The sum of $3$ primes numbers is $100$. One of them exceeds the other by $36$. Find the largest one.
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.