Please show all steps. Thanks. Suppose an integer has the factorization p2 middo

Question

Please show all steps. Thanks.

Suppose an integer has the factorization p2 middot q, where p and q are unique primes. How many positive divisors does this integer have? What is the smallest nonnegative integer with this factorization? The value of the Euler phi function (phi is the Greek letter phi) at the positive integer n is defined to be the number of positive integers less than or equal to n that are relatively prime to n. For example fon n=14, we have {1, 3, 5, 9, 11, 13} are the positive integers less than or equal to 14 which are relatively prime to 14. Thus phi(14) = 6. Find: phi (5) phi (25) phi (10) phi (20)

Explanation / Answer

p2q has 6 divisors: 1, p, q, p2, pq, and p2q. The smallest integer with this factorization is 22*3 = 12 phi(5) = 4 -- the set of rel. prime numbers are {1, 2, 3, 4} phi(25) = 20 -- {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24} phi(10) = 4 -- {1, 3, 7, 9} phi(20) = 8 -- {1, 3, 7, 9, 11, 13, 17, 19}

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