2020 AIME II Problem 1

Question

Find the number of ordered pairs of positive integers

(m, n)

such that

m^2 n = 20^{20}.

Answer

Write

20^{20} = 2^{40} \cdot 5^{20},

and note that once

m

is chosen,

n

is completely determined

You need

m^2

to divide

2^{40} \cdot 5^{20}:

count the possible exponents of

2

and of

5

in

m

Since

20^{20} = 2^{40} \cdot 5^{20},

a valid

m

must have the form

2^a 5^b,

and then

n = 2^{40 - 2a}\,5^{20 - 2b}

is a positive integer exactly when

2a \le 40

and

2b \le 20.

Conversely every such choice works, and

n

is uniquely determined by

m.

So

a

can be any of

0, 1, \ldots, 20

and

b

any of

0, 1, \ldots, 10,

giving

21 \cdot 11 = 231

ordered pairs.

Problem 1 in other years