Quickly break integers into prime factors.
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Divide out small primes in order: all the 2s, then 3s, then 5s, then 7s. Stop when what is left is prime. \(1080 = 2^3 \times 135 = 2^3 \times 3^3 \times 5\).
Divisibility tests make this fast:
What the factorization tells you:
Common slips: stopping at a composite (like 9 or 15), and forgetting the \(+1\) when you count factors.
\(1200 = 2^{4} \times 3 \times 5^{2}\). Each factor picks an exponent from 0 up to the full exponent for every prime, so the count is \((4+1) \times (1+1) \times (2+1) = 30\). Multiplying the exponents without adding 1 to each gives \(8\). Adding the numbers 5, 2, 3 instead of multiplying them gives \(10\). Counting factor pairs instead of factors gives \(15\).
\(72 = 2^{3} \times 3^{2}\). An odd factor uses no 2s at all, so count factors of the odd part \(3^{2}\): \((2+1) = 3\). Counting all the factors gives \(12\). Counting the even factors instead gives \(9\). Multiplying the odd primes' exponents without adding 1 gives \(2\).
\(600 = 2^{3} \times 3 \times 5^{2}\). Because \(x\) and \(y\) share no prime, each prime power (\(2^{3}\), \(3\), \(5^{2}\)) must sit entirely inside one of them. So a full prime power (or any smaller power of that prime) must divide one number. A number built from two or more different primes can be split between \(x\) and \(y\), so it need not divide either one. A prime power higher than the one in 600 need not divide either number. (24 need not divide either number: \(x = 3\) and \(y = 200\) share no factor, and neither is divisible by 24. 16 need not divide either number: \(x = 1\) and \(y = 600\) share no factor, and neither is divisible by 16. 40 need not divide either number: \(x = 8\) and \(y = 75\) share no factor, and neither is divisible by 40.)