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

Since \(46\div4=11.5\), the four integers can never all be equal, so the sum of squared differences can never be zero. The next best thing is to keep every value within 1 of the average, using only 11's and 12's. Four such integers summing to 46 must be two 11's and two 12's (since \(11+11+12+12=46\)). Picking either repeated value as the pivot \(a\), the other three consist of one number equal to \(a\) and two numbers exactly 1 away from it, so the sum of squares works out to \(0^2+1^2+1^2=2\). No arrangement using values further from the average can do better, so the minimum possible value is 2.