Question

If $a,b,c,d$ are four different positive integers selected from $1$ to $25$, then the highest possible value of \(\dfrac{(a+b)+(c+d){(a+b)+(c-d)}\) would be:}

• A. 47
• B. 49
• C. 51
• D. 96
• E. None of the above

Hint:

For ratios like \(\frac{S+d}{S-d}\) with distinct integers in a range, push the “+” term \(d\) to the maximum and keep \(S\) just above \(d\) to minimize the denominator and maximize the fraction.

Answer 1

The Correct Option is C

Let \( S = a + b + c \). Then the expression is \[ R = \frac{S + d}{S - d}. \]

For a fixed \(S\), \(R\) increases with \(d\); for a fixed \(d\), \(R\) decreases as \(S\) grows. Also, to keep \(R > 0\) we need \(S > d\). Hence, to maximize \(R\) choose \(d\) as large as possible and \(S\) just greater than \(d\).

Take \(d = 25\) (largest allowed). We need the smallest \(S = a + b + c\) with three distinct numbers not using 25 such that \(S > 25\). Using \(1, 2, 23\) gives \(S = 26\) (minimal possible \( > 25\)). Then \[ R = \frac{26 + 25}{26 - 25} = \frac{51}{1} = 51. \]

Trying \(d = 24\) gives maximum \[ \frac{25 + 24}{25 - 24} = 49, \] which is smaller.

Final Answer:

\[ \boxed{51} \]