Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1950 AHSME Problems/Problem 20"
(fixed solution sections)
Line 5: Line 5:
 
<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>
  
−
==Solution 1==
+
==Solution==
 +
===Solution 1===
  
 
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>
 
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>
  
−
==Solution 2==
+
===Solution 2===
  
 
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>
 
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>
Line 16: Line 17:
  
 
{{AHSME box|year=1950|num-b=19|num-a=21}}
 
{{AHSME box|year=1950|num-b=19|num-a=21}}
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 13:22, 17 April 2012

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. Notice that no matter what the degree of $x$ of the dividend is, the remainder is always $\boxed{\mathrm{(C)}\ 0.}$

Solution 2

Notice that $1$ is a zero of $x^{13} - 1$. By the factor theorem, since $1$ is a zero, then $x-1$ is a factor of $x^{13} - 1$, and when something is divided by a factor, the remainder is $\textbf{(C)}0$

See Also

1950 AHSME (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
All AHSME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_20&oldid=46423"