If the three numbers are 15a, 10b and b, then 15a+10b+b=100.
So, 15a+11b=100.
So 11b=100-15a.
b=(100-15a)/11.
This can be written:
b=(99-22a)/11+(1+7a)/11.
b=9-2a+(7a+1)/11.
We need integer(s) a such that 7a+1 is a multiple of 11.
a=3, 14, 25, ...
Corresponding values of b are:
b=5, -10, -25, ...
15×3+55=100.
The three numbers could be 45, 50, 5 (most likely answer).
Other solutions would be whole numbers, but negative:
210, -100, -10, for example.