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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Smallest positive Integer Problem | AIME I, 1990 | Question 5

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

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


Integers

Divisibility

Algebra

Check the Answer


Answer: is 432.

AIME I, 1990, Question 5

Elementary Number Theory by David Burton

Try with Hints


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 \(a_1a_2….=75\)

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

Minimizing n third factor =2

and factor 5 raised to least power

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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