CAT 2025 Slot 2 QA Q7 Inequalities & Modulus

Question 7 of the Quantitative Ability (QA) section, CAT 2025 Slot 2, with CAT's official answer and a worked solution.

If \( a , b , c \) and \( d \) are integers such that their sum is 46 , then the minimum possible value of \( ( a - b ) ^ { 2 } + ( a - c ) ^ { 2 } + ( a - d ) ^ { 2 } \) is

TITA: type in the answer. There are no options and no negative marking.

Answer: 2 Inequalities & Modulus TITA

Solution

Write \(b = a+x\), \(c = a+y\), \(d = a+z\). Then \(4a + x + y + z = 46\), and we minimise \(x^2 + y^2 + z^2\).
\(x + y + z = 46 - 4a \equiv 2 \pmod 4\), so the sum cannot be 0 or \(\pm1\); the smallest option is \(x+y+z = \pm 2\).
With sum 2, the least \(x^2+y^2+z^2\) is with \((1,1,0)\): value 2.
Example: \(a = 11, b = 12, c = 12, d = 11\) (sum 46) gives \(1 + 1 + 0 = 2\).