Im pretty sure the original question is:
How many ordered triples of positive integers \((a,b,c)\) have \(\text{GCF}(a, b, c) = 2020\) and \(\text{LCM}(a, b, c) = 2020^2\)
Because, 2020 has factors of 2, 2, 5, 101 and 4020 has factors of 2, 2, 5, 3, 67 and that wouldn't result in an integer.
but the answer to the "real question" would be to:
Note that the prime factorization of 2020 is \(2^2,5,101\) and that the prime factorization of \(2020^2\) is thus \(2^4 \cdot 5^2 \cdot 101^2\)
Thus the exponents of 2 in the prime factorizations of a,b,c must be 2, m, 4 in some order, where \(2 \le m \le 4\)That way, we can ensure that the GCD has two factors of 2 and the LCM has four factors of 2 If m=2 there are 3 distinct shufflings of (2, 4, 4.) If m=3 there are 6 distinct shufflings of (2,3,4) If m=4 there are 3 distinct shufflings of (2,4,4) So there are 3 allowed arrangements of powers of 2 among a,b ,and c .
The exponents of 5 for a,b,c must be some shuffling of (1, 2, 3) or (1, 2, 2) by the same logic, giving us 6 allowed arrangements of powers of 5. Likewise, there are 6 allowed arrangements for the powers of 101 by similar reasoning.
Thus the total number of ordered triples is the product of these three counts: \(12\times6\times6\) = 432 ordered triples.
