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

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X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 27 Jun 2018, 22:03
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\(X= 1^{51}+2^{51}+..............+16^{51}\)
What is the remainder when X is divided by 17?

(a) 0
(b) 3
(c) 10
(d) 13
(e) 16
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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 23 Jul 2018, 06:51
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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.
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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 27 Jun 2018, 23:49
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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
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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 29 Jun 2018, 10:57
Can someone explain the solution to this problem?
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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 30 Jun 2018, 00:07
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PKN wrote:
\(X= 1^{51}+2^{51}+..............+16^{51}\)
What is the remainder when X is divided by 17?

(a) 0
(b) 3
(c) 10
(d) 13
(e) 16


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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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 23 Jul 2018, 08:47
shashankyele wrote:
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.



Nice job, Shashank!!

Welcome to GMAT Club!!



Cheers!
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X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post Updated on: 23 Jul 2018, 11:37
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Originally posted by jxav on 23 Jul 2018, 11:04.
Last edited by jxav on 23 Jul 2018, 11:37, edited 1 time in total.
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X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 23 Jul 2018, 11:09
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jxav wrote:
Hi,

Another solution will be:

X=1^51(1+2+3+...+16)
The sum of the number 1 to 16 will be: (1+16)/2*16=136
Factors of 136=2^3 and 17

Since 17 is one of the prime factors of X, X will be divisible by 17 and the remainder will be 0.

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Hi jxav,

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

How did u arrive at it?
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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 23 Jul 2018, 11:33
PKN wrote:
jxav wrote:
Hi,

Another solution will be:

X=1^51(1+2+3+...+16)
The sum of the number 1 to 16 will be: (1+16)/2*16=136
Factors of 136=2^3 and 17

Since 17 is one of the prime factors of X, X will be divisible by 17 and the remainder will be 0.

Sent from my SM-G935F using GMAT Club Forum mobile app



Hi jxav,

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

How did u arrived at it?


Hi there,

Sorry! I just reviewed my answer, and I think I made a conceptual mistake. My solution is definitely invalid, my bad!
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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 23 Jul 2018, 11:41
jxav wrote:
PKN wrote:
jxav wrote:
Hi,

Another solution will be:

X=1^51(1+2+3+...+16)
The sum of the number 1 to 16 will be: (1+16)/2*16=136
Factors of 136=2^3 and 17

Since 17 is one of the prime factors of X, X will be divisible by 17 and the remainder will be 0.

Sent from my SM-G935F using GMAT Club Forum mobile app



Hi jxav,

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

How did u arrived at it?


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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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided  [#permalink]

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New post 23 Jul 2018, 11:48
jxav wrote:
PKN wrote:
jxav wrote:
Hi,

Another solution will be:

X=1^51(1+2+3+...+16)
The sum of the number 1 to 16 will be: (1+16)/2*16=136
Factors of 136=2^3 and 17

Since 17 is one of the prime factors of X, X will be divisible by 17 and the remainder will be 0.

Sent from my SM-G935F using GMAT Club Forum mobile app



Hi jxav,

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

How did u arrived at it?


Hi there,

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


Cool. No worries! To err is human. I believe all of us make mistakes.
This forum is exclusively meant for this ;) ; learn from mistakes.
_________________

Regards,

PKN

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Re: X= 1^51 + 2^51 + ... + 16^51. What is the remainder when X is divided &nbs [#permalink] 23 Jul 2018, 11:48
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