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, 2022

Combinatorics Difficulty 5.2 AIME, harder Prove it Bulgaria

A positive integer $n$ is called balanced if it is not a prime and for any integer $k$ in the interval $[1, \sqrt{n}-1]$ the number of ways to choose $k$ persons from a group of $n$ people is divisible by $n$ . If $m$ and $n$ are balanced five digit positive integers find the smallest value of the difference $|m-n|$ .

Solution

Answer. 202. As in problem 9.3 we obtain that any 5-digit balanced number is of the forms: $p^2$ or $p(p+2)$ where $p$ and $p+2$ are primes. Since $p(p+2) = (p+1)^2 - 1 < (p+1)^2$ it follows that the smallest positive difference between two 5-digit numbers equals $2p$ where $p$ is the smallest prime number for which $p+2$ is also a prime and $p^2$ and $p(p+2)$ are 5-digit numbers. The smallest 5-digit number is $10000 = 10^4$ , i.e, we want $p \ge 100$ . Direct verification shows that 101 and 103 are both prime numbers. The corresponding 5-digit balanced numbers are $101^2$ and $101 \cdot 103$ with difference $2 \cdot 101 = 202$ .

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