The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is:
Answer: D
\(x+(x+36)+y=100 \\ \Rightarrow 2x+y=64\)
Therefore \(y\) must be even prime, which is 2.
Therefore,
\(2x+2=64 \\ \Rightarrow x=31\)
Third prime number
\(= (x+36)\\= (31+36)\\= 67\)