TITLE: The sum of three primes is 100, one of them exceeds the other by 36, find the largest one
QUESTION [7 upvotes]: The sum of $3$ primes numbers is $100$. One of them exceeds the other by $36$. Find the largest one.

My attempts:
Let $p_1 +p_2  + p_3=100$, also let $p_2=p_1 +36 \implies 2p_1 +p_3=64$, from here $p_1 \neq p_2 \neq2 \implies p_3=2$, hence $p_1 =31 \implies p_2=67$.
Is this make sense, I don't have an answer, please add your answer.

REPLY [3 votes]: As we know that the sum of three odd numbers cannot be even i.e. one of the prime is even or $2$ let one of the other prime is $p_1$ then the third prime is $p_1+36$ now according to question $$2+p_1+p_1+36=100$$ $$2p_1=62$$ $$p_1=31$$ Hence, the primes are 

$$2,31,67$$