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: Write down the five given equations.
\[ B + C + D + E = 4A \quad (ii) \]
\[ C + F = 3A \quad (ii) \]
\[ C + D + E = 2F \quad (iii) \]
\[ F = 2D \quad (iv) \]
\[ E + F = 2C + 1 \quad (vv) \]
All six letters stand for positive integers, and \(A\) is a prime between 12 and 20.

Step 2: Combine (iii) and (iv) to relate C, D and E.
Substituting \(F = 2D\) from (iv) into (iii):
\[ C + D + E = 2(2D) = 4D \]
\[ C + E = 3D \quad (vi) \]

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

Step 4: Substitute (vii) into (vi) to connect C and D alone.
\[ C + (2C - 2D + 1) = 3D \]
\[ 3C + 1 = 5D \quad (viii) \]
So \(D = \dfrac{3C + 1}{5}\), and from (iv), \(F = 2D = \dfrac{2(3C + 1)}{5} = \dfrac{6C + 2}{5} \quad (ix)\)

Step 5: Bring in equation (ii) to connect C and A.
From (ii), \(F = 3A - C\). Setting this equal to (ix):
\[ 3A - C = \frac{6C + 2}{5} \]
\[ 15A - 5C = 6C + 2 \]
\[ 15A = 11C + 2 \quad (xx) \]

Step 6: Test the three primes between 12 and 20.
A prime between 12 and 20 can be 13, 17 or 19. Since \(C = \dfrac{15A - 2}{11}\) must come out as a whole number, check each:
For \(A = 13\): \(\dfrac{15(13) - 2}{11} = \dfrac{193}{11}\), not a whole number.
For \(A = 17\): \(\dfrac{15(17) - 2}{11} = \dfrac{253}{11} = 23\), a whole number.
For \(A = 19\): \(\dfrac{15(19) - 2}{11} = \dfrac{283}{11}\), not a whole number.
So the only value that keeps every letter a whole positive integer is \(A = 17\), which gives \(C = 23\).

Step 7: Verify by finding all six integers.
From (viii): \(D = \dfrac{3(23) + 1}{5} = \dfrac{70}{5} = 14\). From (iv): \(F = 2(14) = 28\). From (vii): \(E = 2(23) - 2(14) + 1 = 46 - 28 + 1 = 19\). From (ii): \(B = 4(17) - 23 - 14 - 19 = 68 - 56 = 12\). Checking (ii): \(C + F = 23 + 28 = 51 = 3(17)\), correct. All six values, \(A = 17\), \(B = 12\), \(C = 23\), \(D = 14\), \(E = 19\), \(F = 28\), are positive integers, confirming the solution is consistent.

Final Answer:
The other options, 21, 19 and 17, do not satisfy equation (xx) for any prime \(A\) between 12 and 20, so they cannot be correct.
\[ \boxed{C = 23} \]