Olympiad Maths Prep

Track / Stage 3 / 72 of 260 #72 of 2000

Problem 72

AMC 10/12, early questions
Number theory Difficulty 3.3 Find the answer

If a,ba,b and cc are positive integers and aa and bb are odd, then 3a+(b−1)2c3^a+(b-1)^2c is
(A) odd for all choices of c(B) even for all choices of c(C) odd if c is even; even if c is odd(D) odd if c is odd; even if c is even(E) odd if c is not a multiple of 3; even if c is a multiple of 3\text{(A) odd for all choices of c} \quad \text{(B) even for all choices of c} \quad\\ \text{(C) odd if c is even; even if c is odd} \quad\\ \text{(D) odd if c is odd; even if c is even} \quad\\ \text{(E) odd if c is not a multiple of 3; even if c is a multiple of 3}

Official solution

Since 3 has no factors of 2, 3a3^a will be odd for all values of aa. Since bb is odd as well, b−1b-1 must be even, so (b−1)2(b-1)^2 must be even. This means that for all choices of cc, (b−1)2c(b-1)^2c must be even because any integer times an even number is still even. Since an odd number plus an even number is odd, 3a+(b−1)2c3^a+(b-1)^2c must be odd for all choices of cc, which corresponds to answer choice A\fbox{A}.

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