For how many natural numbers \(n\) between \(1\) and \(2014\) (both inclusive) is \(\dfrac{8n}{9999 - n}\) an integer?

This note is part of the set Pre-RMO 2014

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## Comments

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TopNewestthe expression equals \(\frac{8}{\frac{9999}{n}-1}\) . now since 8 has only four factors 1,2,4,8 equating the denominator to these values and checking the range we get n=1111

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\(\frac{8n}{ 9999 - n} = \frac{8n - 9999.8}{9999 - n} + \frac{9999.8}{ 9999 - n}\)

\( = -8 + \frac{9999.8}{9999 - n}\)

therefore denominator should divide 9999.8

therefore denominator should be even as well as multiple of 1111

therefore only n = 1111 satisfys it

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Did exactly the same

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If (9999-n)|8n it's also true that (9999-n)|8n+8x(9999-n) which is equal to (9999-n)|3^2x2^3x11x101. Let's call T= (9999-n). For the conditions: (9999-2014)<=T<=(9999-1) or 7985<=T<=9998. The unique possible value of T is 11x101x2^3=8888 from which 9999-n=8888 => n=1111.

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