A number theory problem by Arul Kolla

Number Theory Level 3

It is known that $m, n,$ and $m - n$ are all positive integers. Find $m - n$ if $\LARGE \frac{m^3 - n^3}{{(m-n)}^{3}} = \frac{73}{3}.$

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1 solution

Arul Kolla
Jan 5, 2018

Let $m^3 - n^3 = a$ and ${(m - n)}^{3} = b$ . Then we have $\frac{a}{b} = \frac{73}{3}$ , or $a = \frac{73b}{3}$ (by cross multiplying). For $a$ to be an integer, $b$ must be a multiple of 3. Out of the given options, $b = 1, 8, 27, 64,$ or $125$ . Only one of these, $27$ , is a multiple of 3, so its cube root, $\boxed{3}$ , is our answer.

Going a step further, the solutions for $(m,n)$ are $(10k,7k)$ for any positive integer $k$ .

Brian Charlesworth - 3 years, 5 months ago

A number theory problem by Arul Kolla

1 solution

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