Question

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

• A. 30
• B. 14
• C. 21
• D. Cannot be determined

Hint:

Convert each “equidistant from X” clue into a sum equation (e.g. a+d=72), then find primes satisfying all four sums.

Answer 1

The Correct Option is B

Turning each “equidistant from X” clue into a sum equation converts this into a small system of equations that can be solved directly, without guessing primes blindly.

1. “A and D equidistant from 36” means 36 is exactly halfway between them, so $a+d = 72$.
2. Similarly, $b+c=68$, $a+b=60$, and $c+d=80$.
3. From $a+d=72$ and $a+b=60$: subtracting gives $d - b = 12$.
4. Adding all four sum-equations: $2(a+b+c+d) = 72+68+60+80 = 280$, so $a+b+c+d = 140$, matching the average condition directly and confirming consistency.
5. Using $a+b=60$, try small primes for $a$: with $a=29$, $b=31$ (both prime); then $c = 68-31=37$ and $d=72-29=43$, both prime, and increasing in the right order.

$d - a = 43 - 29 = 14$.

The correct answer is option B, 14.