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CAT 2017 Slot 1 — QA Question 28

Inequality Maximization / MinimizationEasy

If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a - b)2 + (a - c)2 + (a - d)2 is

Answer & solution

Correct answer: 2

Answer: 2

Solution

Easy

The sum of squared differences is smallest when b,c,db,c,d are as close to aa as possible. With integers summing to 3030, cluster all four values around the mean 30/4=7.530/4 = 7.5.

1

Aim for values near the mean. The mean is 30/4=7.530/4 = 7.5, which is not an integer, so the four integers cannot all be equal. The tightest integer cluster summing to 3030 is two 88's and two 77's.

a+b+c+d=30⇒ (a,b,c,d)=(8,8,7,7)(closest integer cluster)\begin{aligned} &a+b+c+d = 30\\ &\Rightarrow\ (a,b,c,d) = (8,8,7,7) \quad\text{(closest integer cluster)} \end{aligned}
2

Evaluate the expression. With a=8a=8, the differences from b,c,db,c,d are 0,1,10,1,1.

(a−b)2+(a−c)2+(a−d)2=(8−8)2+(8−7)2+(8−7)2=0+1+1=2\begin{aligned} &(a-b)^2 + (a-c)^2 + (a-d)^2\\ &= (8-8)^2 + (8-7)^2 + (8-7)^2\\ &= 0 + 1 + 1 = 2 \end{aligned}
3

Why it cannot be smaller. A value of 00 would force a=b=c=da=b=c=d, impossible since 4a=304a=30 has no integer solution. So at least two of b,c,db,c,d must differ from aa, giving a minimum of 22.

minimum=2\begin{aligned} &\text{minimum} = 2 \end{aligned}
22

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CAT 2017 Slot 1 QA Q28: If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a - b) 2 + (a — Solution | TheCATExam