Difference between revisions of "1950 AHSME Problems/Problem 20"

Revision as of 10:18, 30 July 2022

Problem

When $x^{13}+1$ is divided by $x-1$ , the remainder is:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

Solution 1

Use synthetic division, and get that the remainder is $\boxed{\mathrm{(D)}\ 2.}$

Solution 2

By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{(\mathrm{D})\ 2.}$

Solution 3

Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)$ , so $x^{13} - 1$ is divisible by $x-1$ , meaning $(x^{13} - 1) + 2$ leaves a remainder of $\boxed{\mathrm{(D)}\ 2.}$

Video Solution

https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 ===Video Solution===
-https://www.youtube.com/watch?v=z4-bFo2D3TU&list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&index=3&t=3s
+https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS
-(Solution at 41:25 - AMBRIGGS
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