Competition problem — Number theory

Number theory Difficulty 5.4 AIME, harder Prove it Japan

$n$ is a 4-digit integer with ten's place nonzero, and if we take the first 2 digits and the last 2 digits as two 2-digit integers, their product is a divisor of $n$ . Determine all $n$ with this property.

Solution

Let $A$ and $B$ be the first and the last 2 digits of $n$ respectively. We have to determine all $(A, B)$ such that $AB$ divides $100A + B$ .

Since $A$ divides $100A + B$ , $A$ divides $B$ . Let $k = \frac{B}{A}$ . Since $A$ and $B$ are 2-digit numbers, $10 \leq A < \frac{100}{k}$ .

The condition is equivalent to $kA^2 \mid 100A + kA$ and this is equivalent to $kA \mid 100 + k$ . $k$ divides $100 + k$ iff $k$ divides $100$ , and with $k < 10$ we get $k = 1, 2, 4, 5$ .

From $A \mid \frac{100 + k}{k}$ and $10 \leq A < \frac{100}{k}$ we get $(k, A) = (2, 17), (4, 13)$ and we get $n = 1734, 1352$ .

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