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

The correct answer is 2.

Let us solve this step-by-step to understand how to minimize the given expression under the given condition.

We are given four integers $a$, $b$, $c$, and $d$ such that their sum is 46:
$a+b+c+d=46$

We want to find the minimum possible value of the expression:
$S=(a-b{)}^2+(a-c{)}^2+(a-d{)}^2$

To make the sum of non-negative squared terms as small as possible, the values of $a$, $b$, $c$, and $d$ must be chosen as close to each other as possible, with $a$ specifically chosen to be close to the average of the four integers.

Let us calculate the average of the four integers:
$\text{Average}=\frac{46}{4}=11.5$

Since $a$, $b$, $c$, and $d$ are required to be integers, we split 46 into four integers as close to 11.5 as possible. The closest integers to 11.5 are 11 and 12.

To get a sum of 46 using only 11 and 12, we can choose two 11s and two 12s:
$11+11+12+12=46$

Now, let us assign these values to $a$, $b$, $c$, and $d$ to evaluate $S$:

If we set $a=11$, then the remaining integers $b$, $c$, and $d$ must be 11, 12, and 12.
Substituting these values into the expression:
$S=(11-11{)}^2+(11-12{)}^2+(11-12{)}^2$
$S=0^2+(-1{)}^2+(-1{)}^2$
$S=0+1+1=2$

Similarly, if we set $a=12$, the remaining integers are 11, 11, and 12:
$S=(12-11{)}^2+(12-11{)}^2+(12-12{)}^2$
$S=1^2+1^2+0^2=1+1+0=2$

Thus, the minimum possible value of $(a-b{)}^2+(a-c{)}^2+(a-d{)}^2$ is 2.