The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is

### Concept & Logic The sum of three odd numbers is always an odd number. Since the given sum of three prime numbers is 100 (an even number), one of the prime numbers *must* be even. The only even prime number is 2. ### Step-by-Step Solution * **Set up the variables:** Let the three prime numbers be $x, y,$ and $z$. Given: $x + y + z = 100$ * **Identify the even prime:** As established, for the sum to be even, one prime must be 2. Let $x = 2$. Substitute $x$ into the equation: $$2 + y + z = 100$$ $$y + z = 98$$ * **Use the second condition:** We are given that one prime exceeds another by 36. Let this relationship be between $z$ and $y$. $$z - y = 36$$ * **Solve the system of linear equations:** Equation 1: $z + y = 98$ Equation 2: $z - y = 36$ Add the two equations together to eliminate $y$: $$2z = 134$$ $$z = 67$$ * **Find the remaining prime:** Substitute $z = 67$ back into Equation 1: $$67 + y = 98$$ $$y = 31$$ The three prime numbers are 2, 31, and 67. Option (d) contains 67. ### Exam Strategy & Shortcut The realization that one prime *must* be 2 is the key to unlocking this problem in seconds. Once you deduce $y + z = 98$ and $z - y = 36$, you can simply scan the options. You need an option that, when you subtract 36 from it, gives another prime number. Only $67 - 36 = 31$ (which is prime) fits perfectly. ### Common Pitfall Assuming all prime numbers must be odd. Students who forget that 2 is a prime number will find it impossible to solve $x + y + z = 100$ because they will assume $x, y,$ and $z$ are all odd. ### Final Answer **Therefore, the correct answer is 67.**