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Difference between revisions of "1950 AHSME Problems/Problem 20"
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== Problem==
 
== Problem==
  
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When <math>x^{13}-1</math> is divided by <math>x-1</math>, the remainder is:
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When <math>x^{13}+1</math> is divided by <math>x-1</math>, the remainder is:
  
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math>
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math>
  
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==Solution 1==
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==Solution==
 +
===Solution 1===
  
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Use synthetic division. Notice that no matter what the degree of <math>x</math> of the dividend is, the remainder is always <math>\boxed{\mathrm{(C)}\ 0.}</math>
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Using synthetic division, we get that the remainder is <math>\boxed{\textbf{(D)}\ 2}</math>.
  
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==Solution 2==
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===Solution 2===
  
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Notice that <math>1</math> is a zero of <math>x^{13} - 1</math>. By the factor theorem, since <math>1</math> is a zero, then <math>x-1</math> is a factor of <math>x^{13} - 1</math>, and when something is divided by a factor, the remainder is <math>\textbf{(C)}0</math>
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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>
 +
 
 +
===Solution 3===
 +
 
 +
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>
 +
 
 +
===Video Solution===
 +
https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS
  
 
==See Also==
 
==See Also==
  
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{{AHSME box|year=1950|num-b=19|num-a=21}}
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{{AHSME 50p box|year=1950|num-b=19|num-a=21}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

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 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AHSME Problems and Solutions

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