x = 1^51 + 2^51 + ... + 16^51. What is the remainder when x is divided

Question

x = 1^{51} + 2^{51} + ... + 16^{51} 
What is the remainder when x is divided by 17?

(A) 0
(B) 3
(C) 10
(D) 13
(E) 16

sujoykrdatta · Accepted Answer

There is a nice little property: 
 a^n + b^n  is divisible by  (a + b)  if  n is odd

Example:  a^3 + b^3 = (a+b)(a^2 - ab + b^2)  =>  a^3 + b^3  is divisible by  (a + b)

Thus, we have:
 1^{51}+16^{51}  is divisible by (1+16) i.e. 17

2^{51}+15^{51}  is divisible by (2+15) i.e. 17

3^{51}+14^{51}  is divisible by (3+14) i.e. 17

....

8^{51}+9^{51}  is divisible by (8+9) i.e. 17

Thus, the sum of all the terms is divisible by 17

=> Remainder = 0

Answer A

shashankyele · Answer

Hi,

The principle here is, if the value of n is odd, then a^n +b^n will always be divisible by (a+b). Since the exponential power in this case is odd, and the denominator is 17, you can try splitting the terms in the numerator such that they represent 17. For example, 1+16/17, 2+15/17, 3+14/17, 4+13/17, 5+12/17....all the way to 9+8/17. Hence, the numerator is entirely divisible by 17.

sandman13 · Answer

I thought this was a tough one. Maybe someone can suggest a simpler approach.

Ignoring the first (1^51) and last (16^51) terms, we can rewrite other terms in the equation as follows:
2^51 = (17-15)^51
3^51 = (17-14)^51
4^51 = (17-13)^51

Now the remainder of (17-15)^51 will only be determined by the last term (-15)^51 ; that is if you were to open the binomial expression
Let's not calculate the remainder of (-15)^51, because this will be cancelled out by remainder of 15^51, which is also a term in X

All the remainder will cancel each other out, including the two terms which I had excluded in the beginning.

Answer: A

GyMrAT · Answer

The concepts that you need to use for this question are of  Remainder Theorem  & of  negative Remainders .

\frac{16^{51}}{17}  gives a remainder which is same as the remainder of  \frac{(-1)^{51}}{17} .

Lets denote it is R( \frac{16^{51}}{17} ) = R( \frac{(-1)^{51}}{17} ) = - R( \frac{(1)^{51}}{17} )

Similarly, R( \frac{15^{51}}{17} ) = R( \frac{(-2)^{51}}{17} ) = - R( \frac{(2)^{51}}{17} )

R( \frac{14^{51}}{17} ) = R( \frac{(-3)^{51}}{17} ) = - R( \frac{(3)^{51}}{17} )

R( \frac{13^{51}}{17} ) = R( \frac{(-4)^{51}}{17} ) = - R( \frac{(4)^{51}}{17} )

R( \frac{12^{51}}{17} ) = R( \frac{(-5)^{51}}{17} ) = - R( \frac{(5)^{51}}{17} )

R( \frac{11^{51}}{17} ) = R( \frac{(-6)^{51}}{17} ) = - R( \frac{(6)^{51}}{17} )

R( \frac{10^{51}}{17} ) = R( \frac{(-7)^{51}}{17} ) = - R( \frac{(7)^{51}}{17} )

R( \frac{9^{51}}{17} ) = R( \frac{(-8)^{51}}{17} ) = - R( \frac{(8)^{51}}{17} )

Now you can see that each of the remainders of the last 8 terms is going to be equal & opposite in sign to the remainders of the first 8 terms.

They will cancel out & the Answer will be 0.

Although you do not need to do all the above calculations, it is pretty clear after you check for the last two terms.

I will explain the next steps too, for more clarity

R( \frac{X}{17} ) = R( \frac{(1)^{51}}{17} ) + R( \frac{(2)^{51}}{17} ) + R( \frac{(3)^{51}}{17} ) + R( \frac{(4)^{51}}{17} ) + R( \frac{(5)^{51}}{17} ) + R( \frac{(6)^{51}}{17} ) + R( \frac{(7)^{51}}{17} ) + R( \frac{(8)^{51}}{17} ) + R( \frac{(9)^{51}}{17} ) + R( \frac{(10)^{51}}{17} ) + R( \frac{(11)^{51}}{17} ) + R( \frac{(12)^{51}}{17} ) + R( \frac{(13)^{51}}{17} ) + R( \frac{(14)^{51}}{17} ) + R( \frac{(15)^{51}}{17} ) + R( \frac{(16)^{51}}{17} )

R( \frac{X}{17} ) = R( \frac{(1)^{51}}{17} ) + R( \frac{(2)^{51}}{17} ) + R( \frac{(3)^{51}}{17} ) + R( \frac{(4)^{51}}{17} ) + R( \frac{(5)^{51}}{17} ) + R( \frac{(6)^{51}}{17} ) + R( \frac{(7)^{51}}{17} ) + R( \frac{(8)^{51}}{17} ) - R( \frac{(8)^{51}}{17} ) - R( \frac{(7)^{51}}{17} ) - R( \frac{(6)^{51}}{17} ) - R( \frac{(5)^{51}}{17} ) - R( \frac{(4)^{51}}{17} ) - R( \frac{(3)^{51}}{17} ) - R( \frac{(2)^{51}}{17} ) - R( \frac{(1)^{51}}{17} )

Hence, R( \frac{X}{17} ) = 0

Answer A.

Thanks,
GyM

rvgmat12 · Answer

A simple way to solve this :

1+ 2 + 3 + 4 + 5 + 6 + 7 + 8 -8 - 7 - 6 - 5 -4 -3 -2 -1

Remainder must be zero.

Answer A.

Hope it helps

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

x = 1^51 + 2^51 + ... + 16^51. What is the remainder when x is divided

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

x = 1^51 + 2^51 + ... + 16^51. What is the remainder when x is divided

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards