Pre-RMO 2014/13

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

#NumberTheory #Pre-RMO

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Comments

the 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

aman tiwari - 6 years, 7 months ago

$\frac{8n}{ 9999 - n} = \frac{8n - 9999.8}{9999 - n} + \frac{9999.8}{ 9999 - n}$

$= -8 + \frac{9999.8}{9999 - n}$

therefore denominator should divide 9999.8

therefore denominator should be even as well as multiple of 1111

therefore only n = 1111 satisfys it

U Z - 6 years, 8 months ago

Did exactly the same

Aditya Kumar - 5 years ago

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.

Drop TheProblem - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$