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

(Problem)
(Solution 1)
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===Solution 1===
 
===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>
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Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(D)}\ 2.}</math>
  
 
===Solution 2===
 
===Solution 2===

Revision as of 17:50, 11 June 2013

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

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 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AHSME Problems and Solutions