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How many ordered triples of positive integers $(a, b, c)$ have $\text{GCF}(a, b, c) = 2020$ and $\text{LCM}(a, b, c) = 4020?$ 

 Feb 11, 2024
 #1
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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.

 

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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.

 Apr 18, 2024

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