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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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    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$2012-07-20

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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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    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.2012-07-21
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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. $\rm\:d\:|\:37$