Question

The number 42 has the prime factorization 2 · 3 · 7. Thus 42 can be written in four ways as a product of two positive integer factors (without regard to the order of the factors): 1 · 42, 2 · 21, 3 · 14, and 6 · 7. Answer a–d below without regard to the order of the factors.

(a)

List the distinct ways the number 570 can be written as a product of two positive integer factors. (Enter your answer as a comma separated list of products.)

(b)If n = p₁p₂p₃p₄, where the p_i are distinct prime numbers, how many ways can n be written as a product of two positive integer factors?

(c)If n = p₁p₂p₃p₄p₅, where the p_i are distinct prime numbers, how many ways can n be written as a product of two positive integer factors?

(d)If n = p₁p₂    p_k, where the p_i are distinct prime numbers, how many ways can n be written as a product of two positive integer factors?

Answer 1

If the factors of a numbers are listed in ascending order, then the product of factors equidistant from beginning and end is constant, and it equals to the number itself.

This happens simply because f is the factor of N, then N/f is also definitely its factor. This can easily be illustarated by taking any small number. Thus, we can say that the number of ways in which any natural number N can be expressed as the product of two natural numbers, is half of its number of factors F. This is because from F factors, F/2 such pairs can be formed.

F itself has a formula, nased on prime factorization of the number. Let the prime factorization in general be written as

Then its number of factors is obtained as the product of successors of powers of all its prime factors. That is,

Based on this result, all parts of problem can be solved.

For ex, 570 = 2 × 3 × 5 × 19

so number of factors F = (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 16

So required no of ways = 8

If n = p1p2... pk that is, having k distinct primes each with power 1, then F = (1 + 1)(1 + 1)(1 + 1)... k times = 2k

hence, required ways = 2k-1