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Let $N$ be the greatest number that will divide $1305,4665$ and $6905$, leaving the same remainder in each case. Then what is the sum of the digits in $N$?

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  • $\begingroup$ Definitely I have tried and then asked.... I have tried by assuming and then taking H.C.F. $\endgroup$ – harspt92 Sep 20 '14 at 8:56
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Since division by $N$ leaves the same remainder for all three numbers, $N$ leaves zero remainder when divided into the difference of each pair.

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$4665-1305=3360$ so the numbers need to divide $3360$
$6905-4665=2240$ so the numbers need to divide $2240$
The numbers need to divide $1120=2^5\cdot5\cdot7$
Knowing the prime factorization of a number, you can find the sum of all the factors of that number.

Hint 1: The sum of the factors of $p^n$ is $$ 1+p+p^2+p^3+\dots+p^n=\frac{p^{n+1}-1}{p-1} $$ Hint 2: We can find all the factors of $p^3q^2$ in the product $$ (1+p+p^2+p^3)(1+q+q^2) $$

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  • $\begingroup$ Thank you.... My approach was same but I couldn't reach the answer $\endgroup$ – harspt92 Sep 20 '14 at 9:16
  • $\begingroup$ @harspt92: I have added a couple of hints that should finish the job. $\endgroup$ – robjohn Sep 20 '14 at 10:14
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Hint: One way to think of $a$ and $b$ as leaving the same remainder upon division by $m$ is $a\equiv b\pmod{m}$

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  • $\begingroup$ That's not what we have here. We have $N \pmod a = N \pmod b$. $\endgroup$ – TonyK Sep 20 '14 at 11:21
  • $\begingroup$ @TonyK: I took the problem to mean dividing the three given numbers by $N$ and getting the same remainder in each case. The $m$ in my answer corresponds to $N$ in the original problem. $\endgroup$ – paw88789 Sep 20 '14 at 14:07

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