IMO 1985 SL 8

Problem

1b.(TUR 5) Find the smallest positive integer n such that (i) n has exactly 144 distinct positive divisors, and (ii) there are ten consecutive integers among the positive divisors of n.

Solution

Among ten consecutive integers that divide n, there must exist numbers divisible by 23, 32, 5, and 7. Thus the desired number has the form n = 2\alpha13\alpha25\alpha37\alpha411\alpha5 \cdot \cdot \cdot , where \alpha1 \geq3, \alpha2 \geq2, \alpha3 \geq1, \alpha4 \geq1. Since n has (\alpha1 + 1)(\alpha2 + 1)(\alpha3 + 1) \cdot \cdot \cdot distinct factors, and (\alpha1 + 1)(\alpha2 + 1)(\alpha3 + 1)(\alpha4 + 1) \geq48, we must have (\alpha5 + 1) \cdot \cdot \cdot \leq3. Hence at most one \alphaj, j > 4, is positive, and in the minimal n this must be \alpha5. Checking through the possible combinations satisfying (\alpha1 + 1)(\alpha2 + 1) \cdot \cdot \cdot (\alpha5 + 1) = 144 one finds that the minimal n is 25 \cdot 32 \cdot 5 \cdot 7 \cdot 11 = 110880.