Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it Philippines

Problem:

If $2A99561$ is equal to the product when $3 \times (523 + A)$ is multiplied by itself, find the digit $A$ .

Solution

Solution:

We are given that $2A99561 = [3 \times (523 + A)]^2$ , which is equivalent to $2A99561 = 9 \times (523 + A)^2$ .

Since $(523 + A)^2$ is an integer, it follows that $2A99561$ is divisible by $9$ .

By the rule on divisibility by $9$ , after adding all the digits of $2A99561$ , it suffices to find the digit $A$ for which $A + 5$ is divisible by $9$ , which yields $A = 4$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.