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 9 hours ago
greedoid
37.9k114794
asked 10 hours ago
Aamir Khan
455
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 9 hours ago
greedoid
37.9k114794
asked 10 hours ago
Aamir Khan
455
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 9 hours ago
greedoid
37.9k114794
asked 10 hours ago
Aamir Khan
455
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 9 hours ago
greedoid
37.9k114794
asked 10 hours ago
Aamir Khan
455
edited 9 hours ago
greedoid
37.9k114794
asked 10 hours ago
Aamir Khan
455
edited 9 hours ago
greedoid
37.9k114794
edited 9 hours ago
greedoid
37.9k114794
edited 9 hours ago
greedoid
37.9k114794
37.9k114794
asked 10 hours ago
Aamir Khan
455
asked 10 hours ago
Aamir Khan
455
asked 10 hours ago
Aamir Khan
455
455
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 10 hours ago
vadim123
75.5k897189
Thank you so much. This was really very helpful. :)
– Aamir Khan
10 hours ago
My pleasure, glad to help.
– vadim123
10 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
10 hours ago
1
@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
9 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
9 hours ago
|
show 4 more comments
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 10 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 9 hours ago
greedoid
37.9k114794
This answer is of no use?
– greedoid
7 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 $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 5 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 10 hours ago
vadim123
75.5k897189
Thank you so much. This was really very helpful. :)
– Aamir Khan
10 hours ago
My pleasure, glad to help.
– vadim123
10 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
10 hours ago
1
@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
9 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
9 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 10 hours ago
vadim123
75.5k897189
Thank you so much. This was really very helpful. :)
– Aamir Khan
10 hours ago
My pleasure, glad to help.
– vadim123
10 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
10 hours ago
1
@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
9 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
9 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 10 hours ago
vadim123
75.5k897189
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 10 hours ago
vadim123
75.5k897189
answered 10 hours ago
vadim123
75.5k897189
answered 10 hours ago
vadim123
75.5k897189
answered 10 hours ago
vadim123
75.5k897189
75.5k897189
Thank you so much. This was really very helpful. :)
– Aamir Khan
10 hours ago
My pleasure, glad to help.
– vadim123
10 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
10 hours ago
1
@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
9 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
9 hours ago
|
show 4 more comments
Thank you so much. This was really very helpful. :)
– Aamir Khan
10 hours ago
My pleasure, glad to help.
– vadim123
10 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
10 hours ago
1
@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
9 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
9 hours ago
Thank you so much. This was really very helpful. :)
– Aamir Khan
10 hours ago
Thank you so much. This was really very helpful. :)
– Aamir Khan
10 hours ago
My pleasure, glad to help.
– vadim123
10 hours ago
My pleasure, glad to help.
– vadim123
10 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
10 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
10 hours ago
1
1
@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
9 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
9 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
9 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
9 hours ago
|
show 4 more comments
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 10 hours ago
Bill Dubuque
208k29190628
add a comment |
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 10 hours ago
Bill Dubuque
208k29190628
add a comment |
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 10 hours ago
Bill Dubuque
208k29190628
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 10 hours ago
Bill Dubuque
208k29190628
edited 3 hours ago
answered 10 hours ago
Bill Dubuque
208k29190628
answered 10 hours ago
Bill Dubuque
208k29190628
answered 10 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 9 hours ago
greedoid
37.9k114794
This answer is of no use?
– greedoid
7 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 9 hours ago
greedoid
37.9k114794
This answer is of no use?
– greedoid
7 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 9 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 9 hours ago
greedoid
37.9k114794
edited 9 hours ago
answered 9 hours ago
greedoid
37.9k114794
answered 9 hours ago
greedoid
37.9k114794
answered 9 hours ago
greedoid
37.9k114794
37.9k114794
This answer is of no use?
– greedoid
7 hours ago
add a comment |
This answer is of no use?
– greedoid
7 hours ago
This answer is of no use?
– greedoid
7 hours ago
This answer is of no use?
– greedoid
7 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 $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 5 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 $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 5 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 $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 5 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 $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
answered 5 hours ago
steven gregory
17.7k32257
edited 3 hours ago
answered 5 hours ago
steven gregory
17.7k32257
answered 5 hours ago
steven gregory
17.7k32257
answered 5 hours ago
steven gregory
17.7k32257
17.7k32257
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