CATSprint

CAT 2025Slot 2QAQues & Sol

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

1. Concept Used

  • Topic: Minimization of Sum of Squares (Algebra + Number Theory)
  • Key Insight: The sum of squared differences is minimized when the integers are distributed as evenly as possible. If all four integers were equal, every squared difference would be zero — but this requires the sum to be divisible by 4.
  • Formula: $$ (a-b)^{2}+(a-c)^{2}+(a-d)^{2} \geq 0 $$

2. Calculation

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

First, check if all four integers can be equal. If $a = b = c = d$, then $4a = 46 \Rightarrow a = 11.5$, which is not an integer. So the minimum value cannot be zero.

Since $46 = 4 \times 11 + 2$, the remainder when 46 is divided by 4 is 2. This means we cannot split 46 equally among 4 integers. The closest even distribution uses two 11s and two 12s: ${11, 11, 12, 12}$, since $11 + 11 + 12 + 12 = 46$.

Now, to minimize the expression $(a-b)^{2} + (a-c)^{2} + (a-d)^{2}$, we want $a$ to be as close as possible to $b$, $c$, and $d$. Assign $a = 12,\ b = 12,\ c = 11,\ d = 11$:

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

To confirm this is truly the minimum, consider the next-closest distribution ${11, 11, 11, 13}$. Assign $a = 11,\ b = 11,\ c = 11,\ d = 13$:

$$(11 - 11)^{2} + (11 - 11)^{2} + (11 - 13)^{2} = 0 + 0 + 4 = 4$$

This gives 4, which is larger than 2. Similarly, ${10, 12, 12, 12}$ with $a = 12,\ b = 12,\ c = 12,\ d = 10$:

$$(12 - 12)^{2} + (12 - 12)^{2} + (12 - 10)^{2} = 0 + 0 + 4 = 4$$

Again, this gives 4. The distribution ${12, 12, 11, 11}$ clearly yields the smallest possible value of 2.

3. Solution

Answer = 2 ✅

The minimum possible value of $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$ is 2, achieved when $a = 12,\ b = 12,\ c = 11,\ d = 11$ (or any permutation that assigns one of the two 12s to $a$).