Download Problem 10. There are infinitely many positive integers with the

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 .)
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