Olympiad Maths Prep

Track / Stage 4 / 55 of 340 #315 of 2000

Problem 315

AMC 12 late, AIME early
Number theory Difficulty 4.6 Find the answer

Find all integers a,b,ka, b, k such that 2b3a=k(k+1)2^{b} 3^{a}=k(k+1)

Official solution

The only prime divisors of kk and k+1k+1 are 2 and 3. Since they are coprime, one is a power (possibly multiplied by -1) of 2 and the other is a power of 3 (possibly multiplied by -1). We are thus reduced exactly to Exercise 19 which provides the solutions (a,b,k):(0,1,1),(0,1,2),(1,1,2),(1,1,3),(1,2,3),(1,2,4),(2,3,8),(2,3,9)(a, b, k):(0,1,1),(0,1,-2),(1,1,2),(1,1,-3),(1,2,3),(1,2,-4),(2,3,8),(2,3,-9).

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