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Question 28 : 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 [TITA]

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Video Explanation

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Explanatory Answer

Method of solving this CAT Question from Algebra

Given that a , b , c and d are integers such that a + b + c + d = 30
The minimum possible value of (a - b)² + (a - c)² + (a - d)² has to be found.
For that a , b , c , d should be having equal or closer values such that (a – b) = (a - c) = (a – d) = 0 but this is not possible because 30 is not a multiple of 4.
We should find a as close to b , c , d as possible.
⟹ \\frac{30}{4}) = 7.5
So a , b , c , d should take the values of 7 or 8.
Hence taking two 7 and two 8 can work such that a , b , c , d is equal to 7 , 7 , 8 , 8 respectively.
So (a - b)² + (a - c)² + (a - d)² = (7 - 7)² + (7 - 8)² + (7 -8)²
(a - b)² + (a - c)² + (a - d)² = 0 + 1 + 1 = 2
If we are having the difference to be 2, on squaring it becomes 4 so it is not possible.
So let us try for values less than 2 i.e 0 + 0 + 1 but it doesn’t work.
So if a , b , c and d are integers such that a + b + c + d = 30,
the minimum possible value of (a - b)² + (a - c)² + (a - d)² is 2.

The question is "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 [TITA]"

Hence, the answer is 2