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

Step 1: Set up the average equation.
\(a+b+c+d=35\times4=140\).

Step 2: Translate the equidistant conditions into equations.
a, d equidistant from 36: \(a+d=2\times36=72\). b, c equidistant from 34: \(b+c=2\times34=68\). a, b equidistant from 30: \(a+b=2\times30=60\). c, d equidistant from 40: \(c+d=2\times40=80\).

Step 3: Solve the system.
From \(a+b=60\) and \(a+d=72\): \(d-b=12\). From \(a+d=72\) and \(a+b=60\), and using \(b+c=68\): subtracting \(a+b=60\) from \(a+d=72\) gives \(d-b=12\); combined with \(b+c=68\), try prime values.

Step 4: Find the primes by testing.
Testing \(a=29\): \(b=60-29=31\) (prime). \(c=68-31=37\) (prime). \(d=72-29=43\) (prime). Check: \(c+d=37+43=80\) &check;, and \(29+31+37+43=140\) &check;, with all values prime and increasing.

Step 5: Find the difference.
\(d-a=43-29=14\).

Step 6: Final Answer.
The difference between a and d is 14, so option B is correct.