Question

Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including itself and 1. Find n/75.

Answer 1

Answer:

$n=2^4 3^4 5^2 =32400$ and then we have:

$\frac{n}{75}=\frac{2^4 3^4 5^2}{3 5^2}=432$

Step-by-step explanation:

From the info given by the problem we need an integer defined as the smallest positive integer that is a multiple of 75 and have 75 positive integral divisors, and we are assuming that 1 is one possible divisor.

Th first step is find the prime factorization for the number 75 and we see that

$75=3 5^2$

And we know that 3 =2+1 and 5=3+2 and if we replace we got:

$75 = (2+1)(4+1)^2 = (2+1)(4+1)(4+1)$

And in order to find 75 integral divisors we need to satisify this condition:

$n= a^{r_1 -1}_1 a^{r_2 -1}_2 *......$ such that $a_1 *a_2*....=75$

For this case we have two prime factors important 3 and 5. And if we want to minimize n we can use a prime factor like 2. The least common denominator between 2 and 4 is LCM(2,4) =4. So then the need to have the prime factors 2 and 3 elevated at 4 in order to satisfy the condition required, and since 5 is the highest value we need to put the same exponent.

And then the value for n would be given by:

$n=2^4 3^4 5^2 =32400$ and then we have:

$\frac{n}{75}=\frac{2^4 3^4 5^2}{3 5^2}=432$