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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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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]" **

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