If the numbers are 15a, 10b, b, then:
15a+10b+b=100,
15a+11b=100.
So b=(100-15a)/11, which can be written:
b=(99+1-11a-4a)/11=9-a-(4a-1)/11.
What I did here was to break the numbers in this equation down so that 11 would divide into them, and then I just have to work on a simple remainder (4a-1)/11.
Now we just find out what values of a would make this remainder into a whole number. a=3 gives us (12-1)/11=1. Now we can find b=(100-15×3)/11=(100-45)/11=55/11=5.
That gives us the three numbers:
15×3, 10×5, 5, making the numbers: 45, 50 and 5.
Only one of these numbers is in the given list. But the list should contain at least one multiple of 15 and one multiple of 10 to fit in with the question. Since it doesn’t, the list is not valid (wrong list!). 45, 50 and 5 satisfy the conditions and appear to be the only solutions where all the numbers are positive.
Mathematics is not about just using formulas, although they can be useful and are often involved in finding solutions. Compare this with car maintenance: a spanner may be essential in solving a vehicle problem, but it’s only a tool. The hard work is in reasoning out the problem logically step by step, and applying the tools if and when they’re necessary.