# 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.

## 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

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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