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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, 1999 based on Squares and Triangles.

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

But try the problem first…

Source

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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