What is the remainder when the average (arithmetic mean) of the intege

Question

What is the remainder when the average (arithmetic mean) of the integers –x^4, -x^3, x and x^2 is divided by 4?

(1) x – 3 is even
(2) The average (arithmetic mean) of –x^3, x and x^2 is 11

samburslem7 · Answer

When will the solution be posted  Bunuel

rshricnd · Answer

GMATNinja   Bunuel   BrentGMATPrepNow   ExpertsGlobal5  Can someone explain how answer be D not A?

BrentGMATPrepNow · Answer

The GMAT won't ask remainder questions without first specifying that all numbers are positive. 
I sugdest you skip this one.

BrushMyQuant · Answer

We need to find What is the remainder when the average (arithmetic mean) of the integers –x^4, -x^3, x and x^2 is divided by 4

Average = Sum of Terms / Total Number of Terms =  \frac{–x^4 + (-x^3) + x + x^2}{4}  =  \frac{–x^4 -x^3 + x + x^2}{4}  =  \frac{-x^3(x + 1) + x (1+x)}{4}  =  \frac{(x + 1) (-x^3 + x)}{4}  =  \frac{(x + 1) x (1 - x^2)}{4}  =  \frac{x (x + 1)(1-x)(1+x)}{4}  =  \frac{(1-x)*x*(x + 1)^2}{4}  = - \frac{(x-1)*x*(x + 1)^2}{4}

STATEMENT 1 : x – 3 is even

x - Odd = Even
=> x = Even + Odd = Odd
Since x is odd so x-1 and x+1 will become even
Either x-1 is a multiple of 4 and x+1 is a multiple of 2 or x-1 is a multiple of 2 and x+1 is a multiple of 4
in both the cases (x-1)* (x+1)^2  will become a multiple of 4*4 = 16
=> - \frac{(x-1)*x*(x + 1)^2}{4}  will be a multiple of  \frac{16}{4}  = 4
=> Remainder when the average (arithmetic mean) of the integers –x^4, -x^3, x and x^2 is divided by 4 = 0
=>   SUFFICIENT

STATEMENT 2 : The average (arithmetic mean) of –x^3, x and x^2 is 11

\frac{–x^3 + x + x^2}{3}  = 11
=>  x + x^2 - x^3  = 33
=> x is odd
Same as Statement 1
=>   SUFFICIENT

So,  Answer will be D 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

Archit3110 · Answer

target is (arithmetic mean) of the integers –x^4, -x^3, x and x^2 is divided by 4

#1
x-3 is even
x has to be odd
so value of 4 odd integer will give even value only which will be 2^2 at least
plug in x= 1,5,
sufficient divisible by 4
#2
The average (arithmetic mean) of –x^3, x and x^2 is 11
x*(x+1-x^2) =33
x= -3
in target we get yes
sufficient
option D

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What is the remainder when the average (arithmetic mean) of the intege

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