Essentially what they provided was the perimeter of a right triangle as 40. A 3-4-5 right triangle has a perimeter of 12. The closest you could get would be a perimeter of 36: 3*12, but that is not 40. So, the right triangle with a perimeter of 40 cannot be a multiple of 3-4-5. Your answer of 10 makes no sense. The answer is definitely 8, 15, and 17. You can take a^2+b^2=c^2 and a+b+c=40 and come up with ab=800-40c. This tells you that if c is 20, ab is zero and if c is > 20, ab is negative----so c is definitely less than 20. For each choice of c (e.g. 19, 18, 17, 16, etc....) you know that a+b=40-c-----with ab and a+b, you can easily come up with a quadratic equation for a or b and you need to choose the c value that results in a and b being integers. If c is 17, then ab is 120 and a+b is 23 and for b, we have: b^2-23b+120---so b is (23+/-7)/2.