Maths Olympiad Prep

Track / Stage 4 / 143 of 340 #403 of 1964

Problem 403

AMC 12 late, AIME early
Number theory Difficulty 4.8 Find the answer HMMT_11

Let N=5AB37C2N=\overline{5 A B 37 C 2}, where A,B,CA, B, C are digits between 0 and 9, inclusive, and NN is a 7-digit positive integer. If NN is divisible by 792, determine all possible ordered triples (A,B,C)(A, B, C).

A number or a short expression. Spacing and $ signs are ignored.

Official solution

First, note that 792=23×32×11792=2^{3} \times 3^{2} \times 11. So we get that 8N87C2810C+6C=1,5,99N95+A+B+3+7+C+2A+B+C=1,10,1911N115A+B3+7C+2A+BC=11,0\begin{gathered} 8|N \Rightarrow 8| \overline{7 C 2} \Rightarrow 8 \mid 10 C+6 \Rightarrow C=1,5,9 \\ 9|N \Rightarrow 9| 5+A+B+3+7+C+2 \Rightarrow A+B+C=1,10,19 \\ 11|N \Rightarrow 11| 5-A+B-3+7-C+2 \Rightarrow-A+B-C=-11,0 \end{gathered} Adding the last two equations, and noting that they sum to 2B2 B, which must be even, we get that B=4,5B=4,5. Checking values of CC we get possible triplets of (0,5,5),(4,5,1)(0,5,5),(4,5,1), and (6,4,9)(6,4,9).

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