Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on smallest positive integer.
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 \(\frac{n}{75}\).
Integers
Divisibility
Algebra
Answer: is 432.
AIME I, 1990, Question 5
Elementary Number Theory by David Burton
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.