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Number theory Difficulty 5.4 AIME, harder Prove it Romania

Find all positive integers that have exactly 88 positive divisors, among which three are primes of the form aa, bc\overline{bc} and cb\overline{cb}, given that a+bc+cba + \overline{bc} + \overline{cb} is a perfect square and a,b,ca, b, c are digits, with b<cb < c.

Solution

Suppose xx is an integer with the given properties. Then xx is a multiple of y=abccby = a \cdot \overline{bc} \cdot \overline{cb}. Since yy has eight divisors, it follows that x=yx = y.

The numbers bc\overline{bc} and cb\overline{cb} are distinct primes, so bc{13,17,37,79}\overline{bc} \in \{13, 17, 37, 79\}. The condition that a+bc+cba + \overline{bc} + \overline{cb} is a square and aa is a prime forces a=5a = 5, bc=13\overline{bc} = 13 and cb=31\overline{cb} = 31, implying x=2015x = 2015.

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