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:

Hint:

For minimizing sums of squares with a fixed sum constraint, keep the numbers as close together as possible. Here, that meant taking $a,b,c,d$ near the average $\frac{46}{4}=11.5$, and distributing small integer deviations evenly.

Answer 1

Correct Answer: 2

We want to minimize \[ S = (a - b)^2 + (a - c)^2 + (a - d)^2 \] subject to \[ a + b + c + d = 46, \] where $a, b, c, d$ are integers. Step 1: Rewrite in terms of deviations from $a$. Let \[ b = a - \delta_1,\quad c = a - \delta_2,\quad d = a - \delta_3, \] where $\delta_1, \delta_2, \delta_3$ are integers. Then \[ S = \delta_1^2 + \delta_2^2 + \delta_3^2. \] Substitute into the sum constraint: \[ a + (a - \delta_1) + (a - \delta_2) + (a - \delta_3) = 46 \] \[ 4a - (\delta_1 + \delta_2 + \delta_3) = 46 \] \[ \delta_1 + \delta_2 + \delta_3 = 4a - 46. \]
Step 2: Choose $a$ to keep the deviations small. To make $S = \delta_1^2 + \delta_2^2 + \delta_3^2$ as small as possible, we want the sum \[ \delta_1 + \delta_2 + \delta_3 = 4a - 46 \] to be as close to $0$ as possible, because spreading a small total among three integers gives smaller squares. The mean of $a,b,c,d$ is \[ \frac{46}{4} = 11.5, \] so $a$ should be near $11$ or $12$. Case 1: $a = 11$. Then \[ \delta_1 + \delta_2 + \delta_3 = 4(11) - 46 = 44 - 46 = -2. \] To minimize $\delta_1^2 + \delta_2^2 + \delta_3^2$ with sum $-2$, choose values as evenly as possible, e.g. \[ (\delta_1, \delta_2, \delta_3) = (0, -1, -1). \] Then \[ S = 0^2 + (-1)^2 + (-1)^2 = 0 + 1 + 1 = 2. \] Corresponding integers: \[ a = 11,\quad b = 11,\quad c = 12,\quad d = 12 \] (sum $=11+11+12+12=46$, valid). Case 2: $a = 12$. Then \[ \delta_1 + \delta_2 + \delta_3 = 4(12) - 46 = 48 - 46 = 2. \] Even distribution: \[ (\delta_1, \delta_2, \delta_3) = (1, 1, 0). \] Then \[ S = 1^2 + 1^2 + 0^2 = 1 + 1 + 0 = 2. \] Corresponding integers: \[ a = 12,\quad b = 11,\quad c = 11,\quad d = 12 \] (sum $=12+11+11+12=46$, valid).
Step 3: Check that we cannot do better. If $a$ is further from $11.5$, e.g.\ $a=10$ or $a=13$: - For $a = 10$: $\delta_1 + \delta_2 + \delta_3 = 4(10) - 46 = -6$. Best balance is roughly $(-2,-2,-2)$ giving \[ S \approx 2^2 + 2^2 + 2^2 = 12>2. \] - For $a = 13$: $\delta_1 + \delta_2 + \delta_3 = 4(13) - 46 = 6$. Best balance is roughly $(2,2,2)$ giving \[ S \approx 2^2 + 2^2 + 2^2 = 12>2. \] Thus $S$ cannot be smaller than $2$, and we have explicit integer examples achieving $S = 2$. Therefore, the minimum possible value of \[ (a - b)^2 + (a - c)^2 + (a - d)^2 \] is \[ \boxed{2}. \]