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

(Solution 2)
(Solution 2)
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===Solution 2===
 
===Solution 2===
  
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{\textbf{(D)}\ 2.} </math>
+
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{(D)\ 2.} </math>
  
 
==See Also==
 
==See Also==

Revision as of 04:48, 29 November 2014

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

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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