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

Revision as of 17:51, 11 June 2013

When $x^{13}+1$ is divided by $x-1$ , the remainder is:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$

Use synthetic division, and get that the remainder is $\boxed{\mathrm{(D)}\ 2.}$

By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{\textbf{(A)}\ 2.}$

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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 ===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>
+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{(A)}\ 2.} </math>
 ==See Also==