Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Problem on Positive Integer.
Let \(n=2^{31}3^{19}\),find number of positive integer divisors of \(n^{2}\) are less than n but do not divide n.
Integers
Divisibility
Number of divisors
Answer: is 589.
AIME I, 1995, Question 6
Elementary Number Theory by David Burton
Let \(n=p_1^{k_1}p_2^{k_2}\) for some prime \(p_1,p_2\). The factors less than n of \(n^{2}\)
=\(\frac{(2k_1+1)(2k_2+1)-1}{2}\)=\(2k_1k_2+k_1+k_2\)
The number of factors of n less than n=\((k_1+1)(k_2+1)-1\)
=\(k_1k_2+k_1+k_2\)
Required number of factors =(\(2k_1k_2+k_1+k_2\))-(\(k_1k_2+k_1+k_2\))
=\(k_1k_2\)
=\(19 \times 31\)
=589.
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