Answer:
3, 4, 5
Step-by-step explanation:
It is not specified how many numbers we should consider. We assume them to be integers. We also assume that each number appears only once.
In this case, these numbers must be divisors of 60, i.e 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
Their sum is 12, so we can exclude those numbers >10. So these numbers are in the list: 1, 2, 3, 4, 5, 6, 10.
Assume 2 numbers:
a×b=60, a+b=12. a, b <= 10
We gave only 10×6, but sum is not 12.
Assume 3 numbers:
a×b×c=60, a+b+c=12.
10 is out, so the list shrinks to 1,2,3,4,5,6.
We have 2 possible combinations: 3,4,5 and 2,5,6 with product 60. However only the former qualifies, since the later fails the sum condition.
So far, the winning combination: 3, 4, 5.
Assume 4 numbers.
We could have 1,3,4,5 and 1,2,5,6, but both of them fail the sum condition.
Similar for 5 numbers, or more.
