Factors in a different base $ 2b^2!+!9b!+!7,mid, 7b^2!+!9b!+!2$
$begingroup$
Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?
Solution:
But base cannot be negative. Could someone please explain where I am going wrong?
elementary-number-theory divisibility number-systems
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
$endgroup$
add a comment |
$begingroup$
Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?
Solution:
But base cannot be negative. Could someone please explain where I am going wrong?
elementary-number-theory divisibility number-systems
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
$endgroup$
add a comment |
$begingroup$
Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?
Solution:
But base cannot be negative. Could someone please explain where I am going wrong?
elementary-number-theory divisibility number-systems
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
$endgroup$
Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?
Solution:
But base cannot be negative. Could someone please explain where I am going wrong?
elementary-number-theory divisibility number-systems
elementary-number-theory divisibility number-systems
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
edited Dec 31 '18 at 21:51
Maria Mazur
50.3k1361126
50.3k1361126
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
asked Dec 31 '18 at 20:29
Aamir KhanAamir Khan
455
455
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.
If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.
If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.
No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
$endgroup$
$begingroup$
Thank you so much. This was really very helpful. :)
$endgroup$
– Aamir Khan
Dec 31 '18 at 20:37
$begingroup$
My pleasure, glad to help.
$endgroup$
– vadim123
Dec 31 '18 at 20:39
$begingroup$
@AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:26
1
$begingroup$
@BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
$endgroup$
– vadim123
Dec 31 '18 at 21:48
$begingroup$
@vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:58
|
show 4 more comments
$begingroup$
Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$
Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
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add a comment |
$begingroup$
Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$
we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$
so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$
which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
add a comment |
$begingroup$
$$2B^2+9B+7mid 7B^2+9B+2$$
Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.
Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$
Writing this out "subtraction-style", we get
$left.begin{array}{c}
& 7 & 9 & 2 \
-& 2 & 9 & 7 \
hline
phantom{4}
end{array}
right.
implies
left.begin{array}{c}
& 6 & (B+8) & (B+2) \
-& 2 & 9 & 7 \
hline
& 4 & (B-1) & (B-5)
end{array}
right.
$
So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.
We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.
If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.
If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.
No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
$endgroup$
$begingroup$
Thank you so much. This was really very helpful. :)
$endgroup$
– Aamir Khan
Dec 31 '18 at 20:37
$begingroup$
My pleasure, glad to help.
$endgroup$
– vadim123
Dec 31 '18 at 20:39
$begingroup$
@AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:26
1
$begingroup$
@BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
$endgroup$
– vadim123
Dec 31 '18 at 21:48
$begingroup$
@vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:58
|
show 4 more comments
$begingroup$
The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.
If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.
If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.
No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
$endgroup$
$begingroup$
Thank you so much. This was really very helpful. :)
$endgroup$
– Aamir Khan
Dec 31 '18 at 20:37
$begingroup$
My pleasure, glad to help.
$endgroup$
– vadim123
Dec 31 '18 at 20:39
$begingroup$
@AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:26
1
$begingroup$
@BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
$endgroup$
– vadim123
Dec 31 '18 at 21:48
$begingroup$
@vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:58
|
show 4 more comments
$begingroup$
The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.
If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.
If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.
No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
$endgroup$
The long division is the source of the error; you can't have $7/2$ as the quotient. The quotient needs to be an integer, that's what "factor" means.
If the quotient is $2$, then the base is $4$. This is found by solving $7B^2+9B+2=color{red}{ 2}(2B^2+9B+7)$, and discarding the negative root.
If the quotient is $3$, then the base is $19$. This is found by solving $7B^2+9B+2=color{red}{ 3}(2B^2+9B+7)$, and discarding the negative root.
No other quotients make any sense. However, if the base is $4$, then you don't get digits $7$ and $9$. Hence the answer must be $B=19$.
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
answered Dec 31 '18 at 20:35
vadim123vadim123
76.7k899191
76.7k899191
$begingroup$
Thank you so much. This was really very helpful. :)
$endgroup$
– Aamir Khan
Dec 31 '18 at 20:37
$begingroup$
My pleasure, glad to help.
$endgroup$
– vadim123
Dec 31 '18 at 20:39
$begingroup$
@AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:26
1
$begingroup$
@BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
$endgroup$
– vadim123
Dec 31 '18 at 21:48
$begingroup$
@vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:58
|
show 4 more comments
$begingroup$
Thank you so much. This was really very helpful. :)
$endgroup$
– Aamir Khan
Dec 31 '18 at 20:37
$begingroup$
My pleasure, glad to help.
$endgroup$
– vadim123
Dec 31 '18 at 20:39
$begingroup$
@AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:26
1
$begingroup$
@BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
$endgroup$
– vadim123
Dec 31 '18 at 21:48
$begingroup$
@vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:58
$begingroup$
Thank you so much. This was really very helpful. :)
$endgroup$
– Aamir Khan
Dec 31 '18 at 20:37
$begingroup$
Thank you so much. This was really very helpful. :)
$endgroup$
– Aamir Khan
Dec 31 '18 at 20:37
$begingroup$
My pleasure, glad to help.
$endgroup$
– vadim123
Dec 31 '18 at 20:39
$begingroup$
My pleasure, glad to help.
$endgroup$
– vadim123
Dec 31 '18 at 20:39
$begingroup$
@AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:26
$begingroup$
@AamirKhan Beware that the above is not a rigorous solution without proof that those are the only possible quotients. See my answer for one rigorous approach
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:26
1
1
$begingroup$
@BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
$endgroup$
– vadim123
Dec 31 '18 at 21:48
$begingroup$
@BillDubuque, it's not difficult to make it rigorous, to prove that the quotient $q$ can't be bigger than $3$. $(2q-7)B^2+(9q-9)B+(7q-2)$ is strictly positive if $qge 4$ and $B>0$. Hence if $qge 4$, no positive $B$ solves the equation. Similarly, if $q=1$, then the only positive solution is $B=1$, which is not possible.
$endgroup$
– vadim123
Dec 31 '18 at 21:48
$begingroup$
@vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:58
$begingroup$
@vadim You should add the details of a rigorous proof to the answer (I've lost count of the number of times "[this case] doesn't make sense" turned out to be incorrect), so we should not encourage students to handwave like that.
$endgroup$
– Bill Dubuque
Dec 31 '18 at 21:58
|
show 4 more comments
$begingroup$
Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$
Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
$begingroup$
Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$
Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
$begingroup$
Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$
Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
Going $1$ step more with Euclid's algorithm reveals a common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = color{#c00}{dfrac{7b!+!2}{2b!+!7}}inBbb Z , Rightarrow, 7-2 dfrac{color{#c00}{7b!+!2}}{ color{#c00}{2b!+!7}}, =, dfrac{45}{2b!+!7}inBbb Zqquad$$
Therefore $,2b!+!7mid 45 $ so $,b> 9,$(= digit) $,Rightarrow,2b!+!7 = 45,$ $Rightarrow,b=19.$
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
edited Jan 1 at 3:46
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
answered Dec 31 '18 at 21:12
Bill DubuqueBill Dubuque
214k29197660
214k29197660
add a comment |
add a comment |
$begingroup$
Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$
we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$
so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$
which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
add a comment |
$begingroup$
Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$
we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$
so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$
which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
add a comment |
$begingroup$
Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$
we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$
so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$
which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
Since $b+1>0$ and $$(b+1)(2b+7)mid (7b+2)(b+1)implies 2b+7mid 7b+2$$
we have $$2b+7mid (7b+2)-3(2b+7) = b-19$$
so if $b-19> 0$ we have $$2b+7mid b-19 implies 2b+7leq b-19 implies b+26leq 0$$
which is not true. So $bleq 19$. By trial and error we see that $b=4$ and $b=19$ works.
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
edited Dec 31 '18 at 21:49
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
answered Dec 31 '18 at 21:43
Maria MazurMaria Mazur
50.3k1361126
50.3k1361126
add a comment |
add a comment |
$begingroup$
$$2B^2+9B+7mid 7B^2+9B+2$$
Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.
Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$
Writing this out "subtraction-style", we get
$left.begin{array}{c}
& 7 & 9 & 2 \
-& 2 & 9 & 7 \
hline
phantom{4}
end{array}
right.
implies
left.begin{array}{c}
& 6 & (B+8) & (B+2) \
-& 2 & 9 & 7 \
hline
& 4 & (B-1) & (B-5)
end{array}
right.
$
So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.
We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
$endgroup$
add a comment |
$begingroup$
$$2B^2+9B+7mid 7B^2+9B+2$$
Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.
Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$
Writing this out "subtraction-style", we get
$left.begin{array}{c}
& 7 & 9 & 2 \
-& 2 & 9 & 7 \
hline
phantom{4}
end{array}
right.
implies
left.begin{array}{c}
& 6 & (B+8) & (B+2) \
-& 2 & 9 & 7 \
hline
& 4 & (B-1) & (B-5)
end{array}
right.
$
So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.
We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
$endgroup$
add a comment |
$begingroup$
$$2B^2+9B+7mid 7B^2+9B+2$$
Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.
Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$
Writing this out "subtraction-style", we get
$left.begin{array}{c}
& 7 & 9 & 2 \
-& 2 & 9 & 7 \
hline
phantom{4}
end{array}
right.
implies
left.begin{array}{c}
& 6 & (B+8) & (B+2) \
-& 2 & 9 & 7 \
hline
& 4 & (B-1) & (B-5)
end{array}
right.
$
So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.
We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
$endgroup$
$$2B^2+9B+7mid 7B^2+9B+2$$
Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.
Then $[2,9,7]_B mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$
Writing this out "subtraction-style", we get
$left.begin{array}{c}
& 7 & 9 & 2 \
-& 2 & 9 & 7 \
hline
phantom{4}
end{array}
right.
implies
left.begin{array}{c}
& 6 & (B+8) & (B+2) \
-& 2 & 9 & 7 \
hline
& 4 & (B-1) & (B-5)
end{array}
right.
$
So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.
We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
edited Jan 2 at 1:56
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
answered Jan 1 at 1:35
steven gregorysteven gregory
18.5k32359
18.5k32359
add a comment |
add a comment |
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