Is it divisible?

Algebra Level 3

x 9999 + x 8888 + x 7777 + ⋯ + x 1111 + x 0 x^{9999}+x^{8888}+x^{7777}+ \cdots+x^{1111}+x^{0} is divisible by which of the following polynomials ?

x 8 + x 6 + x 4 + x 2 x^{8}+x^{6}+x^{4}+x^{2} x 9 + x 7 + x 5 + x 3 + x x^{9}+x^{7}+x^{5}+x^{3}+x x + 1 x+1 x − 1 x-1

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Piyush Kumar Behera by Piyush Kumar Behera , 20, India

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2 solutions

Tom Engelsman
Feb 9, 2017

The above polynomial factors into:

x 8888 ( x 1111 + 1 ) + x 6666 ( x 1111 + 1 ) + x 4444 ( x 1111 + 1 ) + x 2222 ( x 1111 + 1 ) + ( x 1111 + 1 ) ; x^{8888}(x^{1111} + 1) + x^{6666}(x^{1111} + 1) + x^{4444}(x^{1111} + 1) + x^{2222}(x^{1111} + 1) + (x^{1111} + 1);

or ( x 1111 + 1 ) ( x 8888 + x 6666 + x 4444 + x 2222 + 1 ) ; (x^{1111} + 1)(x^{8888} + x^{6666} + x^{4444} + x^{2222} + 1);

or ( x + 1 ) ( x 1110 − x 1109 + x 1108 − . . . + x 2 − x + 1 ) ( x 8888 + x 6666 + x 4444 + x 2222 + 1 ) . (x + 1)(x^{1110} - x^{1109} + x^{1108} - ... + x^2 - x + 1)(x^{8888} + x^{6666} + x^{4444} + x^{2222} + 1).

Of the above answer choices, only x + 1 x + 1 divides the polynomial neatly.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I expected the same

Piyush Kumar Behera - 4 years, 4 months ago
J D
May 23, 2016

Plug in -1, see that you get zero

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Grouping the terms from the end and the beginning alternately , We can show that the polynomial is divisible by x + 1 x+1 by using the identity of a n + b n a^{n} + b^{n} , we'll get x + 1 x+1 common from all these pairs and hence the polynomial will be divisible by x + 1 x+1 .

Rishabh Tiwari - 4 years, 11 months ago

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