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

Entered answer:

✅ Correct Answer: 2

Since the expression $(a-b)^2 + (a-c)^2 + (a-d)^2$ is a sum of squares, it is minimized when $a, b, c, d$ are as close to each other as possible.

If all four integers were equal:

$a = b = c = d$

$4a = 46$

$a = 11.5$

This is not an integer, so all four values cannot be equal.

The closest integers to $11.5$ are $11$ and $12$ .

Two of the values must be $11$ and two must be $12$ to get the sum $46$ :

$11 + 11 + 12 + 12 = 46$

To minimize $(a-b)^2 + (a-c)^2 + (a-d)^2$ , as many of $b, c, d$ should be equal to $a$ as possible.

Let $a = 11,\ b = 11,\ c = 12,\ d = 12$

$(a-b)^2 + (a-c)^2 + (a-d)^2$

$= (11-11)^2 + (11-12)^2 + (11-12)^2$

$= 0 + 1 + 1$

$= 2$

The minimum possible value is $2$ .

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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