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: Set up the same five equations as before.
\[ B + C + D + E = 4A \quad (ii) \]
\[ C + F = 3A \quad (ii) \]
\[ C + D + E = 2F \quad (iii) \]
\[ F = 2D \quad (iv) \]
\[ E + F = 2C + 1 \quad (vv) \]
We need the value of \(F\) specifically, so it helps to reduce everything to \(F\) and \(A\).

Step 2: Reduce (iii) and (iv) to a relation between C, D, E.
Substituting \(F = 2D\) into (iii): \(C + D + E = 4D\), so \(C + E = 3D \quad (vi)\)

Step 3: Reduce (iv) and (vv) to write E in terms of C and D.
\[ E + 2D = 2C + 1 \]
\[ E = 2C - 2D + 1 \quad (vii) \]

Step 4: Combine (vi) and (vii) to link C and D.
\[ C + (2C - 2D + 1) = 3D \]
\[ 3C + 1 = 5D \]
Since \(F = 2D\), we get \(D = \dfrac{F}{2}\), so \(3C + 1 = \dfrac{5F}{2}\), which gives \(C = \dfrac{5F - 2}{6} \quad (viii)\)

Step 5: Bring in \(C + F = 3A\) to connect F and A.
Substitute (viii): \(\dfrac{5F - 2}{6} + F = 3A\), so \(\dfrac{5F - 2 + 6F}{6} = 3A\), giving \(11F - 2 = 18A\), i.e. \(F = \dfrac{18A + 2}{11} \quad (ix)\)

Step 6: Test the three primes between 12 and 20.
For \(A = 13\): \(F = \dfrac{18(13)+2}{11} = \dfrac{236}{11}\), not whole.
For \(A = 17\): \(F = \dfrac{18(17)+2}{11} = \dfrac{308}{11} = 28\), a whole number.
For \(A = 19\): \(F = \dfrac{18(19)+2}{11} = \dfrac{344}{11}\), not whole.
Only \(A = 17\) gives a whole \(F\), so \(F = 28\).

Step 7: Check the rest stay positive integers.
With \(F = 28\): \(D = 14\), and from (viii) \(C = \dfrac{5(28)-2}{6} = \dfrac{138}{6} = 23\). Then \(E = 2(23) - 2(14) + 1 = 19\), and \(B = 4(17) - 23 - 14 - 19 = 12\). Every value is a positive integer, so \(A = 17\) checks out fully.

Final Answer:
Options 14, 16 and 20 do not satisfy \(F = \dfrac{18A+2}{11}\) for any allowed prime \(A\), so \(F = 28\) is the only consistent value.
\[ \boxed{F = 28} \]