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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.

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    $\begingroup$ It looks fine, so +1. $\endgroup$ Commented Jan 14, 2017 at 14:41
  • $\begingroup$ Actually, how'd you conclude $p_1\ne2$? That step is slightly missing. $\endgroup$ Commented Jan 14, 2017 at 14:43
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    $\begingroup$ @SimpleArt then $p_2$ won't be prime. $\endgroup$
    – mnulb
    Commented Jan 14, 2017 at 14:44
  • $\begingroup$ Ok, then everything looks good :-) $\endgroup$ Commented Jan 14, 2017 at 14:44
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    $\begingroup$ If three primes sum to an even number, then at least one of the primes is 2. $\endgroup$
    – Guangliang
    Commented Jan 14, 2017 at 14:56

3 Answers 3

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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

$$2,31,67$$

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You found correctly that $2p_1+p_3=64$ then you obtain that $p_3$ is even then $p_3=2$ and consequently $p_1=31$. Therefore the result you found was correct.

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I start with @Guangliang, that one of the primes must be $2$. Then the other two sum to $98$ and differ by $36$. Solving the simultaneous equations gives $31$ and $67$.

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