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Problem 10. There are infinitely many positive integers with the following property: When n is divided by 2, the result is a perfect square, when n is divided by 3 the result is a perfect cube, when n is divided by 5, the result is a perfect fifth power, and when n is divided by 7 the result is a perfect seventh power. Find the three smallest such integers. Solution. Let N = pr11 pr22 · · · prmm where p1 , p2 , . . . , pm are distinct primes and r1 , r2 , . . . , rm are nonnegative integers. Then N is a perfect k th power if and only if each rj is a multiple of k. Because n is divisible by each of 2, 3, 5, 7, n must have prime factorization of the form rm n = 2r1 3r2 5r3 7r4 q1r5 · · · qm where q1 , . . . , qm are distinct primes, none of which is equal to 2, 3, 5, or 7. Because we are seeking “small” values of n we will assume initially that the only prime factors of n are 2, 3, 5, 7. Because 21 n is a perfect square, each of r1 − 1, r2 , r3 , r4 must be even. Because 13 n is a perfect cube, each of r1 , r2 − 1, r3 , r4 must be multiples of 3. By similar reasoning, each of r1 , r2 , r3 − 1, r4 is a multiple of 5, and each of r1 , r2 , r3 , r4 − 1 is a multiple of 7. It follows that r1 = 3 · 5 · 7(2k1 + 1) for some nonnegative integer k; note that r1 is an odd multiple of 3, 5, and 7, and r1 −1 is even, as required. By similar reasoning, r2 = 2·5·7(3k2 +1), r3 = 2·3·7(5k3 +3) and r4 = 2 · 3 · 5(7k4 + 4). Thus we consider numbers of the form 2210k1 +105 3210k2 +70 5210k 3 +126 7210k4 +120 . Any number of this form satisfies the “power” requirements and any number that satisfies the conditions of the problem must have such an expression for the first four terms of its prime factorization. It is not hard to check that the three smallest integers with the required property occur with (k1 , k2 , k3 , k4 ) equal to one of (0, 0, 0, 0), (1, 0, 0, 0) and (0, 1, 0, 0). The three smallest such n are 2105 370 5126 7120 , 2315 370 5126 7120 , and 2105 3280 5126 7120 . (Note, the third number in the list is larger than 2525 370 5126 7120 .) 1