Difference between revisions of "1980 AHSME Problems/Problem 28"

(Solution)
(Solution)
Line 10: Line 10:
  
 
== 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 $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals

$\text{(A)} \ 17 \qquad  \text{(B)} \ 20 \qquad  \text{(C)} \ 21 \qquad  \text{(D)} \ 64 \qquad  \text{(E)} \ 65$

Solution

Assume $h(x)=x^2+x+1$ $(x+1)^2n = (h(x)+x)^n = g(x)*h(x) + x^n$

$x^2n = x^2n+x^(2n-1)+x^(2n-2)            -x^(2n-1)-x^(2n-2)-x^(2n-3)          +...$

$x^n = x^n+x^(n-1)+x^(n-2)          -x^(n-1)-x^(n-2)-x^(n-3)   +....$

Therefore, the left term from $x^2n is x^(2n-3u)$

          the left term from $x^n is x^(n-3v)$, 

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 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS