Answer:
and then we have:
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
And we know that 3 =2+1 and 5=3+2 and if we replace we got:
And in order to find 75 integral divisors we need to satisify this condition:
such that
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:
and then we have:
