Question

let a, b, c be positive integers less than 10 such that (100a + 10b + c)^2 = (a+b+c)^5 what is the value of (a*b-c)

Answer 1

lets assume 

A = 9 
B = 9 

99+99 = 198 

as you can see AAC cannot be bigger then 198 so A has to be 1. 

A=1 

11C is a three digit number then B has to be 9 !! 

since 18+81 = 99 --> two digits 

19+91 = 100+10+C 

110 = 110+C 

C=0 

the answer is (E)