The sum of three prime numbers is 100. If one of the numbers is 36 more than another, find one of the numbers.

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Accepted Answer

Let the three prime numbers be $$ x $$, $$ y $$, and $$ y + 36 $$. According to the problem, their sum is:  
$$
x + y + (y + 36) = 100
$$  
Simplify the equation:  
$$
x + 2y + 36 = 100 \implies x + 2y = 64
$$

Assume $$ x = 2 $$, the smallest prime number. Substituting $$ x = 2 $$:  
$$
2 + 2y = 64 \implies 2y = 62 \implies y = 31
$$

Thus, the three prime numbers are $$ x = 2 $$, $$ y = 31 $$, and $$ y + 36 = 67 $$. Among the given options, **67** is one of the numbers.

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