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# Problem on Positive Integer | AIME I, 1995 | Question 6

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

AIME I, 1995, Question 6

Elementary Number Theory by David Burton

## Try with Hints

First hint

Let $n=p_1^{k_1}p_2^{k_2}$ for some prime $p_1,p_2$. The factors less than n of $n^{2}$

=$\frac{(2k_1+1)(2k_2+1)-1}{2}$=$2k_1k_2+k_1+k_2$

Second Hint

The number of factors of n less than n=$(k_1+1)(k_2+1)-1$

=$k_1k_2+k_1+k_2$

Final Step

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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