Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
Assume <math>h(x)=x^2+x+1</math>
+
Let <math>h(x)=x^2+x+1</math>.
−
<math>(x+1)^2n = (h(x)+x)^n = g(x)*h(x) + x^n</math>
 
  
−
<math>x^2n = x^2n+x^{2n-1}+x^{2n-2}
+
Then we have
 +
<cmath>(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,</cmath>
 +
where <math>g(x)</math> is <math>h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}</math> (after expanding <math>(h(x)+x)^n</math> according to the Binomial Theorem.
 +
 
 +
Notice that
 +
<math></math>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}
−
         +...</math>
+
         +...<math>
  
−
<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>
+
   +....<math>
  
−
Therefore, the left term from <math>x^2n</math> is <math>x^{(2n-3u)}</math>
+
Therefore, the left term from </math>x^2n<math> is </math>x^{(2n-3u)}<math>
−
           the left term from <math>x^n</math> is <math>x^{(n-3v)}</math>,  
+
           the left term from </math>x^n<math> is </math>x^{(n-3v)}$,  
  
 
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  
 
                               2n-3u=2 and n-3v=1
 
                               2n-3u=2 and n-3v=1
  
−
The solution will be n=1/2 mod(3). Therefore n=21 is impossible
+
The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible
  
 
~~Wei
 
~~Wei

Revision as of 19:38, 30 October 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

Let $h(x)=x^2+x+1$.

Then we have \[(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,\] where $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem.

Notice that $$ (Error compiling LaTeX. Unknown error_msg)x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x 

          -x^{2n-1}-x^{2n-2}-x^{2n-3}
        +...$$ (Error compiling LaTeX. Unknown error_msg)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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_28&oldid=164332"