Difference between revisions of "1980 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
− | Assume h(x)=x^2+x+1 | + | Assume <math>h(x)=x^2+x+1</math> |
− | (x+1)^2n = (h(x)+x)^n = g(x)*h(x) + x^n | + | <math>(x+1)^2n = (h(x)+x)^n = g(x)*h(x) + x^n</math> |
− | x^2n = x^2n+x^(2n-1)+x^(2n-2) | + | <math>x^2n = x^2n+x^(2n-1)+x^(2n-2) |
-x^(2n-1)-x^(2n-2)-x^(2n-3) | -x^(2n-1)-x^(2n-2)-x^(2n-3) | ||
− | +... | + | +...</math> |
− | x^n = x^n+x^(n-1)+x^(n-2) | + | <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> |
− | Therefore, the left term from x^2n is x^(2n-3u) | + | Therefore, the left term from <math>x^2n is x^(2n-3u)</math> |
− | the left term from x^n is x^(n-3v), | + | the left term from <math>x^n is 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 13:01, 17 June 2021
Problem
The polynomial is not divisible by if equals
Solution
Assume
Therefore, the left term from
the left term from ,
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=1/2 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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