The HCF of three positive integers is 6 and their sum is 120. How many

Question

The HCF of three positive integers is 6 and their sum is 120. How many such triplets exist?

(A) 18
(B) 24
(C) 29
(D) 32
(E) 35

IanStewart · Answer

The question doesn't even say whether we're meant to consider ordered triplets or unordered ones (do we count "6, 6, 108" and "108, 6, 6" as if they were the same, or as if they were different?). From the size of the answer choices, they clearly mean to ask about unordered triples, so 6,6,108 and 108,6,6 should be considered to be the same triplet here, but the question absolutely needs to tell you that.

If our three numbers have a GCD of 6, and sum to 120, then when we divide them all by 6, they will have a GCD of 1 and will sum to 20. Now using GMAT-level math, I don't see a way to arrive at an answer without enumerating the possibilities, and this is not the kind of thing you'd ever need to do on the GMAT. But we can start by imagining our smallest number is 1, then 2, and so on, and count how many triplets we have in each case. Since all three numbers in a triplet cannot be 'above average', one number will need to be at most 20/3, so we can stop once we imagine the smallest value is 6.

1: if '1' is one of our numbers, our GCD will automatically be 1. The distinct triplets are thus the nine possibilities: (1,1,18), (1,2,17), ... (1,9,10)

2: if '2' is one of our numbers, the other two numbers will need to be odd (if they're even, we get a GCD of 2, not 1). So we have the four possibilities (2,3,15), (2,5,13), (2,7,11), (2,9,9). Here, by ensuring our numbers are constantly increasing (or staying the same) within each triplet, we can be sure we aren't counting any triplet twice -- we don't want to count (2, 1, 17) here, for example, because we counted that already when '1' was the smallest value.

3: if '3' is one of our numbers, we'll get a GCD of 1 as long as both of our other numbers are not multiples of 3. But that always happens, because the numbers sum to 20, so the three numbers cannot all be divisible by 3. So we get the six triplets (3,3,14), (3,4,13)... (3,8,9)

4: if '4' is one of our numbers, again the other two numbers must be odd to avoid a GCD of 2. So we have two triplets: (4,5,11), (4,7,9)

5: if '5' is one of our numbers, we must avoid the situation where all three numbers are multiples of 5 (since then the GCD is not 1). So we have two possibilities: (5,6,9) and (5,7,8)

6: lastly, if '6' is the smallest number, the other numbers must be odd, and we have only (6,7,7)

Adding the number of possibilities from each case, we get 9+4+6+2+2+1 = 24. But this is not a realistic GMAT question.

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The HCF of three positive integers is 6 and their sum is 120. How many

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