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The sum of three positive integers is 120. Pairwise, exactly once (out of three possible pairs) are the numbers relatively prime (i.e. they have no common divisor greater than 1). What are the three numbers?

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The common factors of the two pairs with greatest common divisor greater than 1 cannot be 2, 3 or 5 because those divide 120, so the third number would also need to be divisible by that factor. They also cannot be both greater or equal to 11 since 11x13=143 is bigger than 120. So either we have (77, a multiple of 7, a multiple of 11) or we have (91, a multiple of 7, a multiple of 13), and 7x17=119 is already too big. Brute-forcing multiples of 13 in the second case and of 11 in the first case shows us only (77, 21, 22) works.

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