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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: enter image description here

But base cannot be negative. Could someone please explain where I am going wrong?

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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 $B=19$.

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  • $\begingroup$ 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
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    $\begingroup$ @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 $q\ge 4$ and $B>0$. Hence if $q\ge 4$, 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
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Going $1$ step more with Euclid's algorithm reveals a common factor $\,b\!+\!1.\,$ Cancelling it

$$\dfrac{7b^2\!+\!9b\!+\!2}{2b^2\!+\!9b\!+\!7} = \color{#c00}{\dfrac{7b\!+\!2}{2b\!+\!7}}\in\Bbb Z\ \, \Rightarrow\,\ 7-2\ \dfrac{\color{#c00}{7b\!+\!2}}{ \color{#c00}{2b\!+\!7}}\, =\, \dfrac{45}{2b\!+\!7}\in\Bbb Z\qquad$$

Therefore $\,2b\!+\!7\mid 45\ $ so $\,b> 9\,$(= digit) $\,\Rightarrow\,2b\!+\!7 = 45\,$ $\Rightarrow\,b=19.$

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Since $b+1>0$ and $$(b+1)(2b+7)\mid (7b+2)(b+1)\implies 2b+7\mid 7b+2$$

we have $$2b+7\mid (7b+2)-3(2b+7) = b-19$$

so if $b-19> 0$ we have $$2b+7\mid b-19 \implies 2b+7\leq b-19 \implies b+26\leq 0$$

which is not true. So $b\leq 19$. By trial and error we see that $b=4$ and $b=19$ works.

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$$2B^2+9B+7\mid 7B^2+9B+2$$

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 $B=19$.

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