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

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

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

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

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