# 2003 JBMO Problems/Problem 1

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## Problem

Let $n$ be a positive integer. A number $A$ consists of $2n$ digits, each of which is 4; and a number $B$ consists of $n$ digits, each of which is 8. Prove that $A+2B+4$ is a perfect square.

## Solution

Using the definition of base 10, we know that \begin{align*} A &= 4(10^{2n-1} + 10^{2n-2} \cdots 10^1 + 10^0) \\ &= \frac{4 \cdot (10^{2n} - 1)}{9} \\ B &= 8(10^{n-1} + 10^{n-2} \cdots 10^1 + 10^0) \\ &= \frac{8 \cdot (10^{n} - 1)}{9}. \end{align*} Thus, we have \begin{align*} A+2B+4 &= \frac{4 \cdot 10^{2n} - 4 + 16 \cdot 10^n - 16 + 36}{9} \\ &= \frac{4 \cdot (10^n)^2 + 16 \cdot 10^n + 16}{9} \\ &= \left(\frac{2(10^n + 2)}{3}\right)^2. \end{align*} Since we know that $A+2B+4$ is an integer, we confirm that $A+2B+4$ is a perfect square.

## See Also

 2003 JBMO (Problems • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 All JBMO Problems and Solutions
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