Difference between revisions of "1950 AHSME Problems/Problem 20"
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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{( | + | 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> |
==See Also== | ==See Also== |
Revision as of 05:47, 29 November 2014
Problem
When is divided by , the remainder is:
Solution
Solution 1
Use synthetic division, and get that the remainder is
Solution 2
By the remainder theorem, the remainder is equal to the expression when This gives the answer of
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.