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

But try the problem first…

Answer: is 432.

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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