1. Find the number of ordered pairs of positive integers (m, n) such that (m^2)n = 20^20

Question

m²n = 20²⁰

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Answer 1

I am assuming that the n is meant to be an exponent.

Since the two prime factors of m, 5 and 2, have powers 2 and 1, and 1|2 (of course), we only need to consider the single power. As we square it before raising to the n'th power, possible powers of 5 in m (as we then have to double) are possible factors of 10; then, as 10 = 2^1*5^1, there are (1+1)(1+1) = 2 * 2 = 4 factors

(We can list the ordered pairs if you like: (20^1, 10); (20^2, 5); (20^5, 2); (20^10; 1); or

(20, 10); (400, 5); (3200000, 2); (10240000000000, 1)

Answer 2

Note that 20²⁰ factors as 2⁴⁰ 5²⁰. Since m is a positive integer, m² must have even powers for all of its primes. So for instance, you could have m² = 2⁰ 5⁰ = 1 or m² = 2² 5⁰ = 4 or m² = 2⁰ 5² = 25, but you could not have m² = 2¹ 5¹ = 10.

Meanwhile n has no such restriction, n can be whatever we need it to be to "finish the job" of multiplying to 20²⁰. The problem is essentially asking how many choices for m we have, because once we pick m², n is determined.

So how many choices for m²? Well if we were only concerned with a single prime factor this would be trivial. Suppose instead of multiplying to 20²⁰ = 2⁴⁰ 5²⁰ we only wanted to multiply to 2⁴⁰. Then the question is, how many pairs of 2's can you pick? Well you could pick 1 pair, or 2 pairs or 3 pairs... All the way up to 20 pairs, so there would be 20 options.

Similarly, if the question wanted to merely multiply to 5²⁰ we would just need to figure out how many pairs of 5's could be chosen. There are 10 pairs of 5's, and so that would be the solution.

Since we have two distinct prime factors, these two problems run at the same time and independently of each other. You could pick 1 pair of 2's and 3 pairs of 5's or any other combination. But you are still limited to at most 20 pairs of 2's and at most 10 pairs of 5's. This is identical to a problem along the lines of "you have 20 pairs of socks and 10 pairs of shoes, how many footwear combinations do you have?"

The choices are independent, so we multiply the possibilities for each choice. There are 20 * 10 = 200 options for m².

To re-iterate, once we pick m², n is determined. So there are in fact 200 possible ordered pairs of positive integers (m, n) such that m²n = 20²⁰.