Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Problem on Positive Integer.

## Problem on Positive Integer – AIME I, 1995

Let \(n=2^{31}3^{19}\),find number of positive integer divisors of \(n^{2}\) are less than n but do not divide n.

- is 107
- is 589
- is 840
- cannot be determined from the given information

**Key Concepts**

Integers

Divisibility

Number of divisors

## Check the Answer

Answer: is 589.

AIME I, 1995, Question 6

Elementary Number Theory by David Burton

## Try with Hints

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA