100 = 50$\times$2$\times$1$\times$1 not all different
= 10$\times$5$\times$2$\times$1 good! Sum = 18
Is there another way?
Factors of 100 not used: 4, 20, 25
100 = 4$\times$25$\times$1$\times$1 not all different
= 4$\times$5$\times$5$\times$1 not all different. Cannot use 4
= 25$\times$2$\times$2$\times$1 not all different. Cannot use 25
100 = 20$\times$5$\times$1$\times$1 not all different. Cannot use 20
Make a set of numbers that use all the digits from 1 to 9, once and
once only. Add them up. The result is divisible by 9. Add each of
the digits in the new number. What is their sum? Now try some other
possibilities for yourself!