Maths Olympiad Prep

Library / /485 of 860

Number theory Difficulty 5.2 AIME, harder Find the answer

Let NN be a three-digit integer such that the difference between any two positive integer factors of NN is divisible by 3 . Let d(N)d(N) denote the number of positive integers which divide NN. Find the maximum possible value of Nd(N)N \cdot d(N).

A number or a short expression. Spacing and $ signs are ignored.

Solution

We first note that all the prime factors of nn must be 1 modulo 3 (and thus 1 modulo 6 ). The smallest primes with this property are 7,13,19,7,13,19, \ldots Since 74=2401>10007^{4}=2401>1000, the number can have at most 3 prime factors (including repeats). Since 71319=1729>10007 \cdot 13 \cdot 19=1729>1000, the most factors NN can have is 6 . Consider the number 7219=9317^{2} \cdot 19=931, which has 6 factors. For this choice of N,Nd(N)=5586N, N \cdot d(N)=5586. For another NN to do better, it must have at least 6 factors, for otherwise, Nd(N)<10005=5000N \cdot d(N)<1000 \cdot 5=5000. It is easy to verify that 72197^{2} \cdot 19 is the greatest number with 6 prime factors satisfying our conditions, so the answer must be 5586 .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.