Competition problem — Number theory

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Number theory Difficulty 5.5 AIME, harder Prove it Singapore

Two distinct 2-digit prime numbers $p$ , $q$ can be written one after the other in 2 different ways to form two 4-digit numbers. For example, $11$ and $13$ yield $1113$ and $1311$ . If the two 4-digit numbers formed are both divisible by the average value of $p$ and $q$ , find all possible pairs \{ $p, q$ \}.

Solution

The average of $p$ , $q$ is $\frac{p+q}{2}$ . Therefore $2(100p + q) = (p+q)k$ for some integer $k$ . Thus
$200p + 2q = (p+q)k \Rightarrow 198p = (p+q)(k-2).$
Since $p$ , $q$ are primes, $(p+q) \mid 198$ . Now $11 + 13 = 24 \leq p+q \leq 89 + 97 = 186$ . Since $198 = 2 \times 3^2 \times 11$ , the only even factor of $198$ that lies between $24$ and $186$ is $66$ . Thus $p+q = 66$ and there are 4 pairs: $(13, 53)$ , $(19, 47)$ , $(23, 43)$ , $(29, 37)$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.