# Basic maths-calculus-Q18

+1 vote

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 ?

(A) The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

(B) The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

(C) The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

(D) The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

reshown May 24

m is a positive integer, so, lowest possible value of m=1.

Hence, in order to a+b+c+d = 4m+1, when m=1, at least one of a,b,c,d has to be greater than 1.

Take a=b=c=1; d=2

Hence, $a^2+b^2+c^2+d^2 = 7$

Which is equal to $4m^2 + 2m+1$ , for other values of a,b,c,d, $a^2+b^2+c^2+d^2$  will be greater than $4m^2 + 2m+1$