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# 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

Note by Pranshu Gaba
2 years, 9 months ago

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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 · 2 years, 9 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 · 2 years, 9 months ago

Did exactly the same · 1 year, 2 months ago