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Question 78
Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?
Solution
Taking lowest possible positive value of m i.e. 1 . Such that a+b+c+d=5 , so atleast one of them must be grater than 1 ,
take a=b=c=1 and d=2
we get $$a^2 + b^2 + c^2 + d^2 = 7$$ which is equal to $$4m^2+2m+1$$ for other values it is greater than $$4m^2+2m+1$$ . so option B
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