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A number when divided by a divisor leaves $27$ remainder. Twice the number when divided by the same divisor leaves a remainder $3$. Find the divisor.

My attempt: Let, the number be=$n$ and the divisor be=$d$.

Thus, we have $n\equiv 27\pmod d$ and $2n\equiv 3\pmod d$.

Thus we have two congruence equations. Can we, by any chance, apply the Chinese Remainder theorem to find $d$?

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    $\begingroup$ Perhaps. But there is a much easier way. $\endgroup$ – André Nicolas May 2 '16 at 13:25
  • $\begingroup$ @AndréNicolas Hints please..:-).. $\endgroup$ – tatan May 2 '16 at 13:26
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Hint: We have $2n\equiv 54\pmod{d}$.

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  • $\begingroup$ Subtracting the two congruences we have $d\vert -51$...Will $+51$ and $-51$ both be solutions? $\endgroup$ – tatan May 2 '16 at 13:35
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    $\begingroup$ @tatan: Technically yes, but there is probably the implicit understanding that the "divisor" is positive. So I think the expected answer is $51$. It cannot be a proper divisor of $51$, since $51$ has no proper divisors $\gt 27$. $\endgroup$ – André Nicolas May 2 '16 at 13:38
  • $\begingroup$ Also,apparently,the implicit understanding that the remainders are less than the divisor , else we could take 1,3, and 17 as solutions. $\endgroup$ – DanielWainfleet May 2 '16 at 18:26
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    $\begingroup$ @user254665: That the object called the remainder is between $0$ and $|d|-1$ is ordinarily built into the definition of remainder. $\endgroup$ – André Nicolas May 2 '16 at 18:49
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Let the number be $n$, and the divisor be $d$.

$\frac{n}{d} = X + \frac{27}{d}$ and also $\frac{2n}{d} = Y + \frac{3}{d}$ Where $X$ and $Y$ are different or same positive integers, comparing the two

$$2X + \frac{54}{d} = Y + \frac{3}{d}$$ $$2X = Y + \frac{3}{d} - \frac{54}{d}$$ $$2X = Y - \frac{51}{d}$$

Now the value of $d$, must be a factor of $51$, the factors are $1, 3, 17$ or $51$.

It could also be the negative of any of those values.

If the question had stated that $n$ and $d$ are in their lowest term $Y$ is an odd number, because $d$ is also odd $n > 27$ and $n > X*d$ because $\frac{n}{d} = X + \frac{27}{d}$, if $n$ and $d$ were in their lowest term then $n > d$, this is a proper fraction, then $\frac{27}{d}$ must be the improper part, so that $d > 27$

Therefore d would be $51$

If $X$ is odd, $n$ would be even and vice versa.

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