Track / Stage 4 / 234 of 340 #494 of 1964

Problem 494

AMC 12 late, AIME early

Number theory Difficulty 4.9 Find the answer

8.2. Does the equation $x^{9}=2013 y^{10}$ have a solution in natural numbers $x, y$ ?

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Answer: Yes. Solution. We will look for a solution in the form of powers of the number 2013, i.e., $x=2013^{m}, y=2013^{n}$ . Then $2013^{9 m}=2013^{1+10 n}$ , i.e., $9 m-1=10 n$ . We can take $m=9$ (the smallest natural $m$ for which the last digit of the number $9 m$ is 1), then $n=8$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.