1980 AHSME Problems/Problem 28
Problem
The polynomial is not divisible by if equals
Solution 1
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
Solution 2
Notice that the roots of are also the third roots of unity (excluding ). This is fairly easy to prove: multiply both sides by and we get These roots are and .
Now we have Plug in the roots of . Note that However, this will not work if , so cannot be equal to . Hence our answer is .
Solution 3
We start by noting that Let , where .
Thus we have
When , When , which will be divisible by .
When , which will also be divisible by .
Thus , so cannot be divisible by , and the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.