Question 689821
Since the sum of the three numbers is 13, either all three numbers are odd, or only one of them is odd.
{{{60=2*2*3*5}}} (prime factorization of {{{60}}})
By associating (putting in brackets) two of those four factors, that product can be split into three greater than 1 factors several ways:
A {{{(2*2)*3*5=4*3*5}}}
B {{{(2*5)*3*2=10*3*2}}}
C {{{(2*3)*5*2=6*5*2}}}
D {{{(3*5)*2*2=15*2*2}}}
We eliminate option D because the three factors are not all different.
I eliminate option A because I need to count on my fingers to add, and since the three factors include only two odd numbers, their sum will be even, and cannot be 13. (You can add if you want).
I have to add for options B and C:
B {{{10+3+2=15}}} is not the answer
C {{{6+5+2=13}}} is the answer.