Difference between revisions of "1950 AHSME Problems/Problem 20"
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==Solution 2== | ==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> | + | 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> |
==See Also== | ==See Also== | ||
{{AHSME box|year=1950|num-b=19|num-a=21}} | {{AHSME box|year=1950|num-b=19|num-a=21}} | ||
Revision as of 11:41, 29 December 2011
Contents
Problem
When 
is divided by 
, the remainder is:
Solution 1
Use synthetic division. Notice that no matter what the degree of 
of the dividend is, the remainder is always 
Solution 2
Notice that 
is a zero of 
. By the factor theorem, since is a zero, then is a factor of , and when something is divided by a factor, the remainder is 
See Also
| 1950 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
