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Part of 1990 Swedish Mathematical Competition

Problems(1)

sum d_i/\sqrt {n} = sum \sqrt {n}/d_i, for pos. divisors of 1990!

Source: 1990 Swedish Mathematical Competition p1

4/2/2021
Let d1,d2,...,dkd_1, d_2, ... , d_k be the positive divisors of n=1990!n = 1990!. Show that ∑din=∑ndi\sum \frac{d_i}{\sqrt{n}} = \sum \frac{\sqrt{n}}{d_i}.
Sumnumber theoryDivisors