Let a, b and c be the numbers, so a+b+c=100.
a=15n where n is a whole number; b=10c.
15n+10c+c=100; 15n+11c=100. c=(100-15n)/11. When n=3, c=(100-45)/11=5. [The reason for this is that 11 must divide exactly into 100-15n, so 100-15n=11, 22, 33, 44, 55, 66, 77, 88, or 99. Therefore 15 must divide into 89, 78, 67, 56, 45, 34, 23, 12 or 1. Only 45 fits.]
So b=50 and a=45; so the numbers are 45, 50 and 5.