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The average of 4 distinct prime numbers $a$, $b$, $c$, $d$ is $35$ where $a < b < c < d$. Given that $b$ and $c$ are equidistant from $34$ and; $a$ and $b$ are equidistant from $30$ and; $c$ and $d$ are equidistant from $40$; $a$ and $d$ are equidistant from $36$ . The difference between $a$ and $d$ is

a) $30$ b) $14$ c) $21$ d) can't be determined.

My trial: from given conditions:

$$\frac{a+b+c+d}{4}=35$$

$$a+b+c+d=140\tag1$$ $$b+c=2\cdot34=68 \tag2 $$ $$a+b=2\cdot30=60 \tag3 $$ $$c+d=2\cdot40=80 \tag4 $$ $$a+d=2\cdot36=72 \tag5 $$

Solving above equations I couldn't get the value of $d-a$. Can somebody please help me solve this problem? My book says answer says answer is b) $14$ but I didn't get it.

Thank you.

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    $\begingroup$ Well, you know $30<b<34$ so there are aren't a lot of candidates for $b$... $\endgroup$ – lulu Jan 19 '18 at 23:12
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    $\begingroup$ You aren't using the fact that these are prime numbers. $\endgroup$ – saulspatz Jan 19 '18 at 23:14
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As lulu points out, $30 < b < 34$ and $b$ is a prime number, so $b = 31$. Now, $a = 29$, $c = 37$ and $d = 43$, so $d - a = 43 - 29 = 14$.

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