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If a+b+c+d=30, where a,b,c and d are integers, then the minimum possib

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If a+b+c+d=30, where a,b,c and d are integers, then the minimum possib  [#permalink]

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New post 30 Apr 2019, 09:53
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If a+b+c+d=30, where a,b,c and d are integers, then the minimum possible value of \((a - b)^2\) + \((a - c)^2\) + \((a - d)^2\) is

A. 3
B. 2
C. 0
D. 1
E. 5
Manager
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Re: If a+b+c+d=30, where a,b,c and d are integers, then the minimum possib  [#permalink]

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New post 30 Apr 2019, 10:04
Superb question. Simple but tricky.

To minimize the overall value individual values of a-b , a-c, a-d should be minimum and that is possible only when they are same digit or minimum difference.
Values would be 7 8 8 7 which gives answer 2

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Re: If a+b+c+d=30, where a,b,c and d are integers, then the minimum possib  [#permalink]

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New post 30 Apr 2019, 10:50
nick1816 wrote:
If a+b+c+d=30, where a,b,c and d are integers, then the minimum possible value of \((a - b)^2\) + \((a - c)^2\) + \((a - d)^2\) is

A. 3
B. 2
C. 0
D. 1
E. 5


IMO - B

\((a - b)^2\) + \((a - c)^2\) + \((a - d)^2\) - Since in the question the difference of digits is squared, so the min value can be zero. It cant be negative.
Lets assume the min value is zero. It means a=b=c=d
=> 4a = 30
=> a= 30/4 = 15/2 = 7.5 But 'a' cant be a decimal number as the question states that the digits are integers. So, lets assume that a=7 or a=8
In case, a= 7, sum of all digits is 7 x 4 = 28 ( Assuming all digits to be of same value so as to get zero as the difference).
It falls short of 30 by 02. So, we need to increase two 07 into two eights. Hence the numbers are 7,7,8,8

In case, a=8, sum of all digits is 8 x 4 = 32 ( Assuming all digits to be of same value so as to get zero as the difference).
It exceeds 30 by 02. So, we need to decrease two 08 into two seven. Hence the numbers are 8 , 8 , 7 , 7

Now, (8-8)^2 + (8-7)^2 + (8-7)^2 = 0 + 1 + 1 = 2 Answer
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Re: If a+b+c+d=30, where a,b,c and d are integers, then the minimum possib   [#permalink] 30 Apr 2019, 10:50
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