CAT 2025 Slot 2 QA Question & Solution

AlgebraMedium

Question

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

Solution

Given expression: $$(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$$

The given expression has just the sum of squares of the terms. So, the minimum value is either zero or positive.

If we can make all the values equal, we can get zero. But since all the values are integers and the sum 46 is not divisible by 4, we can't make everything equal.

So, the nearest four values are 11, 10, 11, 10.

With this, the minimum value is $$(11-10)^2+(11-11)^2+(11-10)^2 = 2$$