck

edited Jan 28 at 10:19

asked Jan 27 at 22:00

Michał Uraszewski

6116

4

$begingroup$
Can we combine the results of operations? For example, is $(1+3) | 36 = 436$ (where | indicates concatenation)
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
If we need to take a square root, is the two implied?
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
can we use factorial?
$endgroup$
– Omega Krypton
Jan 27 at 23:23

3

$begingroup$
Factorials are most likely out, but what about parentheses, unary minus (like starting with -1) and decimal points?
$endgroup$
– Bass
Jan 28 at 0:25

1

$begingroup$
If decimal is allowed then round would probably valid too?
$endgroup$
– Mukyuu
Jan 28 at 3:31

|
show 3 more comments

Create all the numbers from $1$ to $100$ using the numbers $1$,$3$,$3$, and $6$.

You can only use each number once, except for the $3$, of which you have two.

You can use addition ($x+y$), subtraction ($x-y$), division ($frac{x}{y}$), multiplication ($xtimes y$), exponentiation ($x^y$) and roots ($sqrt[leftroot{-2}uproot{2}x]{y}$).

You can combine numbers like $1$ and $3$ to $13$ etc.

You must use all numbers.
EDIT: no factorials, in squareroots 2 is hidden, no combining results of operations, you can use parentheses and start with negative numbers, no rounding and no decimal points. Good Luck

edited Jan 28 at 10:19

asked Jan 27 at 22:00

Michał Uraszewski

6116

4

$begingroup$
Can we combine the results of operations? For example, is $(1+3) | 36 = 436$ (where | indicates concatenation)
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
If we need to take a square root, is the two implied?
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
can we use factorial?
$endgroup$
– Omega Krypton
Jan 27 at 23:23

3

$begingroup$
Factorials are most likely out, but what about parentheses, unary minus (like starting with -1) and decimal points?
$endgroup$
– Bass
Jan 28 at 0:25

1

$begingroup$
If decimal is allowed then round would probably valid too?
$endgroup$
– Mukyuu
Jan 28 at 3:31

|
show 3 more comments

Create all the numbers from $1$ to $100$ using the numbers $1$,$3$,$3$, and $6$.

You can only use each number once, except for the $3$, of which you have two.

You can use addition ($x+y$), subtraction ($x-y$), division ($frac{x}{y}$), multiplication ($xtimes y$), exponentiation ($x^y$) and roots ($sqrt[leftroot{-2}uproot{2}x]{y}$).

You can combine numbers like $1$ and $3$ to $13$ etc.

You must use all numbers.
EDIT: no factorials, in squareroots 2 is hidden, no combining results of operations, you can use parentheses and start with negative numbers, no rounding and no decimal points. Good Luck

edited Jan 28 at 10:19

asked Jan 27 at 22:00

Michał Uraszewski

6116

Create all the numbers from $1$ to $100$ using the numbers $1$,$3$,$3$, and $6$.

You can only use each number once, except for the $3$, of which you have two.

You can use addition ($x+y$), subtraction ($x-y$), division ($frac{x}{y}$), multiplication ($xtimes y$), exponentiation ($x^y$) and roots ($sqrt[leftroot{-2}uproot{2}x]{y}$).

You can combine numbers like $1$ and $3$ to $13$ etc.

You must use all numbers.
EDIT: no factorials, in squareroots 2 is hidden, no combining results of operations, you can use parentheses and start with negative numbers, no rounding and no decimal points. Good Luck

calculation-puzzle formation-of-numbers

edited Jan 28 at 10:19

asked Jan 27 at 22:00

Michał Uraszewski

6116

edited Jan 28 at 10:19

asked Jan 27 at 22:00

Michał Uraszewski

6116

edited Jan 28 at 10:19

asked Jan 27 at 22:00

Michał Uraszewski

6116

asked Jan 27 at 22:00

Michał Uraszewski

6116

asked Jan 27 at 22:00

Michał Uraszewski

6116

4

$begingroup$
Can we combine the results of operations? For example, is $(1+3) | 36 = 436$ (where | indicates concatenation)
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
If we need to take a square root, is the two implied?
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
can we use factorial?
$endgroup$
– Omega Krypton
Jan 27 at 23:23

3

$begingroup$
Factorials are most likely out, but what about parentheses, unary minus (like starting with -1) and decimal points?
$endgroup$
– Bass
Jan 28 at 0:25

1

$begingroup$
If decimal is allowed then round would probably valid too?
$endgroup$
– Mukyuu
Jan 28 at 3:31

|
show 3 more comments

4

$begingroup$
Can we combine the results of operations? For example, is $(1+3) | 36 = 436$ (where | indicates concatenation)
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
If we need to take a square root, is the two implied?
$endgroup$
– Hugh
Jan 27 at 22:14

1

$begingroup$
can we use factorial?
$endgroup$
– Omega Krypton
Jan 27 at 23:23

3

$begingroup$
Factorials are most likely out, but what about parentheses, unary minus (like starting with -1) and decimal points?
$endgroup$
– Bass
Jan 28 at 0:25

1

$begingroup$
If decimal is allowed then round would probably valid too?
$endgroup$
– Mukyuu
Jan 28 at 3:31

Can we combine the results of operations? For example, is $(1+3) | 36 = 436$ (where | indicates concatenation)

– Hugh
Jan 27 at 22:14

If we need to take a square root, is the two implied?

– Hugh
Jan 27 at 22:14

can we use factorial?

– Omega Krypton
Jan 27 at 23:23

Factorials are most likely out, but what about parentheses, unary minus (like starting with -1) and decimal points?

– Bass
Jan 28 at 0:25

If decimal is allowed then round would probably valid too?

– Mukyuu
Jan 28 at 3:31

|
show 3 more comments

8 Answers
8

active

oldest

votes

These get harder with larger numbers, but here are the first 40 (and a couple of the easier ones after that) with the digits in order:

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50 (getting much harder now, so from now on, only the easier ones)

41: $ $

42: $ (1+3+3)times 6$

43: $ $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51 to 60

51: $ $

52: $ $

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ $

58: $ (1+3)^3-6$

59: $ $

60: $ (1+3times3)times6$

61 to 70

61: $ $

62: $ $

63: $ $

64: $ (1+3/3)^6$

65: $ $

66: $ $

67: $ $

68: $ $

69: $ $

70: $ (1+3)^3+6$

71 to 80

71: $ $

72: $ (1+3)times 3 times 6$

73: $ $

74: $ $

75: $ $

76: $ $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81 to 90

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ $

88: $ $

89: $ $

90: $ $

91 to 100

91: $ $

92: $ $

93: $ $

94: $ $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ $

100: $ $

edited Jan 28 at 13:29

answered Jan 28 at 1:17

Bass

29.2k470178

$begingroup$
Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM=
$endgroup$
– Outman
Jan 28 at 11:19

add a comment |

Here's a solution for most of them. The remaining 21 are impossible.

Most of them only use simple arithmetic.

Some of them use exponentiation.

Some use square roots.

Edit: Using only the binary operators (including digit concatenation), the number of possible combinations is pretty small (arrange three binary operators, then fill in the four numbers in some order) and I double-checked all of them with a computer. Most numbers are solvable with basic arithmetic, some require exponentiation, some require negation or square roots, and the rest are apparently impossible to solve without some extra operation such as rounding.

Accordingly, it is provably impossible to construct the following twenty-one numbers:

41, 44, 46, 47, 56, 68, 69, 74, 77, 79, 83, 85, 86, 89, 90, 91, 92, 95, 97, 98, 100.

The rest can be constructed as follows:

1-25 (complete)

1. (6+1)-(3+3)

2. (3*3)-(6+1)

3. (3*3)-(6*1)

4. 13-(6+3)

5. 36-31

6. 3*(3+1)-6

7. (6+1)+(3-3)

8. (6-3)3-1

9. 6(3-1)-3

10. 3+13-6

11. 6+3+3-1

12. (6+3+3)*1

13. 6+3+3+1

14. (6*3)-(3+1)

15. 13 + (6/3)

16. 13 + (6-3)

17. 33-16

18. (1+3)*3 + 6

19. (6 + 1/3)*3

20. (3*6)+(3-1)

21. (13-6)*3

22. 13+3+6

23. 36-13

24. (6+1)*3 +3

25. 16+(3*3)

26-50 (except 41, 44, 46, 47)

26. 33 - (6+1)

27. 33 - (6*1)

28. 61 - 33

29. 31 - (6/3)

30. (6+3+1)*3

31. 13 + (6*3)

32. 63-31

33. 13*3 - 6

34. 31+(6-3)

35. 6*(3+3) -1

36. (6+3)(3+1)

37. 6(3+3) + 1

38. 33+(6-1)

39. 33+(6*1)

40. 31+3+6

41. round[36 + √(31)]

42. (3+3+1)*6

43. 16 + (3^3)

44. round[(3-√3)^16]

45. (3*13)+6

46. round[3 * (13 + √6)]

47. round[31√3] - 6

48. (3*3 - 1) *6

49. 36+13

50. 63-13

51-75 (except 56, 68, 69, 74)

51. 16*3 + 3

52. 61 - (3*3)

53. (3*3*6)-1

54. 1*3*3*6

55. 61-(3+3)

56. √(3136) (!)

57. (13+6)*3

58. (3+1)^3 - 6

59. 63 - (3+1)

60. (63*1) - 3

61. 61 + (3 - 3)

62. 61 + (3/3)

63. (6+1)*(3*3)

64. 63 + (1^3)

65. 63 + (3-1)

66. 63 + (3*1)

67. 36 + 31

68. round[63 + √31]

69. round[6 √133]

70. 61+(3*3)

71. (6^3)/3 - 1

72. 36 * (3-1)

73. (6^3)/3 + 1

74. round[3 √613]

75. 13*6 - 3

76-100 (except: 77, 79, 83, 85*, 86, 89,90,91,92, 95, 97, 98, 100.)

76. 63+13

77. round[61 * (3-√3)]

78. 13 * √(36)

79. round[6*13 + √3]

80. √(3^6) * 3 - 1

81. 13*6 + 3

82. √(3^6) * 3 + 1

83. round[(3+31)*√6]

84. 3^√(16) + 3

85. -----

86. round[√(6+√3) * 31]

87. (31*3) - 6

88. 61 + (3^3)

89. round[63 * √(3-1)]

90. round[13 * √(6/3)]

91. round[16 * √33]

92. round[3^√(3*6-1)]

93. 31 * (6-3)

94. 63+31

95. round[3*31 + √6]

96. (13+3)*6

97. round[(√3)^6 * √13]

98. round[36 * (1+√3)]

99. (3*31)+6

100. round[3*(31+√6)]

If you're allowed to concatenate the results of operations (e.g. (3+1)|5 = 45 ) then a solution for 85 is :

85. (3*3)|1 - 6

Python:



from itertools import permutations

from math import sqrt, floor, ceil





concat_literal_numbers_only = True



ops = { "+" : lambda a,b: a+b,

        "-" : lambda a,b: a - b,

        "/" : lambda a,b : a/float(b),

        "*" : lambda a,b : a*b,

        "^" : lambda a,b : a**b,

        "C" : lambda a,b : float(str(a) + str(b)),

        "n" : lambda a : -a,

        "s" : lambda a : sqrt(a),

        #"f" : lambda a : floor(a)

}



arity = {"+" : 2,

         "-" : 2,

         "/" : 2,

         "*" : 2,

         "^" : 2,

         "C" : 2,

         "n" : 1,

         "s" : 1,

         "f" : 1,

}





# print ops["/"](1,3)





# args: number of open args available

# nums: available digits to be used

# ops : tuple indicating commands used so far







def evaluate(cmds) :

    """Consume the list of commands in prefix notation, producing a pair (ans, unconsumed_symbols)"""



    x = cmds.pop(0)

    if not ops.get(x) :

        return (x, cmds)

    else :

        args = 

        for y in range(arity[x]) :

            try :

                (a, cmds) = evaluate(cmds)

                args += [a]

            except OverflowError :

                return (None, None)

        return (ops.get(x)(*args), cmds)



def score(ops):

    ret = 0

    ret += ops.count("+")

    ret += 1.1*ops.count("-")

    ret += 2 * ops.count("*")

    ret += 3 * ops.count("/")

    ret += 3 * ops.count("n")

    ret += 4 * ops.count("^")

    ret += 4 * ops.count("s")

    ret += 4 * ops.count("f")

    ret += 4 * ops.count("w")



    # ret += 4 * ops.count("fs")

    # ret += 4 * ops.count("cs")

    return ret





agenda = [{"args" : 1, "nums" : [1,3,3,6], "ops" : }]

seen = {}





only_search_for = None

ret = 



def finish(ops) :

    global ret

    global seen



    ops_tmp = ops[:]

    try :

        n,_ = evaluate(ops_tmp)

    except :

        n = None



    if n is None or not (0  score(ops) :

        seen[n] = ops



        print ops,"t",n









while agenda :

    x = agenda.pop(0)



    if not x["nums"] and not x["args"] : # finished: used up all numbers; no open spaces.

        finish(x["ops"])



    if len(x["nums"]) == x["args"] : # fill in numbers only

        for nums in set(permutations(x["nums"])) :

            finish(x["ops"] + list(nums))



            # print {"args" : 0,

            #        "nums" : ,

            #        "ops" : x["ops"] + list(nums)}



    elif len(x["nums"]) > x["args"] :

        # add new operators









        for op in ops.keys() :

            if arity[op] == 1 and x["ops"] and x["ops"][-1] == op :

                continue # limit repeated unary operations

            if arity[op] == 1 and x["ops"] and arity.get(x["ops"][-1]) == 1 :

                continue # limit repeated unary operations





            if (concat_literal_numbers_only and x["ops"] and (x["ops"][-1] == "C" or (len(x["ops"])>1 and x["ops"][-2] == "C")) and op != "C") :

                continue





            new_x = {"args" : x["args"] + arity[op] - 1,

                     "nums" : x["nums"],

                     "ops" : x["ops"] + [op]}

            agenda = [new_x] + agenda



        if x["args"] == 1 : 

            continue



        for n in set(x["nums"]) :

            new_nums = x["nums"][:]

            new_nums.remove(n)

            new_x = {"args" : x["args"] - 1,

                     "nums" : new_nums,

                     "ops" : x["ops"] + [n]}



            agenda = [new_x] + agenda



# SHOW HOW TO MAKE ALL OF THE NUMBERS

miss = 

for i in range(0+1,100+1) :

    if not seen.get(i) :

        miss += [i]

    print i, "t", seen.get(i, "---")



# SHOW WHICH NUMBERS WERE MISSED

print "missed: ", miss



# IF YOU'RE LOOKING FOR ALL POSSIBLE WAYS TO MAKE SOMETHING, SHOW THEM HERE.

if only_search_for is not None :

    ret = sorted(ret, key=score)

    for x in ret:

        print x

edited Feb 1 at 2:09

answered Jan 28 at 7:27

user326210

1712

$begingroup$
Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely.
$endgroup$
– Bass
Jan 29 at 9:47

$begingroup$
@Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions.
$endgroup$
– user326210
Jan 30 at 1:00

add a comment |

We can generate any integer using only $1$, $3$, $3$ and $6$ with the introduction of one special function.

The function in question is the Logarithm to an arbitrary base $b$ , or $log _{b} (x)$.

To begin, let's discuss square root stacking.
$sqrt{sqrt{a}}$ is equivalent to $sqrt[4]{a}$, and $sqrt{sqrt{sqrt{a}}}$ is equivalent to $sqrt[8]{a}$, which can be rewritten as $a^frac{1}{8}$. This pattern continues indefinitely; $a$ with $n$ square roots stacked to it is equal to $a^frac{1}{2^n}$.

The laws of logarithms state that $log _{b} (x^a) = a cdot log _{b} (x)$. If we take the logarithm to base $b$ or our previous square root stack, we get $log _{b} (a^frac{1}{2^n})$, or $frac{1}{2^n} cdot log _{b}a$. Setting $a$ and $b$ as 3 means that $log _{3} (sqrt{sqrt[...]{3}})$, with $n$ square roots, is equal to $frac{1}{2^n} cdot log _{3}3$, or $2^{-n}$. $(frac{1}{a^x} = a^{-x})$

$sqrt{sqrt{16}} = 2$, and $log _{b}(b^a) = a$. As such,

$0 = -log _{sqrt{sqrt{16}}}(log_{3}(3))$
$1 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{3}))$
$2 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{3}}))$
$3 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{3}}}))$
$4 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{sqrt{3}}}}))$

And so on, such that the amount of square roots is equal to your desired number.

This works for all integers; for negative numbers simply remove the $-$ at the start.

This was inspired by Numberphile's video on the four 4s.

answered Jan 28 at 14:27

user56759

311

add a comment |

Here are some:

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

22: (omega kyrpton did some) $3 * 6 + 3 + 1$

I will do more later.

edited Jan 28 at 15:54

answered Jan 27 at 22:33

Yout Ried

894119

add a comment |

Adding some more...

1-20: (Credits to @YoutRied)

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

21-29

21: $3 * 6 + 3 * 1$

22: $( 1 + 3 ) ! - ( 6 / 3 )$

23: $( 1 + 3 ) ! - ( 6 - 3 )$

24: $( 6 - 3 / 3 - 1 ) !$

25: $1 * 3 ^ 3 - floor(sqrt{6})$

26: $( 6 - 3 ) ^ 3 - 1$

27: $( 6 - 3 ) ^ 3 * 1$

28: $( 6 - 3 ) ^ 3 + 1$

29: $31 - 6 / 3$

41-50: (Credits to @Bass for 42, 45, 49)

41: $ (-1+3!)+36 $

42: $ (1+3+3)times 6$

43: $ 31 + 6 * ceil(sqrt{3})$

44: $floor( 1 * 3 * sqrt{6 ^ 3}) $

45: $ 13times3+6$

46: $ ceil(sqrt{6 ^ 3} + 31)$

47: $ floor(sqrt{sqrt{sqrt{sqrt{sqrt{31!}}}}})+36$

48: $6 * ( 3 * 3 - 1 )$

49: $13+36$

50: $ (6+1)^2 + 3 - 3$

51-60:

51: $( 3 * 6 - 1 ) * 3$

52: $( 3 + 3 + 1 ) * ceil(sqrt{6})$

53: $-1+( 3 * 3 * 6 )$

54: $ 1*3 * 3 * 6 $

55: $1+3*3*6$

56: $61-3!+floor(sqrt{3})$

57: $1*63-3!$

58: $1+63-3!$

59: $floor(sqrt{sqrt{sqrt{sqrt{sqrt{6^{(3-1)}}}}}}*3)$

60: $(1+3*3)*6$

61-70:

61: $63-3+1$

62: $63+1-ceil(sqrt{3})$

63: $63-floor(sqrt{3})+1$

64: $63+ceil(sqrt{3})-1$

65: $63+3-1$

66: $63+3*1$

67: $63+3+1$

68: $61+3!+floor(sqrt{3})$

69: $61+3!+ceil(sqrt{3})$

70: $61+3*3$

71-80

71: $(3+1)!*3-floor(sqrt{sqrt{6}})$

72: $(3+1)*3*6$

73: $(3+1)!*3+floor(sqrt{sqrt{6}})$

74: $(3+1)!*3+floor(sqrt{6})$

75: $(3+1)!*3+ceil(sqrt{6})$

76: $ceil(sqrt{sqrt{sqrt{ceil(sqrt{sqrt{sqrt{sqrt{sqrt{ceil(sqrt{3!!})!}}}}})}}})*(3*6+1)$

76:(much simpler) $13*6-ceil(sqrt{3})$

77: $13*6-floor(sqrt{3})$

78: $13*floor(sqrt{3})*6$

79: $13*6+floor(sqrt{3})$

80: $13*6+ceil(sqrt{3})$

81-90:

81: $(6+3)^{(3-1)}$

82: $(6+3)^{ceil(sqrt{3})}+1$

edited Jan 28 at 8:36

answered Jan 27 at 23:31

Omega Krypton

3,6941338

$begingroup$
Who said you could use factorials?
$endgroup$
– Yout Ried
Jan 28 at 0:20

$begingroup$
What are number 23 (plus you probably can't use factorials and 24? I don't get them.
$endgroup$
– Yout Ried
Jan 28 at 1:08

$begingroup$
Oops forgot a ")" and maybe you're missing a factorial for number 24
$endgroup$
– Yout Ried
Jan 28 at 1:16

add a comment |

Partial answer 1-50 (w/e 41,47):

$1= 1+3+3-6$
$2= 1 + (frac{6}{(3+3)})$
$3= 1^3+(frac{6}{3})$
$4= (frac{6}{3})+3-1$
$5= (frac{6}{3})+3^1$
$6= 6^1+3-3$
$7= 6+1-3+3$
$8= 6 + 3 - 1^3$
$9= 1^3 * (3+6)$
$10= 1^3 +3+6$
$11= 13 - (frac{6}{3})$
$12= 6+3+3^1$
$13= 6+3+3+1$
$14= 6*3 - 3 - 1$
$15= 6*3 - 3^1$
$16= 16 + 3 - 3$
$17= 16 + (frac{3}{3})$
$18= (frac{6*3}{1^3})$
$19= 6*3+1^3$
$20= 6*3+3-1$
$21= 6*3+3^1$
$22= 6*3+3+1$
$23= 36-13$
$24= 6*(3+1^3)$
$25= 16+(3*3)$
$26= 13*(frac{6}{3})$
$27= 33-6^1$
$28= 33-6+1$
$29= 31-(frac{6}{3})$
$30= 6*(3+3-1)$
$31= 13+3*6$
$32= 3^3+6-1$
$33= (frac{33}{1^6})$
$34= 33+1^6$
$35= (3+3)*6-1$
$36= (3+3)^1*6$
$37= 1+(3+3)*6$
$38= 33+6-1$
$39= 33+6^1$
$40= 1+33+6$
$41= $
$42= (1+3+3)*6$
$43= 16+3^3$
$44= round(sqrt{6^3}*3^1)$
$45= 3*3*(6-1)$
$46= ceil(sqrt{6^3}*3)+1$
$47= $
$48= ((3*3)-1)*6$
$49= 16+33$
$50= 63-13$

edited Jan 28 at 3:39

answered Jan 28 at 3:01

Mukyuu

340112

add a comment |

Alrighty, I'm piggybacking off of @OmegaKrypton else and adding some of my own.

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1^3 +3/6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1^3 + 3+6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 3*6+3!-1$ or $ (6-1)^(3!/3)$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ 36-3!-1$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3+16$

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ 16*(3!-3)$

49: $13+36$

50: $ 63-13$

I added a few. It's getting late here; will come back tomorrow.

edited Jan 28 at 6:12

Omega Krypton

3,6941338

answered Jan 28 at 2:58

Brandon_J

1,28427

add a comment |

Expanding on Bass's answer, I added some new numbers.

(I lost track on which numbers I added, though 1-40 is all Bass)

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$
11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3 + 16 $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51: $ 16*3+3 $

52: $ 61 - 3times3$

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ (6times3+1)times3$

58: $ (1+3)^3-6$

59: $ 63 - 3 - 1$

60: $ (1+3times3)times6$

61: $ 63 - 3 + 1$

62: $ 63 - 1^3$

63: $ 63 * 1^3$

64: $ (1+3/3)^6$

65: $ 63 + 3 - 1$

66: $ 1 times (63 + 3) $

67: $ 63 + 3 + 1$

68: $ $

69: $ $

70: $ (1+3)^3+6 $
71: $ 6^3 / 3 - 1 $

72: $ (1+3)times 3 times 6$

73: $ 6^3 / 3 + 1 $

74: $ $

75: $ 3^(3+1) - 6$

76: $ 63+13 $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ 3^(3+1) + 6$

88: $ 61 + 3^3 $

89: $ $

90: $ $

91: $ $

92: $ $

93: $ $

94: $ 33 + 61 $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ 31times3+6$

100: $ $

Only need 41, 44, 46, 47, 50, 56, 68, 69, 74, 77, 79, 83, 84, 85, 86, 89, 90, 91, 92, 93, 94, 95, 97, 98, and 100 now!

edited Jan 30 at 3:49

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

New contributor

$begingroup$
Wow that is A LOT of numbers
$endgroup$
– North
Jan 30 at 3:07

add a comment |

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8 Answers
8

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8 Answers
8

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These get harder with larger numbers, but here are the first 40 (and a couple of the easier ones after that) with the digits in order:

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50 (getting much harder now, so from now on, only the easier ones)

41: $ $

42: $ (1+3+3)times 6$

43: $ $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51 to 60

51: $ $

52: $ $

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ $

58: $ (1+3)^3-6$

59: $ $

60: $ (1+3times3)times6$

61 to 70

61: $ $

62: $ $

63: $ $

64: $ (1+3/3)^6$

65: $ $

66: $ $

67: $ $

68: $ $

69: $ $

70: $ (1+3)^3+6$

71 to 80

71: $ $

72: $ (1+3)times 3 times 6$

73: $ $

74: $ $

75: $ $

76: $ $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81 to 90

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ $

88: $ $

89: $ $

90: $ $

91 to 100

91: $ $

92: $ $

93: $ $

94: $ $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ $

100: $ $

edited Jan 28 at 13:29

answered Jan 28 at 1:17

Bass

29.2k470178

$begingroup$
Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM=
$endgroup$
– Outman
Jan 28 at 11:19

add a comment |

These get harder with larger numbers, but here are the first 40 (and a couple of the easier ones after that) with the digits in order:

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50 (getting much harder now, so from now on, only the easier ones)

41: $ $

42: $ (1+3+3)times 6$

43: $ $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51 to 60

51: $ $

52: $ $

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ $

58: $ (1+3)^3-6$

59: $ $

60: $ (1+3times3)times6$

61 to 70

61: $ $

62: $ $

63: $ $

64: $ (1+3/3)^6$

65: $ $

66: $ $

67: $ $

68: $ $

69: $ $

70: $ (1+3)^3+6$

71 to 80

71: $ $

72: $ (1+3)times 3 times 6$

73: $ $

74: $ $

75: $ $

76: $ $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81 to 90

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ $

88: $ $

89: $ $

90: $ $

91 to 100

91: $ $

92: $ $

93: $ $

94: $ $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ $

100: $ $

edited Jan 28 at 13:29

answered Jan 28 at 1:17

Bass

29.2k470178

$begingroup$
Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM=
$endgroup$
– Outman
Jan 28 at 11:19

add a comment |

These get harder with larger numbers, but here are the first 40 (and a couple of the easier ones after that) with the digits in order:

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50 (getting much harder now, so from now on, only the easier ones)

41: $ $

42: $ (1+3+3)times 6$

43: $ $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51 to 60

51: $ $

52: $ $

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ $

58: $ (1+3)^3-6$

59: $ $

60: $ (1+3times3)times6$

61 to 70

61: $ $

62: $ $

63: $ $

64: $ (1+3/3)^6$

65: $ $

66: $ $

67: $ $

68: $ $

69: $ $

70: $ (1+3)^3+6$

71 to 80

71: $ $

72: $ (1+3)times 3 times 6$

73: $ $

74: $ $

75: $ $

76: $ $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81 to 90

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ $

88: $ $

89: $ $

90: $ $

91 to 100

91: $ $

92: $ $

93: $ $

94: $ $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ $

100: $ $

edited Jan 28 at 13:29

answered Jan 28 at 1:17

Bass

29.2k470178

These get harder with larger numbers, but here are the first 40 (and a couple of the easier ones after that) with the digits in order:

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50 (getting much harder now, so from now on, only the easier ones)

41: $ $

42: $ (1+3+3)times 6$

43: $ $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51 to 60

51: $ $

52: $ $

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ $

58: $ (1+3)^3-6$

59: $ $

60: $ (1+3times3)times6$

61 to 70

61: $ $

62: $ $

63: $ $

64: $ (1+3/3)^6$

65: $ $

66: $ $

67: $ $

68: $ $

69: $ $

70: $ (1+3)^3+6$

71 to 80

71: $ $

72: $ (1+3)times 3 times 6$

73: $ $

74: $ $

75: $ $

76: $ $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81 to 90

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ $

88: $ $

89: $ $

90: $ $

91 to 100

91: $ $

92: $ $

93: $ $

94: $ $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ $

100: $ $

edited Jan 28 at 13:29

answered Jan 28 at 1:17

Bass

29.2k470178

edited Jan 28 at 13:29

answered Jan 28 at 1:17

Bass

29.2k470178

answered Jan 28 at 1:17

Bass

29.2k470178

answered Jan 28 at 1:17

Bass

29.2k470178

$begingroup$
Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM=
$endgroup$
– Outman
Jan 28 at 11:19

add a comment |

$begingroup$
Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM=
$endgroup$
– Outman
Jan 28 at 11:19

Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM=

– Outman
Jan 28 at 11:19

add a comment |

Here's a solution for most of them. The remaining 21 are impossible.

Most of them only use simple arithmetic.

Some of them use exponentiation.

Some use square roots.

Accordingly, it is provably impossible to construct the following twenty-one numbers:

41, 44, 46, 47, 56, 68, 69, 74, 77, 79, 83, 85, 86, 89, 90, 91, 92, 95, 97, 98, 100.

The rest can be constructed as follows:

1-25 (complete)

1. (6+1)-(3+3)

2. (3*3)-(6+1)

3. (3*3)-(6*1)

4. 13-(6+3)

5. 36-31

6. 3*(3+1)-6

7. (6+1)+(3-3)

8. (6-3)3-1

9. 6(3-1)-3

10. 3+13-6

11. 6+3+3-1

12. (6+3+3)*1

13. 6+3+3+1

14. (6*3)-(3+1)

15. 13 + (6/3)

16. 13 + (6-3)

17. 33-16

18. (1+3)*3 + 6

19. (6 + 1/3)*3

20. (3*6)+(3-1)

21. (13-6)*3

22. 13+3+6

23. 36-13

24. (6+1)*3 +3

25. 16+(3*3)

26-50 (except 41, 44, 46, 47)

26. 33 - (6+1)

27. 33 - (6*1)

28. 61 - 33

29. 31 - (6/3)

30. (6+3+1)*3

31. 13 + (6*3)

32. 63-31

33. 13*3 - 6

34. 31+(6-3)

35. 6*(3+3) -1

36. (6+3)(3+1)

37. 6(3+3) + 1

38. 33+(6-1)

39. 33+(6*1)

40. 31+3+6

41. round[36 + √(31)]

42. (3+3+1)*6

43. 16 + (3^3)

44. round[(3-√3)^16]

45. (3*13)+6

46. round[3 * (13 + √6)]

47. round[31√3] - 6

48. (3*3 - 1) *6

49. 36+13

50. 63-13

51-75 (except 56, 68, 69, 74)

51. 16*3 + 3

52. 61 - (3*3)

53. (3*3*6)-1

54. 1*3*3*6

55. 61-(3+3)

56. √(3136) (!)

57. (13+6)*3

58. (3+1)^3 - 6

59. 63 - (3+1)

60. (63*1) - 3

61. 61 + (3 - 3)

62. 61 + (3/3)

63. (6+1)*(3*3)

64. 63 + (1^3)

65. 63 + (3-1)

66. 63 + (3*1)

67. 36 + 31

68. round[63 + √31]

69. round[6 √133]

70. 61+(3*3)

71. (6^3)/3 - 1

72. 36 * (3-1)

73. (6^3)/3 + 1

74. round[3 √613]

75. 13*6 - 3

76-100 (except: 77, 79, 83, 85*, 86, 89,90,91,92, 95, 97, 98, 100.)

76. 63+13

77. round[61 * (3-√3)]

78. 13 * √(36)

79. round[6*13 + √3]

80. √(3^6) * 3 - 1

81. 13*6 + 3

82. √(3^6) * 3 + 1

83. round[(3+31)*√6]

84. 3^√(16) + 3

85. -----

86. round[√(6+√3) * 31]

87. (31*3) - 6

88. 61 + (3^3)

89. round[63 * √(3-1)]

90. round[13 * √(6/3)]

91. round[16 * √33]

92. round[3^√(3*6-1)]

93. 31 * (6-3)

94. 63+31

95. round[3*31 + √6]

96. (13+3)*6

97. round[(√3)^6 * √13]

98. round[36 * (1+√3)]

99. (3*31)+6

100. round[3*(31+√6)]

If you're allowed to concatenate the results of operations (e.g. (3+1)|5 = 45 ) then a solution for 85 is :

85. (3*3)|1 - 6

Python:



from itertools import permutations

from math import sqrt, floor, ceil





concat_literal_numbers_only = True



ops = { "+" : lambda a,b: a+b,

        "-" : lambda a,b: a - b,

        "/" : lambda a,b : a/float(b),

        "*" : lambda a,b : a*b,

        "^" : lambda a,b : a**b,

        "C" : lambda a,b : float(str(a) + str(b)),

        "n" : lambda a : -a,

        "s" : lambda a : sqrt(a),

        #"f" : lambda a : floor(a)

}



arity = {"+" : 2,

         "-" : 2,

         "/" : 2,

         "*" : 2,

         "^" : 2,

         "C" : 2,

         "n" : 1,

         "s" : 1,

         "f" : 1,

}





# print ops["/"](1,3)





# args: number of open args available

# nums: available digits to be used

# ops : tuple indicating commands used so far







def evaluate(cmds) :

    """Consume the list of commands in prefix notation, producing a pair (ans, unconsumed_symbols)"""



    x = cmds.pop(0)

    if not ops.get(x) :

        return (x, cmds)

    else :

        args = 

        for y in range(arity[x]) :

            try :

                (a, cmds) = evaluate(cmds)

                args += [a]

            except OverflowError :

                return (None, None)

        return (ops.get(x)(*args), cmds)



def score(ops):

    ret = 0

    ret += ops.count("+")

    ret += 1.1*ops.count("-")

    ret += 2 * ops.count("*")

    ret += 3 * ops.count("/")

    ret += 3 * ops.count("n")

    ret += 4 * ops.count("^")

    ret += 4 * ops.count("s")

    ret += 4 * ops.count("f")

    ret += 4 * ops.count("w")



    # ret += 4 * ops.count("fs")

    # ret += 4 * ops.count("cs")

    return ret





agenda = [{"args" : 1, "nums" : [1,3,3,6], "ops" : }]

seen = {}





only_search_for = None

ret = 



def finish(ops) :

    global ret

    global seen



    ops_tmp = ops[:]

    try :

        n,_ = evaluate(ops_tmp)

    except :

        n = None



    if n is None or not (0  score(ops) :

        seen[n] = ops



        print ops,"t",n









while agenda :

    x = agenda.pop(0)



    if not x["nums"] and not x["args"] : # finished: used up all numbers; no open spaces.

        finish(x["ops"])



    if len(x["nums"]) == x["args"] : # fill in numbers only

        for nums in set(permutations(x["nums"])) :

            finish(x["ops"] + list(nums))



            # print {"args" : 0,

            #        "nums" : ,

            #        "ops" : x["ops"] + list(nums)}



    elif len(x["nums"]) > x["args"] :

        # add new operators









        for op in ops.keys() :

            if arity[op] == 1 and x["ops"] and x["ops"][-1] == op :

                continue # limit repeated unary operations

            if arity[op] == 1 and x["ops"] and arity.get(x["ops"][-1]) == 1 :

                continue # limit repeated unary operations





            if (concat_literal_numbers_only and x["ops"] and (x["ops"][-1] == "C" or (len(x["ops"])>1 and x["ops"][-2] == "C")) and op != "C") :

                continue





            new_x = {"args" : x["args"] + arity[op] - 1,

                     "nums" : x["nums"],

                     "ops" : x["ops"] + [op]}

            agenda = [new_x] + agenda



        if x["args"] == 1 : 

            continue



        for n in set(x["nums"]) :

            new_nums = x["nums"][:]

            new_nums.remove(n)

            new_x = {"args" : x["args"] - 1,

                     "nums" : new_nums,

                     "ops" : x["ops"] + [n]}



            agenda = [new_x] + agenda



# SHOW HOW TO MAKE ALL OF THE NUMBERS

miss = 

for i in range(0+1,100+1) :

    if not seen.get(i) :

        miss += [i]

    print i, "t", seen.get(i, "---")



# SHOW WHICH NUMBERS WERE MISSED

print "missed: ", miss



# IF YOU'RE LOOKING FOR ALL POSSIBLE WAYS TO MAKE SOMETHING, SHOW THEM HERE.

if only_search_for is not None :

    ret = sorted(ret, key=score)

    for x in ret:

        print x

edited Feb 1 at 2:09

answered Jan 28 at 7:27

user326210

1712

$begingroup$
Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely.
$endgroup$
– Bass
Jan 29 at 9:47

$begingroup$
@Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions.
$endgroup$
– user326210
Jan 30 at 1:00

add a comment |

Here's a solution for most of them. The remaining 21 are impossible.

Most of them only use simple arithmetic.

Some of them use exponentiation.

Some use square roots.

Accordingly, it is provably impossible to construct the following twenty-one numbers:

41, 44, 46, 47, 56, 68, 69, 74, 77, 79, 83, 85, 86, 89, 90, 91, 92, 95, 97, 98, 100.

The rest can be constructed as follows:

1-25 (complete)

1. (6+1)-(3+3)

2. (3*3)-(6+1)

3. (3*3)-(6*1)

4. 13-(6+3)

5. 36-31

6. 3*(3+1)-6

7. (6+1)+(3-3)

8. (6-3)3-1

9. 6(3-1)-3

10. 3+13-6

11. 6+3+3-1

12. (6+3+3)*1

13. 6+3+3+1

14. (6*3)-(3+1)

15. 13 + (6/3)

16. 13 + (6-3)

17. 33-16

18. (1+3)*3 + 6

19. (6 + 1/3)*3

20. (3*6)+(3-1)

21. (13-6)*3

22. 13+3+6

23. 36-13

24. (6+1)*3 +3

25. 16+(3*3)

26-50 (except 41, 44, 46, 47)

26. 33 - (6+1)

27. 33 - (6*1)

28. 61 - 33

29. 31 - (6/3)

30. (6+3+1)*3

31. 13 + (6*3)

32. 63-31

33. 13*3 - 6

34. 31+(6-3)

35. 6*(3+3) -1

36. (6+3)(3+1)

37. 6(3+3) + 1

38. 33+(6-1)

39. 33+(6*1)

40. 31+3+6

41. round[36 + √(31)]

42. (3+3+1)*6

43. 16 + (3^3)

44. round[(3-√3)^16]

45. (3*13)+6

46. round[3 * (13 + √6)]

47. round[31√3] - 6

48. (3*3 - 1) *6

49. 36+13

50. 63-13

51-75 (except 56, 68, 69, 74)

51. 16*3 + 3

52. 61 - (3*3)

53. (3*3*6)-1

54. 1*3*3*6

55. 61-(3+3)

56. √(3136) (!)

57. (13+6)*3

58. (3+1)^3 - 6

59. 63 - (3+1)

60. (63*1) - 3

61. 61 + (3 - 3)

62. 61 + (3/3)

63. (6+1)*(3*3)

64. 63 + (1^3)

65. 63 + (3-1)

66. 63 + (3*1)

67. 36 + 31

68. round[63 + √31]

69. round[6 √133]

70. 61+(3*3)

71. (6^3)/3 - 1

72. 36 * (3-1)

73. (6^3)/3 + 1

74. round[3 √613]

75. 13*6 - 3

76-100 (except: 77, 79, 83, 85*, 86, 89,90,91,92, 95, 97, 98, 100.)

76. 63+13

77. round[61 * (3-√3)]

78. 13 * √(36)

79. round[6*13 + √3]

80. √(3^6) * 3 - 1

81. 13*6 + 3

82. √(3^6) * 3 + 1

83. round[(3+31)*√6]

84. 3^√(16) + 3

85. -----

86. round[√(6+√3) * 31]

87. (31*3) - 6

88. 61 + (3^3)

89. round[63 * √(3-1)]

90. round[13 * √(6/3)]

91. round[16 * √33]

92. round[3^√(3*6-1)]

93. 31 * (6-3)

94. 63+31

95. round[3*31 + √6]

96. (13+3)*6

97. round[(√3)^6 * √13]

98. round[36 * (1+√3)]

99. (3*31)+6

100. round[3*(31+√6)]

If you're allowed to concatenate the results of operations (e.g. (3+1)|5 = 45 ) then a solution for 85 is :

85. (3*3)|1 - 6

Python:



from itertools import permutations

from math import sqrt, floor, ceil





concat_literal_numbers_only = True



ops = { "+" : lambda a,b: a+b,

        "-" : lambda a,b: a - b,

        "/" : lambda a,b : a/float(b),

        "*" : lambda a,b : a*b,

        "^" : lambda a,b : a**b,

        "C" : lambda a,b : float(str(a) + str(b)),

        "n" : lambda a : -a,

        "s" : lambda a : sqrt(a),

        #"f" : lambda a : floor(a)

}



arity = {"+" : 2,

         "-" : 2,

         "/" : 2,

         "*" : 2,

         "^" : 2,

         "C" : 2,

         "n" : 1,

         "s" : 1,

         "f" : 1,

}





# print ops["/"](1,3)





# args: number of open args available

# nums: available digits to be used

# ops : tuple indicating commands used so far







def evaluate(cmds) :

    """Consume the list of commands in prefix notation, producing a pair (ans, unconsumed_symbols)"""



    x = cmds.pop(0)

    if not ops.get(x) :

        return (x, cmds)

    else :

        args = 

        for y in range(arity[x]) :

            try :

                (a, cmds) = evaluate(cmds)

                args += [a]

            except OverflowError :

                return (None, None)

        return (ops.get(x)(*args), cmds)



def score(ops):

    ret = 0

    ret += ops.count("+")

    ret += 1.1*ops.count("-")

    ret += 2 * ops.count("*")

    ret += 3 * ops.count("/")

    ret += 3 * ops.count("n")

    ret += 4 * ops.count("^")

    ret += 4 * ops.count("s")

    ret += 4 * ops.count("f")

    ret += 4 * ops.count("w")



    # ret += 4 * ops.count("fs")

    # ret += 4 * ops.count("cs")

    return ret





agenda = [{"args" : 1, "nums" : [1,3,3,6], "ops" : }]

seen = {}





only_search_for = None

ret = 



def finish(ops) :

    global ret

    global seen



    ops_tmp = ops[:]

    try :

        n,_ = evaluate(ops_tmp)

    except :

        n = None



    if n is None or not (0  score(ops) :

        seen[n] = ops



        print ops,"t",n









while agenda :

    x = agenda.pop(0)



    if not x["nums"] and not x["args"] : # finished: used up all numbers; no open spaces.

        finish(x["ops"])



    if len(x["nums"]) == x["args"] : # fill in numbers only

        for nums in set(permutations(x["nums"])) :

            finish(x["ops"] + list(nums))



            # print {"args" : 0,

            #        "nums" : ,

            #        "ops" : x["ops"] + list(nums)}



    elif len(x["nums"]) > x["args"] :

        # add new operators









        for op in ops.keys() :

            if arity[op] == 1 and x["ops"] and x["ops"][-1] == op :

                continue # limit repeated unary operations

            if arity[op] == 1 and x["ops"] and arity.get(x["ops"][-1]) == 1 :

                continue # limit repeated unary operations





            if (concat_literal_numbers_only and x["ops"] and (x["ops"][-1] == "C" or (len(x["ops"])>1 and x["ops"][-2] == "C")) and op != "C") :

                continue





            new_x = {"args" : x["args"] + arity[op] - 1,

                     "nums" : x["nums"],

                     "ops" : x["ops"] + [op]}

            agenda = [new_x] + agenda



        if x["args"] == 1 : 

            continue



        for n in set(x["nums"]) :

            new_nums = x["nums"][:]

            new_nums.remove(n)

            new_x = {"args" : x["args"] - 1,

                     "nums" : new_nums,

                     "ops" : x["ops"] + [n]}



            agenda = [new_x] + agenda



# SHOW HOW TO MAKE ALL OF THE NUMBERS

miss = 

for i in range(0+1,100+1) :

    if not seen.get(i) :

        miss += [i]

    print i, "t", seen.get(i, "---")



# SHOW WHICH NUMBERS WERE MISSED

print "missed: ", miss



# IF YOU'RE LOOKING FOR ALL POSSIBLE WAYS TO MAKE SOMETHING, SHOW THEM HERE.

if only_search_for is not None :

    ret = sorted(ret, key=score)

    for x in ret:

        print x

edited Feb 1 at 2:09

answered Jan 28 at 7:27

user326210

1712

$begingroup$
Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely.
$endgroup$
– Bass
Jan 29 at 9:47

$begingroup$
@Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions.
$endgroup$
– user326210
Jan 30 at 1:00

add a comment |

Here's a solution for most of them. The remaining 21 are impossible.

Most of them only use simple arithmetic.

Some of them use exponentiation.

Some use square roots.

Accordingly, it is provably impossible to construct the following twenty-one numbers:

41, 44, 46, 47, 56, 68, 69, 74, 77, 79, 83, 85, 86, 89, 90, 91, 92, 95, 97, 98, 100.

The rest can be constructed as follows:

1-25 (complete)

1. (6+1)-(3+3)

2. (3*3)-(6+1)

3. (3*3)-(6*1)

4. 13-(6+3)

5. 36-31

6. 3*(3+1)-6

7. (6+1)+(3-3)

8. (6-3)3-1

9. 6(3-1)-3

10. 3+13-6

11. 6+3+3-1

12. (6+3+3)*1

13. 6+3+3+1

14. (6*3)-(3+1)

15. 13 + (6/3)

16. 13 + (6-3)

17. 33-16

18. (1+3)*3 + 6

19. (6 + 1/3)*3

20. (3*6)+(3-1)

21. (13-6)*3

22. 13+3+6

23. 36-13

24. (6+1)*3 +3

25. 16+(3*3)

26-50 (except 41, 44, 46, 47)

26. 33 - (6+1)

27. 33 - (6*1)

28. 61 - 33

29. 31 - (6/3)

30. (6+3+1)*3

31. 13 + (6*3)

32. 63-31

33. 13*3 - 6

34. 31+(6-3)

35. 6*(3+3) -1

36. (6+3)(3+1)

37. 6(3+3) + 1

38. 33+(6-1)

39. 33+(6*1)

40. 31+3+6

41. round[36 + √(31)]

42. (3+3+1)*6

43. 16 + (3^3)

44. round[(3-√3)^16]

45. (3*13)+6

46. round[3 * (13 + √6)]

47. round[31√3] - 6

48. (3*3 - 1) *6

49. 36+13

50. 63-13

51-75 (except 56, 68, 69, 74)

51. 16*3 + 3

52. 61 - (3*3)

53. (3*3*6)-1

54. 1*3*3*6

55. 61-(3+3)

56. √(3136) (!)

57. (13+6)*3

58. (3+1)^3 - 6

59. 63 - (3+1)

60. (63*1) - 3

61. 61 + (3 - 3)

62. 61 + (3/3)

63. (6+1)*(3*3)

64. 63 + (1^3)

65. 63 + (3-1)

66. 63 + (3*1)

67. 36 + 31

68. round[63 + √31]

69. round[6 √133]

70. 61+(3*3)

71. (6^3)/3 - 1

72. 36 * (3-1)

73. (6^3)/3 + 1

74. round[3 √613]

75. 13*6 - 3

76-100 (except: 77, 79, 83, 85*, 86, 89,90,91,92, 95, 97, 98, 100.)

76. 63+13

77. round[61 * (3-√3)]

78. 13 * √(36)

79. round[6*13 + √3]

80. √(3^6) * 3 - 1

81. 13*6 + 3

82. √(3^6) * 3 + 1

83. round[(3+31)*√6]

84. 3^√(16) + 3

85. -----

86. round[√(6+√3) * 31]

87. (31*3) - 6

88. 61 + (3^3)

89. round[63 * √(3-1)]

90. round[13 * √(6/3)]

91. round[16 * √33]

92. round[3^√(3*6-1)]

93. 31 * (6-3)

94. 63+31

95. round[3*31 + √6]

96. (13+3)*6

97. round[(√3)^6 * √13]

98. round[36 * (1+√3)]

99. (3*31)+6

100. round[3*(31+√6)]

If you're allowed to concatenate the results of operations (e.g. (3+1)|5 = 45 ) then a solution for 85 is :

85. (3*3)|1 - 6

Python:



from itertools import permutations

from math import sqrt, floor, ceil





concat_literal_numbers_only = True



ops = { "+" : lambda a,b: a+b,

        "-" : lambda a,b: a - b,

        "/" : lambda a,b : a/float(b),

        "*" : lambda a,b : a*b,

        "^" : lambda a,b : a**b,

        "C" : lambda a,b : float(str(a) + str(b)),

        "n" : lambda a : -a,

        "s" : lambda a : sqrt(a),

        #"f" : lambda a : floor(a)

}



arity = {"+" : 2,

         "-" : 2,

         "/" : 2,

         "*" : 2,

         "^" : 2,

         "C" : 2,

         "n" : 1,

         "s" : 1,

         "f" : 1,

}





# print ops["/"](1,3)





# args: number of open args available

# nums: available digits to be used

# ops : tuple indicating commands used so far







def evaluate(cmds) :

    """Consume the list of commands in prefix notation, producing a pair (ans, unconsumed_symbols)"""



    x = cmds.pop(0)

    if not ops.get(x) :

        return (x, cmds)

    else :

        args = 

        for y in range(arity[x]) :

            try :

                (a, cmds) = evaluate(cmds)

                args += [a]

            except OverflowError :

                return (None, None)

        return (ops.get(x)(*args), cmds)



def score(ops):

    ret = 0

    ret += ops.count("+")

    ret += 1.1*ops.count("-")

    ret += 2 * ops.count("*")

    ret += 3 * ops.count("/")

    ret += 3 * ops.count("n")

    ret += 4 * ops.count("^")

    ret += 4 * ops.count("s")

    ret += 4 * ops.count("f")

    ret += 4 * ops.count("w")



    # ret += 4 * ops.count("fs")

    # ret += 4 * ops.count("cs")

    return ret





agenda = [{"args" : 1, "nums" : [1,3,3,6], "ops" : }]

seen = {}





only_search_for = None

ret = 



def finish(ops) :

    global ret

    global seen



    ops_tmp = ops[:]

    try :

        n,_ = evaluate(ops_tmp)

    except :

        n = None



    if n is None or not (0  score(ops) :

        seen[n] = ops



        print ops,"t",n









while agenda :

    x = agenda.pop(0)



    if not x["nums"] and not x["args"] : # finished: used up all numbers; no open spaces.

        finish(x["ops"])



    if len(x["nums"]) == x["args"] : # fill in numbers only

        for nums in set(permutations(x["nums"])) :

            finish(x["ops"] + list(nums))



            # print {"args" : 0,

            #        "nums" : ,

            #        "ops" : x["ops"] + list(nums)}



    elif len(x["nums"]) > x["args"] :

        # add new operators









        for op in ops.keys() :

            if arity[op] == 1 and x["ops"] and x["ops"][-1] == op :

                continue # limit repeated unary operations

            if arity[op] == 1 and x["ops"] and arity.get(x["ops"][-1]) == 1 :

                continue # limit repeated unary operations





            if (concat_literal_numbers_only and x["ops"] and (x["ops"][-1] == "C" or (len(x["ops"])>1 and x["ops"][-2] == "C")) and op != "C") :

                continue





            new_x = {"args" : x["args"] + arity[op] - 1,

                     "nums" : x["nums"],

                     "ops" : x["ops"] + [op]}

            agenda = [new_x] + agenda



        if x["args"] == 1 : 

            continue



        for n in set(x["nums"]) :

            new_nums = x["nums"][:]

            new_nums.remove(n)

            new_x = {"args" : x["args"] - 1,

                     "nums" : new_nums,

                     "ops" : x["ops"] + [n]}



            agenda = [new_x] + agenda



# SHOW HOW TO MAKE ALL OF THE NUMBERS

miss = 

for i in range(0+1,100+1) :

    if not seen.get(i) :

        miss += [i]

    print i, "t", seen.get(i, "---")



# SHOW WHICH NUMBERS WERE MISSED

print "missed: ", miss



# IF YOU'RE LOOKING FOR ALL POSSIBLE WAYS TO MAKE SOMETHING, SHOW THEM HERE.

if only_search_for is not None :

    ret = sorted(ret, key=score)

    for x in ret:

        print x

edited Feb 1 at 2:09

answered Jan 28 at 7:27

user326210

1712

Here's a solution for most of them. The remaining 21 are impossible.

Most of them only use simple arithmetic.

Some of them use exponentiation.

Some use square roots.

Accordingly, it is provably impossible to construct the following twenty-one numbers:

41, 44, 46, 47, 56, 68, 69, 74, 77, 79, 83, 85, 86, 89, 90, 91, 92, 95, 97, 98, 100.

The rest can be constructed as follows:

1-25 (complete)

1. (6+1)-(3+3)

2. (3*3)-(6+1)

3. (3*3)-(6*1)

4. 13-(6+3)

5. 36-31

6. 3*(3+1)-6

7. (6+1)+(3-3)

8. (6-3)3-1

9. 6(3-1)-3

10. 3+13-6

11. 6+3+3-1

12. (6+3+3)*1

13. 6+3+3+1

14. (6*3)-(3+1)

15. 13 + (6/3)

16. 13 + (6-3)

17. 33-16

18. (1+3)*3 + 6

19. (6 + 1/3)*3

20. (3*6)+(3-1)

21. (13-6)*3

22. 13+3+6

23. 36-13

24. (6+1)*3 +3

25. 16+(3*3)

26-50 (except 41, 44, 46, 47)

26. 33 - (6+1)

27. 33 - (6*1)

28. 61 - 33

29. 31 - (6/3)

30. (6+3+1)*3

31. 13 + (6*3)

32. 63-31

33. 13*3 - 6

34. 31+(6-3)

35. 6*(3+3) -1

36. (6+3)(3+1)

37. 6(3+3) + 1

38. 33+(6-1)

39. 33+(6*1)

40. 31+3+6

41. round[36 + √(31)]

42. (3+3+1)*6

43. 16 + (3^3)

44. round[(3-√3)^16]

45. (3*13)+6

46. round[3 * (13 + √6)]

47. round[31√3] - 6

48. (3*3 - 1) *6

49. 36+13

50. 63-13

51-75 (except 56, 68, 69, 74)

51. 16*3 + 3

52. 61 - (3*3)

53. (3*3*6)-1

54. 1*3*3*6

55. 61-(3+3)

56. √(3136) (!)

57. (13+6)*3

58. (3+1)^3 - 6

59. 63 - (3+1)

60. (63*1) - 3

61. 61 + (3 - 3)

62. 61 + (3/3)

63. (6+1)*(3*3)

64. 63 + (1^3)

65. 63 + (3-1)

66. 63 + (3*1)

67. 36 + 31

68. round[63 + √31]

69. round[6 √133]

70. 61+(3*3)

71. (6^3)/3 - 1

72. 36 * (3-1)

73. (6^3)/3 + 1

74. round[3 √613]

75. 13*6 - 3

76-100 (except: 77, 79, 83, 85*, 86, 89,90,91,92, 95, 97, 98, 100.)

76. 63+13

77. round[61 * (3-√3)]

78. 13 * √(36)

79. round[6*13 + √3]

80. √(3^6) * 3 - 1

81. 13*6 + 3

82. √(3^6) * 3 + 1

83. round[(3+31)*√6]

84. 3^√(16) + 3

85. -----

86. round[√(6+√3) * 31]

87. (31*3) - 6

88. 61 + (3^3)

89. round[63 * √(3-1)]

90. round[13 * √(6/3)]

91. round[16 * √33]

92. round[3^√(3*6-1)]

93. 31 * (6-3)

94. 63+31

95. round[3*31 + √6]

96. (13+3)*6

97. round[(√3)^6 * √13]

98. round[36 * (1+√3)]

99. (3*31)+6

100. round[3*(31+√6)]

If you're allowed to concatenate the results of operations (e.g. (3+1)|5 = 45 ) then a solution for 85 is :

85. (3*3)|1 - 6

Python:



from itertools import permutations

from math import sqrt, floor, ceil





concat_literal_numbers_only = True



ops = { "+" : lambda a,b: a+b,

        "-" : lambda a,b: a - b,

        "/" : lambda a,b : a/float(b),

        "*" : lambda a,b : a*b,

        "^" : lambda a,b : a**b,

        "C" : lambda a,b : float(str(a) + str(b)),

        "n" : lambda a : -a,

        "s" : lambda a : sqrt(a),

        #"f" : lambda a : floor(a)

}



arity = {"+" : 2,

         "-" : 2,

         "/" : 2,

         "*" : 2,

         "^" : 2,

         "C" : 2,

         "n" : 1,

         "s" : 1,

         "f" : 1,

}





# print ops["/"](1,3)





# args: number of open args available

# nums: available digits to be used

# ops : tuple indicating commands used so far







def evaluate(cmds) :

    """Consume the list of commands in prefix notation, producing a pair (ans, unconsumed_symbols)"""



    x = cmds.pop(0)

    if not ops.get(x) :

        return (x, cmds)

    else :

        args = 

        for y in range(arity[x]) :

            try :

                (a, cmds) = evaluate(cmds)

                args += [a]

            except OverflowError :

                return (None, None)

        return (ops.get(x)(*args), cmds)



def score(ops):

    ret = 0

    ret += ops.count("+")

    ret += 1.1*ops.count("-")

    ret += 2 * ops.count("*")

    ret += 3 * ops.count("/")

    ret += 3 * ops.count("n")

    ret += 4 * ops.count("^")

    ret += 4 * ops.count("s")

    ret += 4 * ops.count("f")

    ret += 4 * ops.count("w")



    # ret += 4 * ops.count("fs")

    # ret += 4 * ops.count("cs")

    return ret





agenda = [{"args" : 1, "nums" : [1,3,3,6], "ops" : }]

seen = {}





only_search_for = None

ret = 



def finish(ops) :

    global ret

    global seen



    ops_tmp = ops[:]

    try :

        n,_ = evaluate(ops_tmp)

    except :

        n = None



    if n is None or not (0  score(ops) :

        seen[n] = ops



        print ops,"t",n









while agenda :

    x = agenda.pop(0)



    if not x["nums"] and not x["args"] : # finished: used up all numbers; no open spaces.

        finish(x["ops"])



    if len(x["nums"]) == x["args"] : # fill in numbers only

        for nums in set(permutations(x["nums"])) :

            finish(x["ops"] + list(nums))



            # print {"args" : 0,

            #        "nums" : ,

            #        "ops" : x["ops"] + list(nums)}



    elif len(x["nums"]) > x["args"] :

        # add new operators









        for op in ops.keys() :

            if arity[op] == 1 and x["ops"] and x["ops"][-1] == op :

                continue # limit repeated unary operations

            if arity[op] == 1 and x["ops"] and arity.get(x["ops"][-1]) == 1 :

                continue # limit repeated unary operations





            if (concat_literal_numbers_only and x["ops"] and (x["ops"][-1] == "C" or (len(x["ops"])>1 and x["ops"][-2] == "C")) and op != "C") :

                continue





            new_x = {"args" : x["args"] + arity[op] - 1,

                     "nums" : x["nums"],

                     "ops" : x["ops"] + [op]}

            agenda = [new_x] + agenda



        if x["args"] == 1 : 

            continue



        for n in set(x["nums"]) :

            new_nums = x["nums"][:]

            new_nums.remove(n)

            new_x = {"args" : x["args"] - 1,

                     "nums" : new_nums,

                     "ops" : x["ops"] + [n]}



            agenda = [new_x] + agenda



# SHOW HOW TO MAKE ALL OF THE NUMBERS

miss = 

for i in range(0+1,100+1) :

    if not seen.get(i) :

        miss += [i]

    print i, "t", seen.get(i, "---")



# SHOW WHICH NUMBERS WERE MISSED

print "missed: ", miss



# IF YOU'RE LOOKING FOR ALL POSSIBLE WAYS TO MAKE SOMETHING, SHOW THEM HERE.

if only_search_for is not None :

    ret = sorted(ret, key=score)

    for x in ret:

        print x

edited Feb 1 at 2:09

answered Jan 28 at 7:27

user326210

1712

edited Feb 1 at 2:09

answered Jan 28 at 7:27

user326210

1712

answered Jan 28 at 7:27

user326210

1712

answered Jan 28 at 7:27

user326210

1712

$begingroup$
Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely.
$endgroup$
– Bass
Jan 29 at 9:47

$begingroup$
@Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions.
$endgroup$
– user326210
Jan 30 at 1:00

add a comment |

$begingroup$
Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely.
$endgroup$
– Bass
Jan 29 at 9:47

$begingroup$
@Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions.
$endgroup$
– user326210
Jan 30 at 1:00

Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely.

– Bass
Jan 29 at 9:47

@Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions.

– user326210
Jan 30 at 1:00

add a comment |

We can generate any integer using only $1$, $3$, $3$ and $6$ with the introduction of one special function.

The function in question is the Logarithm to an arbitrary base $b$ , or $log _{b} (x)$.

To begin, let's discuss square root stacking.
$sqrt{sqrt{a}}$ is equivalent to $sqrt[4]{a}$, and $sqrt{sqrt{sqrt{a}}}$ is equivalent to $sqrt[8]{a}$, which can be rewritten as $a^frac{1}{8}$. This pattern continues indefinitely; $a$ with $n$ square roots stacked to it is equal to $a^frac{1}{2^n}$.

The laws of logarithms state that $log _{b} (x^a) = a cdot log _{b} (x)$. If we take the logarithm to base $b$ or our previous square root stack, we get $log _{b} (a^frac{1}{2^n})$, or $frac{1}{2^n} cdot log _{b}a$. Setting $a$ and $b$ as 3 means that $log _{3} (sqrt{sqrt[...]{3}})$, with $n$ square roots, is equal to $frac{1}{2^n} cdot log _{3}3$, or $2^{-n}$. $(frac{1}{a^x} = a^{-x})$

$sqrt{sqrt{16}} = 2$, and $log _{b}(b^a) = a$. As such,

$0 = -log _{sqrt{sqrt{16}}}(log_{3}(3))$
$1 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{3}))$
$2 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{3}}))$
$3 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{3}}}))$
$4 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{sqrt{3}}}}))$

And so on, such that the amount of square roots is equal to your desired number.

This works for all integers; for negative numbers simply remove the $-$ at the start.

This was inspired by Numberphile's video on the four 4s.

answered Jan 28 at 14:27

user56759

311

add a comment |

We can generate any integer using only $1$, $3$, $3$ and $6$ with the introduction of one special function.

The function in question is the Logarithm to an arbitrary base $b$ , or $log _{b} (x)$.

To begin, let's discuss square root stacking.
$sqrt{sqrt{a}}$ is equivalent to $sqrt[4]{a}$, and $sqrt{sqrt{sqrt{a}}}$ is equivalent to $sqrt[8]{a}$, which can be rewritten as $a^frac{1}{8}$. This pattern continues indefinitely; $a$ with $n$ square roots stacked to it is equal to $a^frac{1}{2^n}$.

The laws of logarithms state that $log _{b} (x^a) = a cdot log _{b} (x)$. If we take the logarithm to base $b$ or our previous square root stack, we get $log _{b} (a^frac{1}{2^n})$, or $frac{1}{2^n} cdot log _{b}a$. Setting $a$ and $b$ as 3 means that $log _{3} (sqrt{sqrt[...]{3}})$, with $n$ square roots, is equal to $frac{1}{2^n} cdot log _{3}3$, or $2^{-n}$. $(frac{1}{a^x} = a^{-x})$

$sqrt{sqrt{16}} = 2$, and $log _{b}(b^a) = a$. As such,

$0 = -log _{sqrt{sqrt{16}}}(log_{3}(3))$
$1 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{3}))$
$2 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{3}}))$
$3 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{3}}}))$
$4 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{sqrt{3}}}}))$

And so on, such that the amount of square roots is equal to your desired number.

This works for all integers; for negative numbers simply remove the $-$ at the start.

This was inspired by Numberphile's video on the four 4s.

answered Jan 28 at 14:27

user56759

311

add a comment |

We can generate any integer using only $1$, $3$, $3$ and $6$ with the introduction of one special function.

The function in question is the Logarithm to an arbitrary base $b$ , or $log _{b} (x)$.

To begin, let's discuss square root stacking.
$sqrt{sqrt{a}}$ is equivalent to $sqrt[4]{a}$, and $sqrt{sqrt{sqrt{a}}}$ is equivalent to $sqrt[8]{a}$, which can be rewritten as $a^frac{1}{8}$. This pattern continues indefinitely; $a$ with $n$ square roots stacked to it is equal to $a^frac{1}{2^n}$.

The laws of logarithms state that $log _{b} (x^a) = a cdot log _{b} (x)$. If we take the logarithm to base $b$ or our previous square root stack, we get $log _{b} (a^frac{1}{2^n})$, or $frac{1}{2^n} cdot log _{b}a$. Setting $a$ and $b$ as 3 means that $log _{3} (sqrt{sqrt[...]{3}})$, with $n$ square roots, is equal to $frac{1}{2^n} cdot log _{3}3$, or $2^{-n}$. $(frac{1}{a^x} = a^{-x})$

$sqrt{sqrt{16}} = 2$, and $log _{b}(b^a) = a$. As such,

$0 = -log _{sqrt{sqrt{16}}}(log_{3}(3))$
$1 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{3}))$
$2 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{3}}))$
$3 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{3}}}))$
$4 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{sqrt{3}}}}))$

And so on, such that the amount of square roots is equal to your desired number.

This works for all integers; for negative numbers simply remove the $-$ at the start.

This was inspired by Numberphile's video on the four 4s.

answered Jan 28 at 14:27

user56759

311

We can generate any integer using only $1$, $3$, $3$ and $6$ with the introduction of one special function.

The function in question is the Logarithm to an arbitrary base $b$ , or $log _{b} (x)$.

To begin, let's discuss square root stacking.
$sqrt{sqrt{a}}$ is equivalent to $sqrt[4]{a}$, and $sqrt{sqrt{sqrt{a}}}$ is equivalent to $sqrt[8]{a}$, which can be rewritten as $a^frac{1}{8}$. This pattern continues indefinitely; $a$ with $n$ square roots stacked to it is equal to $a^frac{1}{2^n}$.

The laws of logarithms state that $log _{b} (x^a) = a cdot log _{b} (x)$. If we take the logarithm to base $b$ or our previous square root stack, we get $log _{b} (a^frac{1}{2^n})$, or $frac{1}{2^n} cdot log _{b}a$. Setting $a$ and $b$ as 3 means that $log _{3} (sqrt{sqrt[...]{3}})$, with $n$ square roots, is equal to $frac{1}{2^n} cdot log _{3}3$, or $2^{-n}$. $(frac{1}{a^x} = a^{-x})$

$sqrt{sqrt{16}} = 2$, and $log _{b}(b^a) = a$. As such,

$0 = -log _{sqrt{sqrt{16}}}(log_{3}(3))$
$1 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{3}))$
$2 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{3}}))$
$3 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{3}}}))$
$4 = -log _{sqrt{sqrt{16}}}(log_{3}(sqrt{sqrt{sqrt{sqrt{3}}}}))$

And so on, such that the amount of square roots is equal to your desired number.

This works for all integers; for negative numbers simply remove the $-$ at the start.

This was inspired by Numberphile's video on the four 4s.

answered Jan 28 at 14:27

user56759

311

answered Jan 28 at 14:27

user56759

311

answered Jan 28 at 14:27

user56759

311

answered Jan 28 at 14:27

user56759

311

add a comment |

Here are some:

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

22: (omega kyrpton did some) $3 * 6 + 3 + 1$

I will do more later.

edited Jan 28 at 15:54

answered Jan 27 at 22:33

Yout Ried

894119

add a comment |

Here are some:

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

22: (omega kyrpton did some) $3 * 6 + 3 + 1$

I will do more later.

edited Jan 28 at 15:54

answered Jan 27 at 22:33

Yout Ried

894119

add a comment |

Here are some:

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

22: (omega kyrpton did some) $3 * 6 + 3 + 1$

I will do more later.

edited Jan 28 at 15:54

answered Jan 27 at 22:33

Yout Ried

894119

Here are some:

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

22: (omega kyrpton did some) $3 * 6 + 3 + 1$

I will do more later.

edited Jan 28 at 15:54

answered Jan 27 at 22:33

Yout Ried

894119

edited Jan 28 at 15:54

answered Jan 27 at 22:33

Yout Ried

894119

answered Jan 27 at 22:33

Yout Ried

894119

answered Jan 27 at 22:33

Yout Ried

894119

add a comment |

Adding some more...

1-20: (Credits to @YoutRied)

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

21-29

21: $3 * 6 + 3 * 1$

22: $( 1 + 3 ) ! - ( 6 / 3 )$

23: $( 1 + 3 ) ! - ( 6 - 3 )$

24: $( 6 - 3 / 3 - 1 ) !$

25: $1 * 3 ^ 3 - floor(sqrt{6})$

26: $( 6 - 3 ) ^ 3 - 1$

27: $( 6 - 3 ) ^ 3 * 1$

28: $( 6 - 3 ) ^ 3 + 1$

29: $31 - 6 / 3$

41-50: (Credits to @Bass for 42, 45, 49)

41: $ (-1+3!)+36 $

42: $ (1+3+3)times 6$

43: $ 31 + 6 * ceil(sqrt{3})$

44: $floor( 1 * 3 * sqrt{6 ^ 3}) $

45: $ 13times3+6$

46: $ ceil(sqrt{6 ^ 3} + 31)$

47: $ floor(sqrt{sqrt{sqrt{sqrt{sqrt{31!}}}}})+36$

48: $6 * ( 3 * 3 - 1 )$

49: $13+36$

50: $ (6+1)^2 + 3 - 3$

51-60:

51: $( 3 * 6 - 1 ) * 3$

52: $( 3 + 3 + 1 ) * ceil(sqrt{6})$

53: $-1+( 3 * 3 * 6 )$

54: $ 1*3 * 3 * 6 $

55: $1+3*3*6$

56: $61-3!+floor(sqrt{3})$

57: $1*63-3!$

58: $1+63-3!$

59: $floor(sqrt{sqrt{sqrt{sqrt{sqrt{6^{(3-1)}}}}}}*3)$

60: $(1+3*3)*6$

61-70:

61: $63-3+1$

62: $63+1-ceil(sqrt{3})$

63: $63-floor(sqrt{3})+1$

64: $63+ceil(sqrt{3})-1$

65: $63+3-1$

66: $63+3*1$

67: $63+3+1$

68: $61+3!+floor(sqrt{3})$

69: $61+3!+ceil(sqrt{3})$

70: $61+3*3$

71-80

71: $(3+1)!*3-floor(sqrt{sqrt{6}})$

72: $(3+1)*3*6$

73: $(3+1)!*3+floor(sqrt{sqrt{6}})$

74: $(3+1)!*3+floor(sqrt{6})$

75: $(3+1)!*3+ceil(sqrt{6})$

76: $ceil(sqrt{sqrt{sqrt{ceil(sqrt{sqrt{sqrt{sqrt{sqrt{ceil(sqrt{3!!})!}}}}})}}})*(3*6+1)$

76:(much simpler) $13*6-ceil(sqrt{3})$

77: $13*6-floor(sqrt{3})$

78: $13*floor(sqrt{3})*6$

79: $13*6+floor(sqrt{3})$

80: $13*6+ceil(sqrt{3})$

81-90:

81: $(6+3)^{(3-1)}$

82: $(6+3)^{ceil(sqrt{3})}+1$

edited Jan 28 at 8:36

answered Jan 27 at 23:31

Omega Krypton

3,6941338

$begingroup$
Who said you could use factorials?
$endgroup$
– Yout Ried
Jan 28 at 0:20

$begingroup$
What are number 23 (plus you probably can't use factorials and 24? I don't get them.
$endgroup$
– Yout Ried
Jan 28 at 1:08

$begingroup$
Oops forgot a ")" and maybe you're missing a factorial for number 24
$endgroup$
– Yout Ried
Jan 28 at 1:16

add a comment |

Adding some more...

1-20: (Credits to @YoutRied)

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

21-29

21: $3 * 6 + 3 * 1$

22: $( 1 + 3 ) ! - ( 6 / 3 )$

23: $( 1 + 3 ) ! - ( 6 - 3 )$

24: $( 6 - 3 / 3 - 1 ) !$

25: $1 * 3 ^ 3 - floor(sqrt{6})$

26: $( 6 - 3 ) ^ 3 - 1$

27: $( 6 - 3 ) ^ 3 * 1$

28: $( 6 - 3 ) ^ 3 + 1$

29: $31 - 6 / 3$

41-50: (Credits to @Bass for 42, 45, 49)

41: $ (-1+3!)+36 $

42: $ (1+3+3)times 6$

43: $ 31 + 6 * ceil(sqrt{3})$

44: $floor( 1 * 3 * sqrt{6 ^ 3}) $

45: $ 13times3+6$

46: $ ceil(sqrt{6 ^ 3} + 31)$

47: $ floor(sqrt{sqrt{sqrt{sqrt{sqrt{31!}}}}})+36$

48: $6 * ( 3 * 3 - 1 )$

49: $13+36$

50: $ (6+1)^2 + 3 - 3$

51-60:

51: $( 3 * 6 - 1 ) * 3$

52: $( 3 + 3 + 1 ) * ceil(sqrt{6})$

53: $-1+( 3 * 3 * 6 )$

54: $ 1*3 * 3 * 6 $

55: $1+3*3*6$

56: $61-3!+floor(sqrt{3})$

57: $1*63-3!$

58: $1+63-3!$

59: $floor(sqrt{sqrt{sqrt{sqrt{sqrt{6^{(3-1)}}}}}}*3)$

60: $(1+3*3)*6$

61-70:

61: $63-3+1$

62: $63+1-ceil(sqrt{3})$

63: $63-floor(sqrt{3})+1$

64: $63+ceil(sqrt{3})-1$

65: $63+3-1$

66: $63+3*1$

67: $63+3+1$

68: $61+3!+floor(sqrt{3})$

69: $61+3!+ceil(sqrt{3})$

70: $61+3*3$

71-80

71: $(3+1)!*3-floor(sqrt{sqrt{6}})$

72: $(3+1)*3*6$

73: $(3+1)!*3+floor(sqrt{sqrt{6}})$

74: $(3+1)!*3+floor(sqrt{6})$

75: $(3+1)!*3+ceil(sqrt{6})$

76: $ceil(sqrt{sqrt{sqrt{ceil(sqrt{sqrt{sqrt{sqrt{sqrt{ceil(sqrt{3!!})!}}}}})}}})*(3*6+1)$

76:(much simpler) $13*6-ceil(sqrt{3})$

77: $13*6-floor(sqrt{3})$

78: $13*floor(sqrt{3})*6$

79: $13*6+floor(sqrt{3})$

80: $13*6+ceil(sqrt{3})$

81-90:

81: $(6+3)^{(3-1)}$

82: $(6+3)^{ceil(sqrt{3})}+1$

edited Jan 28 at 8:36

answered Jan 27 at 23:31

Omega Krypton

3,6941338

$begingroup$
Who said you could use factorials?
$endgroup$
– Yout Ried
Jan 28 at 0:20

$begingroup$
What are number 23 (plus you probably can't use factorials and 24? I don't get them.
$endgroup$
– Yout Ried
Jan 28 at 1:08

$begingroup$
Oops forgot a ")" and maybe you're missing a factorial for number 24
$endgroup$
– Yout Ried
Jan 28 at 1:16

add a comment |

Adding some more...

1-20: (Credits to @YoutRied)

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

21-29

21: $3 * 6 + 3 * 1$

22: $( 1 + 3 ) ! - ( 6 / 3 )$

23: $( 1 + 3 ) ! - ( 6 - 3 )$

24: $( 6 - 3 / 3 - 1 ) !$

25: $1 * 3 ^ 3 - floor(sqrt{6})$

26: $( 6 - 3 ) ^ 3 - 1$

27: $( 6 - 3 ) ^ 3 * 1$

28: $( 6 - 3 ) ^ 3 + 1$

29: $31 - 6 / 3$

41-50: (Credits to @Bass for 42, 45, 49)

41: $ (-1+3!)+36 $

42: $ (1+3+3)times 6$

43: $ 31 + 6 * ceil(sqrt{3})$

44: $floor( 1 * 3 * sqrt{6 ^ 3}) $

45: $ 13times3+6$

46: $ ceil(sqrt{6 ^ 3} + 31)$

47: $ floor(sqrt{sqrt{sqrt{sqrt{sqrt{31!}}}}})+36$

48: $6 * ( 3 * 3 - 1 )$

49: $13+36$

50: $ (6+1)^2 + 3 - 3$

51-60:

51: $( 3 * 6 - 1 ) * 3$

52: $( 3 + 3 + 1 ) * ceil(sqrt{6})$

53: $-1+( 3 * 3 * 6 )$

54: $ 1*3 * 3 * 6 $

55: $1+3*3*6$

56: $61-3!+floor(sqrt{3})$

57: $1*63-3!$

58: $1+63-3!$

59: $floor(sqrt{sqrt{sqrt{sqrt{sqrt{6^{(3-1)}}}}}}*3)$

60: $(1+3*3)*6$

61-70:

61: $63-3+1$

62: $63+1-ceil(sqrt{3})$

63: $63-floor(sqrt{3})+1$

64: $63+ceil(sqrt{3})-1$

65: $63+3-1$

66: $63+3*1$

67: $63+3+1$

68: $61+3!+floor(sqrt{3})$

69: $61+3!+ceil(sqrt{3})$

70: $61+3*3$

71-80

71: $(3+1)!*3-floor(sqrt{sqrt{6}})$

72: $(3+1)*3*6$

73: $(3+1)!*3+floor(sqrt{sqrt{6}})$

74: $(3+1)!*3+floor(sqrt{6})$

75: $(3+1)!*3+ceil(sqrt{6})$

76: $ceil(sqrt{sqrt{sqrt{ceil(sqrt{sqrt{sqrt{sqrt{sqrt{ceil(sqrt{3!!})!}}}}})}}})*(3*6+1)$

76:(much simpler) $13*6-ceil(sqrt{3})$

77: $13*6-floor(sqrt{3})$

78: $13*floor(sqrt{3})*6$

79: $13*6+floor(sqrt{3})$

80: $13*6+ceil(sqrt{3})$

81-90:

81: $(6+3)^{(3-1)}$

82: $(6+3)^{ceil(sqrt{3})}+1$

edited Jan 28 at 8:36

answered Jan 27 at 23:31

Omega Krypton

3,6941338

Adding some more...

1-20: (Credits to @YoutRied)

1: $3 + 3 - 6 + 1$

2: $3 * 3 - (6 + 1)$

3: $3 * 3 * 1 - 6$

4: $3 * 3 - 6 + 1$

5: $(3 * 6) / 3 - 1$

6: $(3 * 6) / 3 * 1$

7: $(3 * 6) / 3 + 1$

8: $3 * 3 - 1 ^ 6$

9: $(3 * 6) / (3 - 1)$

10: $3 * 3 + 1 ^ 6$

11: $36 / 3 - 1$

12: $36 / 3 * 1$

13: $36 / 3 + 1$

14: $3 * 6 - (3 + 1)$

15: $3 * 6 - (3 * 1)$

16: $3 * 6 - (3 - 1)$

17: $3 * 6 - 1 ^ 3$

18: $3 * 6 * 1 ^ 3$

19: $3 * 6 + 1 ^ 3$

20: $3 * 6 + 3 - 1$

21-29

21: $3 * 6 + 3 * 1$

22: $( 1 + 3 ) ! - ( 6 / 3 )$

23: $( 1 + 3 ) ! - ( 6 - 3 )$

24: $( 6 - 3 / 3 - 1 ) !$

25: $1 * 3 ^ 3 - floor(sqrt{6})$

26: $( 6 - 3 ) ^ 3 - 1$

27: $( 6 - 3 ) ^ 3 * 1$

28: $( 6 - 3 ) ^ 3 + 1$

29: $31 - 6 / 3$

41-50: (Credits to @Bass for 42, 45, 49)

41: $ (-1+3!)+36 $

42: $ (1+3+3)times 6$

43: $ 31 + 6 * ceil(sqrt{3})$

44: $floor( 1 * 3 * sqrt{6 ^ 3}) $

45: $ 13times3+6$

46: $ ceil(sqrt{6 ^ 3} + 31)$

47: $ floor(sqrt{sqrt{sqrt{sqrt{sqrt{31!}}}}})+36$

48: $6 * ( 3 * 3 - 1 )$

49: $13+36$

50: $ (6+1)^2 + 3 - 3$

51-60:

51: $( 3 * 6 - 1 ) * 3$

52: $( 3 + 3 + 1 ) * ceil(sqrt{6})$

53: $-1+( 3 * 3 * 6 )$

54: $ 1*3 * 3 * 6 $

55: $1+3*3*6$

56: $61-3!+floor(sqrt{3})$

57: $1*63-3!$

58: $1+63-3!$

59: $floor(sqrt{sqrt{sqrt{sqrt{sqrt{6^{(3-1)}}}}}}*3)$

60: $(1+3*3)*6$

61-70:

61: $63-3+1$

62: $63+1-ceil(sqrt{3})$

63: $63-floor(sqrt{3})+1$

64: $63+ceil(sqrt{3})-1$

65: $63+3-1$

66: $63+3*1$

67: $63+3+1$

68: $61+3!+floor(sqrt{3})$

69: $61+3!+ceil(sqrt{3})$

70: $61+3*3$

71-80

71: $(3+1)!*3-floor(sqrt{sqrt{6}})$

72: $(3+1)*3*6$

73: $(3+1)!*3+floor(sqrt{sqrt{6}})$

74: $(3+1)!*3+floor(sqrt{6})$

75: $(3+1)!*3+ceil(sqrt{6})$

76: $ceil(sqrt{sqrt{sqrt{ceil(sqrt{sqrt{sqrt{sqrt{sqrt{ceil(sqrt{3!!})!}}}}})}}})*(3*6+1)$

76:(much simpler) $13*6-ceil(sqrt{3})$

77: $13*6-floor(sqrt{3})$

78: $13*floor(sqrt{3})*6$

79: $13*6+floor(sqrt{3})$

80: $13*6+ceil(sqrt{3})$

81-90:

81: $(6+3)^{(3-1)}$

82: $(6+3)^{ceil(sqrt{3})}+1$

edited Jan 28 at 8:36

answered Jan 27 at 23:31

Omega Krypton

3,6941338

edited Jan 28 at 8:36

answered Jan 27 at 23:31

Omega Krypton

3,6941338

answered Jan 27 at 23:31

Omega Krypton

3,6941338

answered Jan 27 at 23:31

Omega Krypton

3,6941338

$begingroup$
Who said you could use factorials?
$endgroup$
– Yout Ried
Jan 28 at 0:20

$begingroup$
What are number 23 (plus you probably can't use factorials and 24? I don't get them.
$endgroup$
– Yout Ried
Jan 28 at 1:08

$begingroup$
Oops forgot a ")" and maybe you're missing a factorial for number 24
$endgroup$
– Yout Ried
Jan 28 at 1:16

add a comment |

$begingroup$
Who said you could use factorials?
$endgroup$
– Yout Ried
Jan 28 at 0:20

$begingroup$
What are number 23 (plus you probably can't use factorials and 24? I don't get them.
$endgroup$
– Yout Ried
Jan 28 at 1:08

$begingroup$
Oops forgot a ")" and maybe you're missing a factorial for number 24
$endgroup$
– Yout Ried
Jan 28 at 1:16

Who said you could use factorials?

– Yout Ried
Jan 28 at 0:20

What are number 23 (plus you probably can't use factorials and 24? I don't get them.

– Yout Ried
Jan 28 at 1:08

Oops forgot a ")" and maybe you're missing a factorial for number 24

– Yout Ried
Jan 28 at 1:16

add a comment |

Partial answer 1-50 (w/e 41,47):

$1= 1+3+3-6$
$2= 1 + (frac{6}{(3+3)})$
$3= 1^3+(frac{6}{3})$
$4= (frac{6}{3})+3-1$
$5= (frac{6}{3})+3^1$
$6= 6^1+3-3$
$7= 6+1-3+3$
$8= 6 + 3 - 1^3$
$9= 1^3 * (3+6)$
$10= 1^3 +3+6$
$11= 13 - (frac{6}{3})$
$12= 6+3+3^1$
$13= 6+3+3+1$
$14= 6*3 - 3 - 1$
$15= 6*3 - 3^1$
$16= 16 + 3 - 3$
$17= 16 + (frac{3}{3})$
$18= (frac{6*3}{1^3})$
$19= 6*3+1^3$
$20= 6*3+3-1$
$21= 6*3+3^1$
$22= 6*3+3+1$
$23= 36-13$
$24= 6*(3+1^3)$
$25= 16+(3*3)$
$26= 13*(frac{6}{3})$
$27= 33-6^1$
$28= 33-6+1$
$29= 31-(frac{6}{3})$
$30= 6*(3+3-1)$
$31= 13+3*6$
$32= 3^3+6-1$
$33= (frac{33}{1^6})$
$34= 33+1^6$
$35= (3+3)*6-1$
$36= (3+3)^1*6$
$37= 1+(3+3)*6$
$38= 33+6-1$
$39= 33+6^1$
$40= 1+33+6$
$41= $
$42= (1+3+3)*6$
$43= 16+3^3$
$44= round(sqrt{6^3}*3^1)$
$45= 3*3*(6-1)$
$46= ceil(sqrt{6^3}*3)+1$
$47= $
$48= ((3*3)-1)*6$
$49= 16+33$
$50= 63-13$

edited Jan 28 at 3:39

answered Jan 28 at 3:01

Mukyuu

340112

add a comment |

Partial answer 1-50 (w/e 41,47):

$1= 1+3+3-6$
$2= 1 + (frac{6}{(3+3)})$
$3= 1^3+(frac{6}{3})$
$4= (frac{6}{3})+3-1$
$5= (frac{6}{3})+3^1$
$6= 6^1+3-3$
$7= 6+1-3+3$
$8= 6 + 3 - 1^3$
$9= 1^3 * (3+6)$
$10= 1^3 +3+6$
$11= 13 - (frac{6}{3})$
$12= 6+3+3^1$
$13= 6+3+3+1$
$14= 6*3 - 3 - 1$
$15= 6*3 - 3^1$
$16= 16 + 3 - 3$
$17= 16 + (frac{3}{3})$
$18= (frac{6*3}{1^3})$
$19= 6*3+1^3$
$20= 6*3+3-1$
$21= 6*3+3^1$
$22= 6*3+3+1$
$23= 36-13$
$24= 6*(3+1^3)$
$25= 16+(3*3)$
$26= 13*(frac{6}{3})$
$27= 33-6^1$
$28= 33-6+1$
$29= 31-(frac{6}{3})$
$30= 6*(3+3-1)$
$31= 13+3*6$
$32= 3^3+6-1$
$33= (frac{33}{1^6})$
$34= 33+1^6$
$35= (3+3)*6-1$
$36= (3+3)^1*6$
$37= 1+(3+3)*6$
$38= 33+6-1$
$39= 33+6^1$
$40= 1+33+6$
$41= $
$42= (1+3+3)*6$
$43= 16+3^3$
$44= round(sqrt{6^3}*3^1)$
$45= 3*3*(6-1)$
$46= ceil(sqrt{6^3}*3)+1$
$47= $
$48= ((3*3)-1)*6$
$49= 16+33$
$50= 63-13$

edited Jan 28 at 3:39

answered Jan 28 at 3:01

Mukyuu

340112

add a comment |

Partial answer 1-50 (w/e 41,47):

$1= 1+3+3-6$
$2= 1 + (frac{6}{(3+3)})$
$3= 1^3+(frac{6}{3})$
$4= (frac{6}{3})+3-1$
$5= (frac{6}{3})+3^1$
$6= 6^1+3-3$
$7= 6+1-3+3$
$8= 6 + 3 - 1^3$
$9= 1^3 * (3+6)$
$10= 1^3 +3+6$
$11= 13 - (frac{6}{3})$
$12= 6+3+3^1$
$13= 6+3+3+1$
$14= 6*3 - 3 - 1$
$15= 6*3 - 3^1$
$16= 16 + 3 - 3$
$17= 16 + (frac{3}{3})$
$18= (frac{6*3}{1^3})$
$19= 6*3+1^3$
$20= 6*3+3-1$
$21= 6*3+3^1$
$22= 6*3+3+1$
$23= 36-13$
$24= 6*(3+1^3)$
$25= 16+(3*3)$
$26= 13*(frac{6}{3})$
$27= 33-6^1$
$28= 33-6+1$
$29= 31-(frac{6}{3})$
$30= 6*(3+3-1)$
$31= 13+3*6$
$32= 3^3+6-1$
$33= (frac{33}{1^6})$
$34= 33+1^6$
$35= (3+3)*6-1$
$36= (3+3)^1*6$
$37= 1+(3+3)*6$
$38= 33+6-1$
$39= 33+6^1$
$40= 1+33+6$
$41= $
$42= (1+3+3)*6$
$43= 16+3^3$
$44= round(sqrt{6^3}*3^1)$
$45= 3*3*(6-1)$
$46= ceil(sqrt{6^3}*3)+1$
$47= $
$48= ((3*3)-1)*6$
$49= 16+33$
$50= 63-13$

edited Jan 28 at 3:39

answered Jan 28 at 3:01

Mukyuu

340112

Partial answer 1-50 (w/e 41,47):

$1= 1+3+3-6$
$2= 1 + (frac{6}{(3+3)})$
$3= 1^3+(frac{6}{3})$
$4= (frac{6}{3})+3-1$
$5= (frac{6}{3})+3^1$
$6= 6^1+3-3$
$7= 6+1-3+3$
$8= 6 + 3 - 1^3$
$9= 1^3 * (3+6)$
$10= 1^3 +3+6$
$11= 13 - (frac{6}{3})$
$12= 6+3+3^1$
$13= 6+3+3+1$
$14= 6*3 - 3 - 1$
$15= 6*3 - 3^1$
$16= 16 + 3 - 3$
$17= 16 + (frac{3}{3})$
$18= (frac{6*3}{1^3})$
$19= 6*3+1^3$
$20= 6*3+3-1$
$21= 6*3+3^1$
$22= 6*3+3+1$
$23= 36-13$
$24= 6*(3+1^3)$
$25= 16+(3*3)$
$26= 13*(frac{6}{3})$
$27= 33-6^1$
$28= 33-6+1$
$29= 31-(frac{6}{3})$
$30= 6*(3+3-1)$
$31= 13+3*6$
$32= 3^3+6-1$
$33= (frac{33}{1^6})$
$34= 33+1^6$
$35= (3+3)*6-1$
$36= (3+3)^1*6$
$37= 1+(3+3)*6$
$38= 33+6-1$
$39= 33+6^1$
$40= 1+33+6$
$41= $
$42= (1+3+3)*6$
$43= 16+3^3$
$44= round(sqrt{6^3}*3^1)$
$45= 3*3*(6-1)$
$46= ceil(sqrt{6^3}*3)+1$
$47= $
$48= ((3*3)-1)*6$
$49= 16+33$
$50= 63-13$

edited Jan 28 at 3:39

answered Jan 28 at 3:01

Mukyuu

340112

edited Jan 28 at 3:39

answered Jan 28 at 3:01

Mukyuu

340112

answered Jan 28 at 3:01

Mukyuu

340112

answered Jan 28 at 3:01

Mukyuu

340112

add a comment |

Alrighty, I'm piggybacking off of @OmegaKrypton else and adding some of my own.

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1^3 +3/6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1^3 + 3+6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 3*6+3!-1$ or $ (6-1)^(3!/3)$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ 36-3!-1$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3+16$

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ 16*(3!-3)$

49: $13+36$

50: $ 63-13$

I added a few. It's getting late here; will come back tomorrow.

edited Jan 28 at 6:12

Omega Krypton

3,6941338

answered Jan 28 at 2:58

Brandon_J

1,28427

add a comment |

Alrighty, I'm piggybacking off of @OmegaKrypton else and adding some of my own.

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1^3 +3/6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1^3 + 3+6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 3*6+3!-1$ or $ (6-1)^(3!/3)$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ 36-3!-1$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3+16$

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ 16*(3!-3)$

49: $13+36$

50: $ 63-13$

I added a few. It's getting late here; will come back tomorrow.

edited Jan 28 at 6:12

Omega Krypton

3,6941338

answered Jan 28 at 2:58

Brandon_J

1,28427

add a comment |

Alrighty, I'm piggybacking off of @OmegaKrypton else and adding some of my own.

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1^3 +3/6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1^3 + 3+6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 3*6+3!-1$ or $ (6-1)^(3!/3)$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ 36-3!-1$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3+16$

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ 16*(3!-3)$

49: $13+36$

50: $ 63-13$

I added a few. It's getting late here; will come back tomorrow.

edited Jan 28 at 6:12

Omega Krypton

3,6941338

answered Jan 28 at 2:58

Brandon_J

1,28427

Alrighty, I'm piggybacking off of @OmegaKrypton else and adding some of my own.

1 to 10

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1^3 +3/6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1^3 + 3+6$

11 to 20

11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21 to 30

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 3*6+3!-1$ or $ (6-1)^(3!/3)$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ 36-3!-1$

30: $ (-1+3+3)times 6$

31 to 40

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41 to 50

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3+16$

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ 16*(3!-3)$

49: $13+36$

50: $ 63-13$

I added a few. It's getting late here; will come back tomorrow.

edited Jan 28 at 6:12

Omega Krypton

3,6941338

answered Jan 28 at 2:58

Brandon_J

1,28427

edited Jan 28 at 6:12

Omega Krypton

3,6941338

edited Jan 28 at 6:12

Omega Krypton

3,6941338

edited Jan 28 at 6:12

Omega Krypton

3,6941338

answered Jan 28 at 2:58

Brandon_J

1,28427

answered Jan 28 at 2:58

Brandon_J

1,28427

answered Jan 28 at 2:58

Brandon_J

1,28427

add a comment |

Expanding on Bass's answer, I added some new numbers.

(I lost track on which numbers I added, though 1-40 is all Bass)

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$
11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3 + 16 $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51: $ 16*3+3 $

52: $ 61 - 3times3$

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ (6times3+1)times3$

58: $ (1+3)^3-6$

59: $ 63 - 3 - 1$

60: $ (1+3times3)times6$

61: $ 63 - 3 + 1$

62: $ 63 - 1^3$

63: $ 63 * 1^3$

64: $ (1+3/3)^6$

65: $ 63 + 3 - 1$

66: $ 1 times (63 + 3) $

67: $ 63 + 3 + 1$

68: $ $

69: $ $

70: $ (1+3)^3+6 $
71: $ 6^3 / 3 - 1 $

72: $ (1+3)times 3 times 6$

73: $ 6^3 / 3 + 1 $

74: $ $

75: $ 3^(3+1) - 6$

76: $ 63+13 $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ 3^(3+1) + 6$

88: $ 61 + 3^3 $

89: $ $

90: $ $

91: $ $

92: $ $

93: $ $

94: $ 33 + 61 $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ 31times3+6$

100: $ $

Only need 41, 44, 46, 47, 50, 56, 68, 69, 74, 77, 79, 83, 84, 85, 86, 89, 90, 91, 92, 93, 94, 95, 97, 98, and 100 now!

edited Jan 30 at 3:49

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

New contributor

$begingroup$
Wow that is A LOT of numbers
$endgroup$
– North
Jan 30 at 3:07

add a comment |

Expanding on Bass's answer, I added some new numbers.

(I lost track on which numbers I added, though 1-40 is all Bass)

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$
11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3 + 16 $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51: $ 16*3+3 $

52: $ 61 - 3times3$

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ (6times3+1)times3$

58: $ (1+3)^3-6$

59: $ 63 - 3 - 1$

60: $ (1+3times3)times6$

61: $ 63 - 3 + 1$

62: $ 63 - 1^3$

63: $ 63 * 1^3$

64: $ (1+3/3)^6$

65: $ 63 + 3 - 1$

66: $ 1 times (63 + 3) $

67: $ 63 + 3 + 1$

68: $ $

69: $ $

70: $ (1+3)^3+6 $
71: $ 6^3 / 3 - 1 $

72: $ (1+3)times 3 times 6$

73: $ 6^3 / 3 + 1 $

74: $ $

75: $ 3^(3+1) - 6$

76: $ 63+13 $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ 3^(3+1) + 6$

88: $ 61 + 3^3 $

89: $ $

90: $ $

91: $ $

92: $ $

93: $ $

94: $ 33 + 61 $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ 31times3+6$

100: $ $

Only need 41, 44, 46, 47, 50, 56, 68, 69, 74, 77, 79, 83, 84, 85, 86, 89, 90, 91, 92, 93, 94, 95, 97, 98, and 100 now!

edited Jan 30 at 3:49

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

New contributor

$begingroup$
Wow that is A LOT of numbers
$endgroup$
– North
Jan 30 at 3:07

add a comment |

Expanding on Bass's answer, I added some new numbers.

(I lost track on which numbers I added, though 1-40 is all Bass)

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$
11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3 + 16 $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51: $ 16*3+3 $

52: $ 61 - 3times3$

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ (6times3+1)times3$

58: $ (1+3)^3-6$

59: $ 63 - 3 - 1$

60: $ (1+3times3)times6$

61: $ 63 - 3 + 1$

62: $ 63 - 1^3$

63: $ 63 * 1^3$

64: $ (1+3/3)^6$

65: $ 63 + 3 - 1$

66: $ 1 times (63 + 3) $

67: $ 63 + 3 + 1$

68: $ $

69: $ $

70: $ (1+3)^3+6 $
71: $ 6^3 / 3 - 1 $

72: $ (1+3)times 3 times 6$

73: $ 6^3 / 3 + 1 $

74: $ $

75: $ 3^(3+1) - 6$

76: $ 63+13 $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ 3^(3+1) + 6$

88: $ 61 + 3^3 $

89: $ $

90: $ $

91: $ $

92: $ $

93: $ $

94: $ 33 + 61 $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ 31times3+6$

100: $ $

Only need 41, 44, 46, 47, 50, 56, 68, 69, 74, 77, 79, 83, 84, 85, 86, 89, 90, 91, 92, 93, 94, 95, 97, 98, and 100 now!

edited Jan 30 at 3:49

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

New contributor

Expanding on Bass's answer, I added some new numbers.

(I lost track on which numbers I added, though 1-40 is all Bass)

1: $1 + 3 + 3 - 6$

2: $(1 + 3) times 3 / 6$

3: $1*3 * 3 - 6$

4: $13 - 3 - 6$

5: $-1^{33} +6$

6: $1times3-3+6$

7: $ 1 + 3 -3 +6$

8: $ 1+3/3 + 6$

9: $ 1^3 times (3+6)$

10: $ 1 + sqrt[3]3^6$
11: $ sqrt{1+3}+3+6$

12: $1times 3 + 3 + 6$

13: $1 + 3+3+6$

14: $-1 + 3times 3+6$

15: $-1times3 + 3times 6$

16: $1 - 3 + 3 times 6$

17: $ -1^3 +3times 6$

18: $ (1+3)*3+6 $

19: $13 + sqrt{36}$

20: $-1 + 3^3 - 6$

21: $ 1 * 3^3 - 6 $

22: $ 13 + 3 + 6$

23: $ -13+36 $

24: $ (1+3)timessqrt{36}$

25: $ 1 - 3 + sqrt3^6$

26: $ -1+33-6$

27: $ 1*33-6 $

28: $ 1+33-6$

29: $ -1 + 3 + sqrt3^6$

30: $ (-1+3+3)times 6$

31: $ 13+3*6 $

32: $ -1+3^3+6$

33: $ 13*3-6 $

34: $ 1+3^3+6$

35: $ -1+(3+3)times6 $

36: $ 1times(3+3)times 6$

37: $ 1^3+36$

38: $ sqrt{1+3}+36$

39: $ 1times3 + 36$

40: $ 1+33+6$

41: $ $

42: $ (1+3+3)times 6$

43: $ 3^3 + 16 $

44: $ $

45: $ 13times3+6$

46: $ $

47: $ $

48: $ (-1 + 3 times 3) times 6 $

49: $13+36$

50: $ $

51: $ 16*3+3 $

52: $ 61 - 3times3$

53: $ -1 +3 times 3 times 6$

54: $ 1times 3 times 3 times 6$

55: $ 1 + 3 times 3 times 6$

56: $ $

57: $ (6times3+1)times3$

58: $ (1+3)^3-6$

59: $ 63 - 3 - 1$

60: $ (1+3times3)times6$

61: $ 63 - 3 + 1$

62: $ 63 - 1^3$

63: $ 63 * 1^3$

64: $ (1+3/3)^6$

65: $ 63 + 3 - 1$

66: $ 1 times (63 + 3) $

67: $ 63 + 3 + 1$

68: $ $

69: $ $

70: $ (1+3)^3+6 $
71: $ 6^3 / 3 - 1 $

72: $ (1+3)times 3 times 6$

73: $ 6^3 / 3 + 1 $

74: $ $

75: $ 3^(3+1) - 6$

76: $ 63+13 $

77: $ $

78: $ 13 times sqrt{36}$

79: $ $

80: $ -1 + 3timessqrt3^6$

81: $ 1times3timessqrt3^6$

82: $ 1+3timessqrt3^6$

83: $ $

84: $ $

85: $ $

86: $ $

87: $ 3^(3+1) + 6$

88: $ 61 + 3^3 $

89: $ $

90: $ $

91: $ $

92: $ $

93: $ $

94: $ 33 + 61 $

95: $ $

96: $ (13+3)times6 $

97: $ $

98: $ $

99: $ 31times3+6$

100: $ $

Only need 41, 44, 46, 47, 50, 56, 68, 69, 74, 77, 79, 83, 84, 85, 86, 89, 90, 91, 92, 93, 94, 95, 97, 98, and 100 now!

edited Jan 30 at 3:49

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

New contributor

edited Jan 30 at 3:49

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

New contributor

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

answered Jan 30 at 2:56

Embodiment of Ignorance

1096

New contributor

Embodiment of Ignorance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$begingroup$
Wow that is A LOT of numbers
$endgroup$
– North
Jan 30 at 3:07

add a comment |

$begingroup$
Wow that is A LOT of numbers
$endgroup$
– North
Jan 30 at 3:07

Wow that is A LOT of numbers

– North
Jan 30 at 3:07

add a comment |

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