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

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==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>
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1950|num-b=19|num-a=21}}
 
{{AHSME box|year=1950|num-b=19|num-a=21}}

Revision as of 10:41, 29 December 2011

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 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 (ProblemsAnswer KeyResources)
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