How many positive three digit integers can be formed such that the hundreds digit is equal to the sum of the tens digit and ones digit?

start with the 100's place

0 = 0+0

1 = 1+0 = 0+1

2 = 1+1=2+0=0+2

3 = 3+0 = 2+1 = 1+2 = 0+3

...

with the 100's digit, the number of such numbers is one greater than that number

0 : 1

1 : 2

...

therefore there are 1+2+3+4+5+6+7+8+9+10 = 5*11 = 55 such numbers

unless you don't count 000 as a three digit number, then there are 54 such numbers.