Factors in a different base $ 2b^2!+!9b!+!7,mid, 7b^2!+!9b!+!2$
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 5 hours ago
greedoid
37.9k114794
asked 7 hours ago
Aamir Khan
405
add a comment |
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 5 hours ago
greedoid
37.9k114794
asked 7 hours ago
Aamir Khan
405
add a comment |
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 5 hours ago
greedoid
37.9k114794
asked 7 hours ago
Aamir Khan
405
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 5 hours ago
greedoid
37.9k114794
asked 7 hours ago
Aamir Khan
405
edited 5 hours ago
greedoid
37.9k114794
asked 7 hours ago
Aamir Khan
405
edited 5 hours ago
greedoid
37.9k114794
edited 5 hours ago
greedoid
37.9k114794
edited 5 hours ago
greedoid
37.9k114794
37.9k114794
asked 7 hours ago
Aamir Khan
405
asked 7 hours ago
Aamir Khan
405
asked 7 hours ago
Aamir Khan
405
405
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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 7 hours ago
vadim123
75.4k896187
Thank you so much. This was really very helpful. :)
– Aamir Khan
6 hours ago
My pleasure, glad to help.
– vadim123
6 hours ago
@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
– Bill Dubuque
6 hours ago
@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.
– vadim123
5 hours ago
@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.
– Bill Dubuque
5 hours ago
|
show 4 more comments
Going $1$ step more with Euclid's algorithm shows they have common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = dfrac{7b!+!2}{2b!+!7}inBbb Z Rightarrow 7-dfrac{2,(7b!+!2)}{ 2b+7} = dfrac{45}{2b+7}inBbb Zqquad$$
Hence $,2b!+!7mid 45 $ so $ 2b!+!7 = 45,$ or $15,$ (by $b>1),,$ so $,b = 19,$ or $4$.
answered 6 hours ago
Bill Dubuque
208k29190628
add a comment |
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 5 hours ago
greedoid
37.9k114794
This answer is of no use?
– greedoid
3 hours ago
add a comment |
$$2B^2+9B+7mid 7B^2+9B+2$$
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 $[4,B-1,B-5]_B = 2[2,9,7] = [4,18,14]$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 2 hours ago
steven gregory
17.7k32257
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
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 7 hours ago
vadim123
75.4k896187
Thank you so much. This was really very helpful. :)
– Aamir Khan
6 hours ago
My pleasure, glad to help.
– vadim123
6 hours ago
@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
– Bill Dubuque
6 hours ago
@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.
– vadim123
5 hours ago
@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.
– Bill Dubuque
5 hours ago
|
show 4 more comments
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 7 hours ago
vadim123
75.4k896187
Thank you so much. This was really very helpful. :)
– Aamir Khan
6 hours ago
My pleasure, glad to help.
– vadim123
6 hours ago
@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
– Bill Dubuque
6 hours ago
@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.
– vadim123
5 hours ago
@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.
– Bill Dubuque
5 hours ago
|
show 4 more comments
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 7 hours ago
vadim123
75.4k896187
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 7 hours ago
vadim123
75.4k896187
answered 7 hours ago
vadim123
75.4k896187
answered 7 hours ago
vadim123
75.4k896187
answered 7 hours ago
vadim123
75.4k896187
75.4k896187
Thank you so much. This was really very helpful. :)
– Aamir Khan
6 hours ago
My pleasure, glad to help.
– vadim123
6 hours ago
@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
– Bill Dubuque
6 hours ago
@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.
– vadim123
5 hours ago
@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.
– Bill Dubuque
5 hours ago
|
show 4 more comments
Thank you so much. This was really very helpful. :)
– Aamir Khan
6 hours ago
My pleasure, glad to help.
– vadim123
6 hours ago
@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
– Bill Dubuque
6 hours ago
@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.
– vadim123
5 hours ago
@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.
– Bill Dubuque
5 hours ago
Thank you so much. This was really very helpful. :)
– Aamir Khan
6 hours ago
Thank you so much. This was really very helpful. :)
– Aamir Khan
6 hours ago
My pleasure, glad to help.
– vadim123
6 hours ago
My pleasure, glad to help.
– vadim123
6 hours ago
@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
– Bill Dubuque
6 hours ago
@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
– Bill Dubuque
6 hours ago
@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.
– vadim123
5 hours ago
@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.
– vadim123
5 hours ago
@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.
– Bill Dubuque
5 hours ago
@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.
– Bill Dubuque
5 hours ago
|
show 4 more comments
Going $1$ step more with Euclid's algorithm shows they have common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = dfrac{7b!+!2}{2b!+!7}inBbb Z Rightarrow 7-dfrac{2,(7b!+!2)}{ 2b+7} = dfrac{45}{2b+7}inBbb Zqquad$$
Hence $,2b!+!7mid 45 $ so $ 2b!+!7 = 45,$ or $15,$ (by $b>1),,$ so $,b = 19,$ or $4$.
answered 6 hours ago
Bill Dubuque
208k29190628
add a comment |
Going $1$ step more with Euclid's algorithm shows they have common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = dfrac{7b!+!2}{2b!+!7}inBbb Z Rightarrow 7-dfrac{2,(7b!+!2)}{ 2b+7} = dfrac{45}{2b+7}inBbb Zqquad$$
Hence $,2b!+!7mid 45 $ so $ 2b!+!7 = 45,$ or $15,$ (by $b>1),,$ so $,b = 19,$ or $4$.
answered 6 hours ago
Bill Dubuque
208k29190628
add a comment |
Going $1$ step more with Euclid's algorithm shows they have common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = dfrac{7b!+!2}{2b!+!7}inBbb Z Rightarrow 7-dfrac{2,(7b!+!2)}{ 2b+7} = dfrac{45}{2b+7}inBbb Zqquad$$
Hence $,2b!+!7mid 45 $ so $ 2b!+!7 = 45,$ or $15,$ (by $b>1),,$ so $,b = 19,$ or $4$.
answered 6 hours ago
Bill Dubuque
208k29190628
Going $1$ step more with Euclid's algorithm shows they have common factor $,b!+!1.,$ Cancelling it
$$dfrac{7b^2!+!9b!+!2}{2b^2!+!9b!+!7} = dfrac{7b!+!2}{2b!+!7}inBbb Z Rightarrow 7-dfrac{2,(7b!+!2)}{ 2b+7} = dfrac{45}{2b+7}inBbb Zqquad$$
Hence $,2b!+!7mid 45 $ so $ 2b!+!7 = 45,$ or $15,$ (by $b>1),,$ so $,b = 19,$ or $4$.
answered 6 hours ago
Bill Dubuque
208k29190628
edited 5 hours ago
answered 6 hours ago
Bill Dubuque
208k29190628
answered 6 hours ago
Bill Dubuque
208k29190628
answered 6 hours ago
Bill Dubuque
208k29190628
208k29190628
add a comment |
add a comment |
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 5 hours ago
greedoid
37.9k114794
This answer is of no use?
– greedoid
3 hours ago
add a comment |
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 5 hours ago
greedoid
37.9k114794
This answer is of no use?
– greedoid
3 hours ago
add a comment |
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 5 hours ago
greedoid
37.9k114794
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 5 hours ago
greedoid
37.9k114794
edited 5 hours ago
answered 5 hours ago
greedoid
37.9k114794
answered 5 hours ago
greedoid
37.9k114794
answered 5 hours ago
greedoid
37.9k114794
37.9k114794
This answer is of no use?
– greedoid
3 hours ago
add a comment |
This answer is of no use?
– greedoid
3 hours ago
This answer is of no use?
– greedoid
3 hours ago
This answer is of no use?
– greedoid
3 hours ago
add a comment |
$$2B^2+9B+7mid 7B^2+9B+2$$
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 $[4,B-1,B-5]_B = 2[2,9,7] = [4,18,14]$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 2 hours ago
steven gregory
17.7k32257
add a comment |
$$2B^2+9B+7mid 7B^2+9B+2$$
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 $[4,B-1,B-5]_B = 2[2,9,7] = [4,18,14]$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 2 hours ago
steven gregory
17.7k32257
add a comment |
$$2B^2+9B+7mid 7B^2+9B+2$$
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 $[4,B-1,B-5]_B = 2[2,9,7] = [4,18,14]$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 2 hours ago
steven gregory
17.7k32257
$$2B^2+9B+7mid 7B^2+9B+2$$
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 $[4,B-1,B-5]_B = 2[2,9,7] = [4,18,14]$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 2 hours ago
steven gregory
17.7k32257
answered 2 hours ago
steven gregory
17.7k32257
answered 2 hours ago
steven gregory
17.7k32257
answered 2 hours ago
steven gregory
17.7k32257
17.7k32257
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