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

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.

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
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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

Can someone explain the solution to this problem?

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
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Nice job, Shashank!!

Welcome to GMAT Club!!



Cheers!
GyM
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this post is deleted

Hi jxav,

Could you please explain :
X=1^51(1+2+3+...+16)

How did u arrive at it?
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Hi there,

Sorry! I just reviewed my answer, and I think I made a conceptual mistake. My solution is definitely invalid, my bad!

Don't worry! happens to the best of us!


Cheers
GyM
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Cool. No worries! To err is human. I believe all of us make mistakes.
This forum is exclusively meant for this ; learn from mistakes.
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(a power n + b power n ) is divisible by a+n if n is odd

add 1+2
3+4 and so on til 15+16
we get 105 , it is divisible by 17

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

why did you add and subtract?

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15 ^51 = (17-2)^51

All terms except last will be divisible by 17

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