CAT 2025

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Minimization of Sum of Squares (Algebra + Number Theory)
• Formula: $$ (a-b)^{2}+(a-c)^{2}+(a-d)^{2} \geq 0 $$

The key insight is that the minimum value of a sum of squares is zero — but only when all differences are zero, i.e., when ( a = b = c = d ). This is only possible when the total sum is divisible by 4.

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2. Calculation

We are given that ( a + b + c + d = 46 ), where ( a, b, c, d ) are integers, and we want to minimize $( (a-b)^{2} + (a-c)^{2} + (a-d)^{2} )$.

Step 1: Check if all four values can be equal.

If ( a = b = c = d ), then $( 4a = 46 \Rightarrow a = 11.5 )$, which is not an integer. So the expression cannot be zero.

Step 2: Find the closest integer distribution.

Since $( 46 = 4 \times 11 + 2 )$, we can distribute the sum as evenly as possible among four integers. The best we can do is use two values of 11 and two values of 12, or equivalently, use $( {11, 11, 12, 12} )$ — but we want to minimize differences specifically involving ( a ).

Let us assign $( a = 12 ), ( b = 12 ), ( c = 11 ), ( d = 11 )$. Then: $$a + b + c + d = 12 + 12 + 11 + 11 = 46 \checkmark$$

Step 3: Evaluate the expression.

$$(a - b)^{2} + (a - c)^{2} + (a - d)^{2}$$ $$= (12 - 12)^{2} + (12 - 11)^{2} + (12 - 11)^{2}$$ $$= 0 + 1 + 1 = \mathbf{2}$$

Step 4: Verify this is the minimum.

The only other symmetric integer split near equal values would be ( {11, 11, 11, 13} ) or ( {10, 12, 12, 12} ). Assigning ( a = 11 ), ( b = 11 ), ( c = 11 ), ( d = 13 ): $$(11-11)^2 + (11-11)^2 + (11-13)^2 = 0 + 0 + 4 = 4$$ This gives 4, which is larger than 2. So the minimum is confirmed to be 2.

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3. Solution

Answer = 2 ✅

The minimum possible value of ( (a-b)^{2}+(a-c)^{2}+(a-d)^{2} ) is 2, achieved when the four integers are as equal as possible, e.g., ( {12, 12, 11, 11} ).