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# 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

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.