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I am trying to solve this problem

A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

I have established two relations here but dont know how to proceed , any suggestions would be appreciated .

$x= dq + 24 $ where d=divisor

$2x= dk + 11 $ where d=divisor

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    $\begingroup$ Shouldn't it be a different $q$ in the second equation? Maybe it's better to write it as $x \equiv 24 \bmod d$ and $2x \equiv 11 \bmod d$ $\endgroup$ – draks ... Jul 20 '12 at 14:00
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As you wrote, corrected: $$\begin{align*}2x=&dk+11\\x=&dq+24\end{align*}$$ Substracting second eq. from first one and comparing the result with second eq.: $$x=d(k-q)-13=dq+24\Longrightarrow d(k-2q)=37\Longrightarrow d=37$$ as $\,37\,$ is prime.

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  • $\begingroup$ How did you assume that d=37 when $37$ is actually equally to $d(k-2q)$ ? $\endgroup$ – Rajeshwar Jul 20 '12 at 23:44
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    $\begingroup$ I didn't "assume": since $\,d\mid 37\,$, then $\,d=1\,\,\,or\,\,\,d=37\,$ , right? $\endgroup$ – DonAntonio Jul 20 '12 at 23:55
  • $\begingroup$ Yes that makes sense $\endgroup$ – Rajeshwar Jul 20 '12 at 23:57
  • $\begingroup$ Another way that I am trying to solve this - If I multiply equation (ii) with $2$ I get $2x=2dq+48$. After that I compare the two equations $2dq+48=dk+11$ which implies $d(2q-k) = -37$ .Does the negative sign here make my work wrong and invalidate it ? $\endgroup$ – Rajeshwar Jul 21 '12 at 0:27
  • $\begingroup$ Not at all: we work here with integer numbers, so positive or negative it doesn't usually mind (unless there's some specific requirement). It is customary to take positive divisors whenever possible, though. $\endgroup$ – DonAntonio Jul 21 '12 at 0:34
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So $x \equiv 24 \bmod d$ and $2x \equiv 11 \bmod d$. Subtracting the second from the first you get $$-24 \equiv -x \equiv (24 -11) = 13 \bmod d.$$ and therefore $d | (24+13) = 37$. Note that $37$ is prime, so $d = 37$.

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Note that the quotients won't be the same. Use $q_1$ and $q_2$. Another thing worth noting is that you can sub in $x$ from the first expression into the second. Once you've done that, the rest falls out immediately once you've gathered your $d$ terms on one side and the rest on the other--note that we can't have $d=1$.

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Hint $\rm\ mod\ d\!:\ x \equiv 24\:\Rightarrow\: 48\equiv 2x \equiv 11 \:\Rightarrow\: 37 \equiv 0,\:$ i.e. $\rm\:d\:|\:37$

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