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 which of the following must be true?

• A. D is the lowest integer and D = 14
• B. C is the greatest integer and C = 23
• C. B is the lowest integer and B = 12
• D. F is the greatest integer and F = 24

Hint:

Solve the system fully first, then sort all six values from smallest to largest before checking each statement.

Answer 1

The Correct Option is C

This question checks whether you correctly solved the system in the first place and then correctly ranked the six values by size. Solving the five equations with $A$ a prime between 12 and 20 gives the unique set $A = 17$, $B = 12$, $C = 23$, $D = 14$, $E = 19$, $F = 28$. Sorted from smallest to largest, this is $B = 12$, $D = 14$, $A = 17$, $E = 19$, $C = 23$, $F = 28$.

1. D is the lowest integer and D = 14: The value 14 for $D$ is right, but $D$ is not the smallest, since $B = 12$ is smaller. This option fails on the "lowest" claim.
2. C is the greatest integer and C = 23: The value 23 for $C$ is right, but $C$ is not the largest, since $F = 28$ is bigger. This option fails on the "greatest" claim.
3. B is the lowest integer and B = 12: $B = 12$ is correct and, checking the sorted list, $B$ really is the smallest of all six values. Both parts of this statement hold.
4. F is the greatest integer and F = 24: $F$ is correctly identified as the largest value, but its actual value is 28, not 24, so this option fails on the numeric part.

Only the third statement survives both checks: $B$ is the lowest integer, and $B = 12$.

Let's summarize:

• The sorted order of the six values is $B < D < A < E < C < F$.
• Three of the four options get the numeric value right but the ranking wrong, or vice versa; only one option gets both right.

So the correct statement is that B is the lowest integer and B = 12.