Difference between revisions of "1980 AHSME Problems/Problem 28"
m (→Solution) |
m (→Solution) |
||
| 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
![\[(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,\]](http://latex.artofproblemsolving.com/6/2/f/62fb179abbba076535939a467d039a9b373eed26.png)
where 
is 
(after expanding 
according to the Binomial Theorem.
Notice that
![\[x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x -x^{2n-1}-x^{2n-2}-x^{2n-3} +...\]](http://latex.artofproblemsolving.com/6/0/3/60335eb9fb695aa6aa7905fb73f1942f48205574.png)
Therefore, the left term from 
is 
the left term fromis
,
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
