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 \(C\)?

• A. 23
• B. 21
• C. 19
• D. 17

Hint:

Combine the equations to get a single relation between A and C, then test the three primes between 12 and 20 to see which one gives whole numbers.

Answer 1

The Correct Option is A

Step 1: List the only candidates for A.
Since $A$ is a prime strictly between 12 and 20, the only options are $A = 13$, $A = 17$, or $A = 19$. Instead of deriving one master formula, try each candidate directly in the system and see which one gives whole positive integers throughout.

Step 2: Reduce the system to two equations in C and D first.
From $F = 2D$ (given) and $C + D + E = 2F$, we get $C + D + E = 4D$, so $E = 3D - C$.
From $E + F = 2C + 1$, substituting $E = 3D - C$ and $F = 2D$: $3D - C + 2D = 2C + 1$, which simplifies to $5D = 3C + 1$, i.e. $D = \dfrac{3C+1}{5}$. This relation does not involve $A$ yet, so it must hold no matter which $A$ turns out to be correct.

Step 3: Bring in $A$ through $C + F = 3A$.
Since $F = 2D = \dfrac{2(3C+1)}{5}$, equation $C + F = 3A$ becomes $C + \dfrac{6C+2}{5} = 3A$, i.e. $\dfrac{11C + 2}{5} = 3A$, so $11C + 2 = 15A$.

Step 4: Plug in each candidate value of A and check for a whole-number C.
If $A = 13$: $11C = 15(13) - 2 = 193$, so $C = 193/11$, not a whole number; reject $A = 13$.
If $A = 17$: $11C = 15(17) - 2 = 253$, so $C = 253/11 = 23$, a whole number; keep $A = 17$.
If $A = 19$: $11C = 15(19) - 2 = 283$, so $C = 283/11$, not a whole number; reject $A = 19$.
Only $A = 17$ survives, forcing $C = 23$.

Step 5: Confirm every other letter comes out positive and whole.
With $C = 23$: $D = \dfrac{3(23)+1}{5} = 14$, $F = 2D = 28$, $E = 3D - C = 42 - 23 = 19$, and from $B + C + D + E = 4A$: $B = 4(17) - 23 - 14 - 19 = 12$. Every one of $A = 17, B = 12, C = 23, D = 14, E = 19, F = 28$ is a positive integer, so the trial value $A = 17$ is fully consistent.

Final Answer:
Testing all three candidate primes directly shows only $A = 17$ works, giving $C = 23$.
\[ \boxed{C = 23} \]