how can it be possible to add three odd numbers and the answer is even
$begingroup$
From the days I started to learn Maths, I've have been taught that
Adding Odd times Odd numbers the Answer always would be Odd
e.g 3 + 5 + 1 = 9
OK, but look at this question
This question was solved and the answer was 30, how it was possible? Need a valid explanation please.
arithmetic recreational-mathematics puzzle
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
$endgroup$
|
show 2 more comments
$begingroup$
From the days I started to learn Maths, I've have been taught that
Adding Odd times Odd numbers the Answer always would be Odd
e.g 3 + 5 + 1 = 9
OK, but look at this question
This question was solved and the answer was 30, how it was possible? Need a valid explanation please.
arithmetic recreational-mathematics puzzle
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
$endgroup$
6
$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55
$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05
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I don't understand the downvotes, either. Added therecreational-mathematicsandpuzzletags where the question may be better received.
$endgroup$
– dxiv
Aug 24 '17 at 3:11
$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12
2
$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48
|
show 2 more comments
$begingroup$
From the days I started to learn Maths, I've have been taught that
Adding Odd times Odd numbers the Answer always would be Odd
e.g 3 + 5 + 1 = 9
OK, but look at this question
This question was solved and the answer was 30, how it was possible? Need a valid explanation please.
arithmetic recreational-mathematics puzzle
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
$endgroup$
From the days I started to learn Maths, I've have been taught that
Adding Odd times Odd numbers the Answer always would be Odd
e.g 3 + 5 + 1 = 9
OK, but look at this question
This question was solved and the answer was 30, how it was possible? Need a valid explanation please.
arithmetic recreational-mathematics puzzle
arithmetic recreational-mathematics puzzle
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
edited Aug 24 '17 at 3:09
dxiv
57.8k648101
57.8k648101
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
asked Aug 24 '17 at 2:52
Arshad AliArshad Ali
13514
13514
6
$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55
$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05
$begingroup$
I don't understand the downvotes, either. Added therecreational-mathematicsandpuzzletags where the question may be better received.
$endgroup$
– dxiv
Aug 24 '17 at 3:11
$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12
2
$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48
|
show 2 more comments
6
$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55
$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05
$begingroup$
I don't understand the downvotes, either. Added therecreational-mathematicsandpuzzletags where the question may be better received.
$endgroup$
– dxiv
Aug 24 '17 at 3:11
$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12
2
$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48
6
6
$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55
$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55
$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05
$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05
$begingroup$
I don't understand the downvotes, either. Added the
recreational-mathematics and puzzle tags where the question may be better received.$endgroup$
– dxiv
Aug 24 '17 at 3:11
$begingroup$
I don't understand the downvotes, either. Added the
recreational-mathematics and puzzle tags where the question may be better received.$endgroup$
– dxiv
Aug 24 '17 at 3:11
$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12
$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12
2
2
$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48
$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48
|
show 2 more comments
8 Answers
8
active
oldest
votes
$begingroup$
You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.
You should not believe everything you read in a photo on the internet.
answered Aug 24 '17 at 2:55
60056005
36.2k751125
$endgroup$
$begingroup$
That's so kind of you!
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:00
add a comment |
$begingroup$
Clearly impossible in base 10. Can it be done in another number base?
In base 5: $11+11+3=30$
Actually there are many other possibilities!
Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.
It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
$endgroup$
$begingroup$
May be! this one makes sense a lot.
$endgroup$
– Arshad Ali
Aug 24 '17 at 17:56
$begingroup$
If you do $1+11+13$ you won't even need to use a number twice.
$endgroup$
– Jyrki Lahtonen
Aug 24 '17 at 18:21
$begingroup$
Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
$endgroup$
– bluemaster
Aug 24 '17 at 23:52
add a comment |
$begingroup$
Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
$endgroup$
add a comment |
$begingroup$
If this is a riddle, I would do : 13,1 + 7,9 + 9
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
$endgroup$
$begingroup$
You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
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– cgiovanardi
Aug 24 '17 at 18:28
1
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Imitation is the sincerest form of flattery, as the saying goes ;-)
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– dxiv
Aug 26 '17 at 0:35
add a comment |
$begingroup$
you can also repeat the numbers
Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
decimal separator).
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
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$begingroup$
Ooh! so Strange...
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– Arshad Ali
Aug 24 '17 at 3:02
2
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@ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
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– Ross Millikan
Aug 24 '17 at 3:14
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@RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
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– dxiv
Aug 24 '17 at 3:19
1
$begingroup$
Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
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– bluemaster
Aug 25 '17 at 7:50
add a comment |
$begingroup$
What about
6+9+15=30?
they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
$endgroup$
add a comment |
$begingroup$
Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$
Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
$$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.
Edit:
Change of base can also be made to work if number repetition were eliminated.
$$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
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parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
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– bluemaster
Aug 25 '17 at 7:29
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@bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
$endgroup$
– Eric Towers
Aug 25 '17 at 13:27
add a comment |
$begingroup$
It’s $$3_{3} + 3_{3} +3_{3} = 30.$$
Where $3_{3}$ is read as $3$ base $3$.
edited Nov 27 '18 at 1:34
dantopa
6,46942243
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
$endgroup$
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.
You should not believe everything you read in a photo on the internet.
answered Aug 24 '17 at 2:55
60056005
36.2k751125
$endgroup$
$begingroup$
That's so kind of you!
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:00
add a comment |
$begingroup$
You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.
You should not believe everything you read in a photo on the internet.
answered Aug 24 '17 at 2:55
60056005
36.2k751125
$endgroup$
$begingroup$
That's so kind of you!
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:00
add a comment |
$begingroup$
You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.
You should not believe everything you read in a photo on the internet.
answered Aug 24 '17 at 2:55
60056005
36.2k751125
$endgroup$
You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.
You should not believe everything you read in a photo on the internet.
answered Aug 24 '17 at 2:55
60056005
36.2k751125
answered Aug 24 '17 at 2:55
60056005
36.2k751125
answered Aug 24 '17 at 2:55
60056005
36.2k751125
answered Aug 24 '17 at 2:55
60056005
36.2k751125
36.2k751125
$begingroup$
That's so kind of you!
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:00
add a comment |
$begingroup$
That's so kind of you!
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:00
$begingroup$
That's so kind of you!
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:00
$begingroup$
That's so kind of you!
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:00
add a comment |
$begingroup$
Clearly impossible in base 10. Can it be done in another number base?
In base 5: $11+11+3=30$
Actually there are many other possibilities!
Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.
It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
$endgroup$
$begingroup$
May be! this one makes sense a lot.
$endgroup$
– Arshad Ali
Aug 24 '17 at 17:56
$begingroup$
If you do $1+11+13$ you won't even need to use a number twice.
$endgroup$
– Jyrki Lahtonen
Aug 24 '17 at 18:21
$begingroup$
Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
$endgroup$
– bluemaster
Aug 24 '17 at 23:52
add a comment |
$begingroup$
Clearly impossible in base 10. Can it be done in another number base?
In base 5: $11+11+3=30$
Actually there are many other possibilities!
Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.
It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
$endgroup$
$begingroup$
May be! this one makes sense a lot.
$endgroup$
– Arshad Ali
Aug 24 '17 at 17:56
$begingroup$
If you do $1+11+13$ you won't even need to use a number twice.
$endgroup$
– Jyrki Lahtonen
Aug 24 '17 at 18:21
$begingroup$
Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
$endgroup$
– bluemaster
Aug 24 '17 at 23:52
add a comment |
$begingroup$
Clearly impossible in base 10. Can it be done in another number base?
In base 5: $11+11+3=30$
Actually there are many other possibilities!
Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.
It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
$endgroup$
Clearly impossible in base 10. Can it be done in another number base?
In base 5: $11+11+3=30$
Actually there are many other possibilities!
Another possibility: fill the boxes with $binom{5}{3}$,15 and 5.
It holds that $binom{5}{3}+15+5=30$ (base 10). Notice that $binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
edited Aug 25 '17 at 12:02
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
answered Aug 24 '17 at 13:34
bluemasterbluemaster
1,638519
1,638519
$begingroup$
May be! this one makes sense a lot.
$endgroup$
– Arshad Ali
Aug 24 '17 at 17:56
$begingroup$
If you do $1+11+13$ you won't even need to use a number twice.
$endgroup$
– Jyrki Lahtonen
Aug 24 '17 at 18:21
$begingroup$
Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
$endgroup$
– bluemaster
Aug 24 '17 at 23:52
add a comment |
$begingroup$
May be! this one makes sense a lot.
$endgroup$
– Arshad Ali
Aug 24 '17 at 17:56
$begingroup$
If you do $1+11+13$ you won't even need to use a number twice.
$endgroup$
– Jyrki Lahtonen
Aug 24 '17 at 18:21
$begingroup$
Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
$endgroup$
– bluemaster
Aug 24 '17 at 23:52
$begingroup$
May be! this one makes sense a lot.
$endgroup$
– Arshad Ali
Aug 24 '17 at 17:56
$begingroup$
May be! this one makes sense a lot.
$endgroup$
– Arshad Ali
Aug 24 '17 at 17:56
$begingroup$
If you do $1+11+13$ you won't even need to use a number twice.
$endgroup$
– Jyrki Lahtonen
Aug 24 '17 at 18:21
$begingroup$
If you do $1+11+13$ you won't even need to use a number twice.
$endgroup$
– Jyrki Lahtonen
Aug 24 '17 at 18:21
$begingroup$
Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
$endgroup$
– bluemaster
Aug 24 '17 at 23:52
$begingroup$
Or 1+11+15=30 (base 7), just to have another instance without repetition, in another base :).
$endgroup$
– bluemaster
Aug 24 '17 at 23:52
add a comment |
$begingroup$
Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
$endgroup$
add a comment |
$begingroup$
Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
$endgroup$
add a comment |
$begingroup$
Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
$endgroup$
Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$fbox{ } + fbox{15} + fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
answered Aug 25 '17 at 3:52
Robert SoupeRobert Soupe
11k21950
11k21950
add a comment |
add a comment |
$begingroup$
If this is a riddle, I would do : 13,1 + 7,9 + 9
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
$endgroup$
$begingroup$
You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
$endgroup$
– cgiovanardi
Aug 24 '17 at 18:28
1
$begingroup$
Imitation is the sincerest form of flattery, as the saying goes ;-)
$endgroup$
– dxiv
Aug 26 '17 at 0:35
add a comment |
$begingroup$
If this is a riddle, I would do : 13,1 + 7,9 + 9
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
$endgroup$
$begingroup$
You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
$endgroup$
– cgiovanardi
Aug 24 '17 at 18:28
1
$begingroup$
Imitation is the sincerest form of flattery, as the saying goes ;-)
$endgroup$
– dxiv
Aug 26 '17 at 0:35
add a comment |
$begingroup$
If this is a riddle, I would do : 13,1 + 7,9 + 9
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
$endgroup$
If this is a riddle, I would do : 13,1 + 7,9 + 9
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
answered Aug 24 '17 at 17:30
cgiovanardicgiovanardi
749411
749411
$begingroup$
You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
$endgroup$
– cgiovanardi
Aug 24 '17 at 18:28
1
$begingroup$
Imitation is the sincerest form of flattery, as the saying goes ;-)
$endgroup$
– dxiv
Aug 26 '17 at 0:35
add a comment |
$begingroup$
You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
$endgroup$
– cgiovanardi
Aug 24 '17 at 18:28
1
$begingroup$
Imitation is the sincerest form of flattery, as the saying goes ;-)
$endgroup$
– dxiv
Aug 26 '17 at 0:35
$begingroup$
You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
$endgroup$
– cgiovanardi
Aug 24 '17 at 18:28
$begingroup$
You didn't like my answer? What about this: (1+3)+19+11 or (1+1)+13+15? These questions are usually asked to check if people think out of the box or not.
$endgroup$
– cgiovanardi
Aug 24 '17 at 18:28
1
1
$begingroup$
Imitation is the sincerest form of flattery, as the saying goes ;-)
$endgroup$
– dxiv
Aug 26 '17 at 0:35
$begingroup$
Imitation is the sincerest form of flattery, as the saying goes ;-)
$endgroup$
– dxiv
Aug 26 '17 at 0:35
add a comment |
$begingroup$
you can also repeat the numbers
Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
decimal separator).
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
$endgroup$
$begingroup$
Ooh! so Strange...
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:02
2
$begingroup$
@ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:14
$begingroup$
@RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
$endgroup$
– dxiv
Aug 24 '17 at 3:19
1
$begingroup$
Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
$endgroup$
– bluemaster
Aug 25 '17 at 7:50
add a comment |
$begingroup$
you can also repeat the numbers
Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
decimal separator).
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
$endgroup$
$begingroup$
Ooh! so Strange...
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:02
2
$begingroup$
@ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:14
$begingroup$
@RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
$endgroup$
– dxiv
Aug 24 '17 at 3:19
1
$begingroup$
Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
$endgroup$
– bluemaster
Aug 25 '17 at 7:50
add a comment |
$begingroup$
you can also repeat the numbers
Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
decimal separator).
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
$endgroup$
you can also repeat the numbers
Wonder if that means $,11,5+13,5+5=30,$ (where the $,,,$ comma works as
decimal separator).
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
answered Aug 24 '17 at 3:01
dxivdxiv
57.8k648101
57.8k648101
$begingroup$
Ooh! so Strange...
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:02
2
$begingroup$
@ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:14
$begingroup$
@RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
$endgroup$
– dxiv
Aug 24 '17 at 3:19
1
$begingroup$
Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
$endgroup$
– bluemaster
Aug 25 '17 at 7:50
add a comment |
$begingroup$
Ooh! so Strange...
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:02
2
$begingroup$
@ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:14
$begingroup$
@RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
$endgroup$
– dxiv
Aug 24 '17 at 3:19
1
$begingroup$
Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
$endgroup$
– bluemaster
Aug 25 '17 at 7:50
$begingroup$
Ooh! so Strange...
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:02
$begingroup$
Ooh! so Strange...
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:02
2
2
$begingroup$
@ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:14
$begingroup$
@ArshadAli: this is the kind of unfair solution I was talking about. I enjoy puzzles, as in the riddle I wrote in my first comment. They are not fair at a final exam.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:14
$begingroup$
@RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
$endgroup$
– dxiv
Aug 24 '17 at 3:19
$begingroup$
@RossMillikan Completely agree with your comment. That said, I have no idea what a "UPSC final exam" might be. The first page of google hits brings up mainly links to this very puzzle, which is a bit odd in itself.
$endgroup$
– dxiv
Aug 24 '17 at 3:19
1
1
$begingroup$
Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
$endgroup$
– bluemaster
Aug 25 '17 at 7:50
$begingroup$
Most likely, UPSC = Union Public Service Commission, notorious for their difficult entrance exams for public employees in India. Check here for some previous exams.
$endgroup$
– bluemaster
Aug 25 '17 at 7:50
add a comment |
$begingroup$
What about
6+9+15=30?
they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
$endgroup$
add a comment |
$begingroup$
What about
6+9+15=30?
they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
$endgroup$
add a comment |
$begingroup$
What about
6+9+15=30?
they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
$endgroup$
What about
6+9+15=30?
they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
answered Aug 25 '17 at 11:20
AndrewAndrew
1071212
1071212
add a comment |
add a comment |
$begingroup$
Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$
Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
$$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.
Edit:
Change of base can also be made to work if number repetition were eliminated.
$$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
$endgroup$
$begingroup$
parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
$endgroup$
– bluemaster
Aug 25 '17 at 7:29
$begingroup$
@bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
$endgroup$
– Eric Towers
Aug 25 '17 at 13:27
add a comment |
$begingroup$
Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$
Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
$$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.
Edit:
Change of base can also be made to work if number repetition were eliminated.
$$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
$endgroup$
$begingroup$
parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
$endgroup$
– bluemaster
Aug 25 '17 at 7:29
$begingroup$
@bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
$endgroup$
– Eric Towers
Aug 25 '17 at 13:27
add a comment |
$begingroup$
Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$
Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
$$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.
Edit:
Change of base can also be made to work if number repetition were eliminated.
$$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
$endgroup$
Change the base: $$11_9 + 11_9 + 11_9 = 10+10+10 = 30 text{.}$$
Parentheses are in the list of usable box contents. Commas too. But you're upper limited to one pair of parens and to seven commas.
$$(15+15,15)+15 = 30 \ (15+15,15+15) = 30 text{.}$$ Here, the parens represent the GCD.
Edit:
Change of base can also be made to work if number repetition were eliminated.
$$ 11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30 text{.} $$
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
edited Aug 31 '17 at 14:25
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
answered Aug 25 '17 at 4:18
Eric TowersEric Towers
32.4k22268
32.4k22268
$begingroup$
parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
$endgroup$
– bluemaster
Aug 25 '17 at 7:29
$begingroup$
@bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
$endgroup$
– Eric Towers
Aug 25 '17 at 13:27
add a comment |
$begingroup$
parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
$endgroup$
– bluemaster
Aug 25 '17 at 7:29
$begingroup$
@bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
$endgroup$
– Eric Towers
Aug 25 '17 at 13:27
$begingroup$
parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
$endgroup$
– bluemaster
Aug 25 '17 at 7:29
$begingroup$
parentheses and commas are in the list for usable symbols, but not the + sign. Your second solution uses an extra plus sign inside the first box, not in the list of allowed symbols :).
$endgroup$
– bluemaster
Aug 25 '17 at 7:29
$begingroup$
@bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
$endgroup$
– Eric Towers
Aug 25 '17 at 13:27
$begingroup$
@bluemaster : Meh. It was late enough I didn't notice. But, ... [edited]
$endgroup$
– Eric Towers
Aug 25 '17 at 13:27
add a comment |
$begingroup$
It’s $$3_{3} + 3_{3} +3_{3} = 30.$$
Where $3_{3}$ is read as $3$ base $3$.
edited Nov 27 '18 at 1:34
dantopa
6,46942243
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
$endgroup$
add a comment |
$begingroup$
It’s $$3_{3} + 3_{3} +3_{3} = 30.$$
Where $3_{3}$ is read as $3$ base $3$.
edited Nov 27 '18 at 1:34
dantopa
6,46942243
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
$endgroup$
add a comment |
$begingroup$
It’s $$3_{3} + 3_{3} +3_{3} = 30.$$
Where $3_{3}$ is read as $3$ base $3$.
edited Nov 27 '18 at 1:34
dantopa
6,46942243
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
$endgroup$
It’s $$3_{3} + 3_{3} +3_{3} = 30.$$
Where $3_{3}$ is read as $3$ base $3$.
edited Nov 27 '18 at 1:34
dantopa
6,46942243
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
edited Nov 27 '18 at 1:34
dantopa
6,46942243
edited Nov 27 '18 at 1:34
dantopa
6,46942243
edited Nov 27 '18 at 1:34
dantopa
6,46942243
6,46942243
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
answered Nov 27 '18 at 0:39
Darshan JainDarshan Jain
71
71
add a comment |
add a comment |
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6
$begingroup$
Do you believe the statement that somebody solved it fairly? Why? There is an old riddle that you have three bags and thirty stones. You need to put the stones in the bags so each bag contains an odd number of stones. One solution is to put fifteen stones into each of two bags, then put one of those two bags into the third. That doesn't work here.
$endgroup$
– Ross Millikan
Aug 24 '17 at 2:55
$begingroup$
Why -ve 1, please comment
$endgroup$
– Arshad Ali
Aug 24 '17 at 3:05
$begingroup$
I don't understand the downvotes, either. Added the
recreational-mathematicsandpuzzletags where the question may be better received.$endgroup$
– dxiv
Aug 24 '17 at 3:11
$begingroup$
I didn't downvote. I am surprised that this got an upvote. It also got two downvotes (as of this comment). People seem to agree with my question of why you believe the statement that somebody solved it because of the reason you gave.
$endgroup$
– Ross Millikan
Aug 24 '17 at 3:12
2
$begingroup$
It's possible there is a misunderstanding about "XYZ could solve it". It may mean that XYZ correctly answered the question "can you solve this <equation>".. Presumably his solution stated "no, we cannot solve <equation>" for the reason you gave.
$endgroup$
– Bill Dubuque
Aug 24 '17 at 3:48