If the product of three consecutive integers is 120, then the sum of the integers is · ABC26GN1725

$120=2 \times 2 \times 2 \times 3 \times 5=(2 \times 2) \times 5 \times(2 \times 3)=4 \times 5 \times 6$. Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6. Required sum $=4+5+6=15$.