Difference between revisions of "1980 AHSME Problems/Problem 28"
m (→Solution) |
m (→Solution) |
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Line 17: | Line 17: | ||
Notice that | Notice that | ||
− | < | + | <cmath>x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x |
-x^{2n-1}-x^{2n-2}-x^{2n-3} | -x^{2n-1}-x^{2n-2}-x^{2n-3} | ||
− | +...< | + | +...</cmath> |
− | < | + | <math>x^n = x^n+x^{n-1}+x^{n-2} |
-x^{n-1}-x^{n-2}-x^{n-3} | -x^{n-1}-x^{n-2}-x^{n-3} | ||
− | +....<math> | + | +....</math> |
− | Therefore, the left term from < | + | Therefore, the left term from <math>x^2n</math> is <math>x^{(2n-3u)}</math> |
− | the left term from < | + | the left term from <math>x^n</math> is <math>x^{(n-3v)}</math>, |
If divisible by h(x), we need 2n-3u=1 and n-3v=2 or | If divisible by h(x), we need 2n-3u=1 and n-3v=2 or |
Revision as of 18:38, 30 October 2021
Problem
The polynomial is not divisible by if equals
Solution
Let .
Then we have where is (after expanding according to the Binomial Theorem.
Notice that
Therefore, the left term from is
the left term from is ,
If divisible by h(x), we need 2n-3u=1 and n-3v=2 or
2n-3u=2 and n-3v=1
The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible
~~Wei
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 |
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