Difference between revisions of "1980 AHSME Problems/Problem 28"
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\text{(E)} \ 65 </math> | \text{(E)} \ 65 </math> | ||
− | == Solution == | + | == Solution 1== |
Let <math>h(x)=x^2+x+1</math>. | Let <math>h(x)=x^2+x+1</math>. | ||
Then we have | Then we have | ||
<cmath>(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,</cmath> | <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. | + | 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 | Notice that | ||
Line 34: | Line 34: | ||
~~Wei | ~~Wei | ||
+ | |||
+ | ==Solution 2== | ||
+ | Notice that the roots of <math>w^2+w+1=0</math> are also the third roots of unity (excluding <math>w=1</math>). This is fairly easy to prove: multiply both sides by <math>w-1</math> and we get <cmath>(w-1)(w^2+w+1) = w^3 - 1 = 0.</cmath> These roots are <math>w = e^{i \pi /3}</math> and <math>w = e^{2i \pi /3}</math>. | ||
+ | |||
+ | Now we have | ||
+ | <cmath>\begin{align*} | ||
+ | w^{2n} + 1 + (w+1)^{2n} &= w^{2n} + 1 + (-w^2)^{2n} \ | ||
+ | &= w^{4n} + w^{2n} + 1\ | ||
+ | &= 0. | ||
+ | \end{align*}</cmath> | ||
+ | Plug in the roots of <math>w^2+w+1=0</math>. Note that | ||
+ | <cmath>e^{2i \pi /3} + 1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i + 1 = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i \pi /3}.</cmath> | ||
+ | However, this will not work if <math>n=3m</math>, so <math>n</math> cannot be equal to <math>21</math>. Hence our answer is <math>\textrm{(C)}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We start by noting that <cmath>x + 1 \equiv -x^2 \mod (x^2+x+1).</cmath> | ||
+ | Let <math>n = 3k+r</math>, where <math>r \in \{ 0,1,2 \}</math>. | ||
+ | |||
+ | Thus we have <cmath>x^{4n} + x^{2n} + 1 \equiv x^{4r} + x^{2r} + 1 \mod (x^3 -1).</cmath> | ||
+ | |||
+ | When <math>r = 0</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv 3 \mod (x^3 -1).</cmath> | ||
+ | When <math>r = 1</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),</cmath> which will be divisible by <math>x^2+x+1</math>. | ||
+ | |||
+ | When <math>r = 2</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),</cmath> which will also be divisible by <math>x^2+x+1</math>. | ||
+ | |||
+ | Thus <math>r \ne 0</math>, so <math>n</math> cannot be divisible by <math>3</math>, and the answer is <math>\textrm{(C)}</math>. | ||
== See also == | == See also == |
Latest revision as of 19:04, 30 October 2021
Contents
[hide]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 | |
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All AHSME Problems and Solutions |
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