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

(Solution 2)
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===Solution 1===
 
===Solution 1===
  
Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(D)}\ 2.}</math>
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Using synthetic division, we get that the remainder is <math>\boxed{\textbf{(D)}\ 2}</math>.
  
 
===Solution 2===
 
===Solution 2===
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By the remainder theorem, the remainder is equal to the expression <math>x^{13}+1</math> when <math>x=1.</math> This gives the answer of <math> \boxed{(\mathrm{D})\ 2.} </math>
 
By the remainder theorem, the remainder is equal to the expression <math>x^{13}+1</math> when <math>x=1.</math> This gives the answer of <math> \boxed{(\mathrm{D})\ 2.} </math>
  
EDIT: This solution is invalid because x can't be 1.
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===Solution 3===
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Note that <math>x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)</math>, so <math>x^{13} - 1</math> is divisible by <math>x-1</math>, meaning <math>(x^{13} - 1) + 2</math> leaves a remainder of <math>\boxed{\mathrm{(D)}\ 2.}</math>
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===Video Solution===
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https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS
  
 
==See Also==
 
==See Also==

Latest revision as of 21:14, 6 February 2023

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

Using synthetic division, we get that the remainder is $\boxed{\textbf{(D)}\ 2}$.

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

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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