Question

A, B, C, D, E and F are six positive integers such that
\(B + C + D + E = 4A\)
\(C + F = 3A\)
\(C + D + E = 2F\)
\(F = 2D\)
\(E + F = 2C + 1\)
If \(A\) is a prime number between 12 and 20, then what is the value of \(F\)?

• A. 14
• B. 16
• C. 20
• D. 28

Hint:

Express every letter in terms of F and A, combine them into one equation, then test the three candidate primes for A.

Answer 1

The Correct Option is D

Step 1: Express every letter in terms of F and A right away.
From $F = 2D$, get $D = F/2$. From $C + F = 3A$, get $C = 3A - F$. From $E + F = 2C + 1$, substitute $C$: $E = 2(3A - F) + 1 - F = 6A - 3F + 1$.

Step 2: Plug all three into $C + D + E = 2F$ at once.
$(3A - F) + \dfrac{F}{2} + (6A - 3F + 1) = 2F$
Collect the $A$ terms: $3A + 6A = 9A$. Collect the $F$ terms: $-F + \dfrac{F}{2} - 3F = -\dfrac{7F}{2}$. So the left side becomes $9A + 1 - \dfrac{7F}{2}$.
$9A + 1 - \dfrac{7F}{2} = 2F$

Step 3: Solve this single equation for F in terms of A.
Multiply every term by 2: $18A + 2 - 7F = 4F$, so $18A + 2 = 11F$, giving $F = \dfrac{18A + 2}{11}$.

Step 4: Try the three candidate primes for A.
$A = 13$: $F = \dfrac{236}{11}$, not whole, reject.
$A = 17$: $F = \dfrac{308}{11} = 28$, whole, keep.
$A = 19$: $F = \dfrac{344}{11}$, not whole, reject.
Only $A = 17$ gives a whole $F$, namely $F = 28$.

Step 5: Back-fill and confirm all six letters.
With $A = 17$, $F = 28$: $D = 14$, $C = 3(17) - 28 = 23$, $E = 6(17) - 3(28) + 1 = 102 - 84 + 1 = 19$, and from $B + C + D + E = 4A$: $B = 68 - 23 - 14 - 19 = 12$. All six values, $17, 12, 23, 14, 19, 28$, are positive integers, so the system is fully consistent.

Final Answer:
$F = 28$
\[ \boxed{28} \]