Difference between revisions of "1950 AHSME Problems/Problem 20"
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===Solution 1=== | ===Solution 1=== | ||
− | + | Using synthetic division, we get that the remainder is <math>\boxed{\textbf{(D)}\ 2}</math>. | |
===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{\ | + | 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{(\mathrm{D})\ 2.} </math> |
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+ | ===Solution 3=== | ||
+ | |||
+ | Note that <math>x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)</math>, so <math>x^{13} - 1</math> is divisible by <math>x-1</math>, meaning <math>(x^{13} - 1) + 2</math> leaves a remainder of <math>\boxed{\mathrm{(D)}\ 2.}</math> | ||
+ | |||
+ | ===Video Solution=== | ||
+ | https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS | ||
==See Also== | ==See Also== | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:14, 6 February 2023
Contents
[hide]Problem
When is divided by , the remainder is:
Solution
Solution 1
Using synthetic division, we get that the remainder is .
Solution 2
By the remainder theorem, the remainder is equal to the expression when This gives the answer of
Solution 3
Note that , so is divisible by , meaning leaves a remainder of
Video Solution
https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.