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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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Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(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>
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 +
===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>
  
 
==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}}

Revision as of 21:55, 13 May 2016

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 $\boxed{\mathrm{(D)}\ 2.}$

See Also

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

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