Question

The sum of three non-zero prime numbers is $100$. One of them exceeds the other by $36$. Find the largest number.

• A. $73$
• B. $91$
• C. $67$
• D. $57$

Answer 1

The correct option is C $67$

As we know that the sum of three odd numbers cannot be even.

$\therefore$ one of the prime is even

Since $2$ is the only prime number which is even.

$\therefore$ one of the three prime numbers is $2$.

Let one of the other prime numbers is $p_1$ then the third prime number will be $p_1+36$.

Now according to question,

$2+p_1+p_1+36=100$

$2p_1=62$

$p1=31$

Hence, the three prime numbers are $2,31$ and $67$ and the largest among them is $67$.

Hence, $67$ is the correct answer.