Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on smallest positive integer.

Smallest positive Integer Problem – AIME I, 1990

Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1 and itself. Find \(\frac{n}{75}\).

  • is 107
  • is 432
  • is 840
  • cannot be determined from the given information

Key Concepts




Check the Answer

But try the problem first…

Answer: is 432.

Suggested Reading

AIME I, 1990, Question 5

Elementary Number Theory by David Burton

Try with Hints

First hint

75=\(3 \times 5^{2}\)=(2+1)(4+1)(4=1)

or, \(n=p_1^{a_1-1}p_2^{a_2-1}…..\) such that \(a_1a_2….=75\)

Second Hint

or, 75|n with two prime factors 3 and 5

Minimizing n third factor =2

and factor 5 raised to least power

Final Step

or, \(n=(2)^{4}(3)^{4}(5)^{2}\)

and \(\frac{n}{75}=(2)^{4}(3)^{4}(5)^{2}\)=(16)(27)=432.

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