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April 28, 2020

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

Check the Answer


Answer: is 589.

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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