Difference between revisions of "2021 AMC 10B Problems/Problem 15"
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==Solution 2== | ==Solution 2== | ||
− | Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, | + | Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, and notice that this is also a solution the equation <math>x^2-x-1=0</math>, i.e. we have <math>x^2=x+1</math>. Repeatedly using this on the given (you can also just note Fibonacci numbers), |
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
(x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\ | (x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\ | ||
Line 19: | Line 19: | ||
&=(-2x^7+3x^6)+x^3 \\ | &=(-2x^7+3x^6)+x^3 \\ | ||
&=(x^6-2x^5)+x^3 \\ | &=(x^6-2x^5)+x^3 \\ | ||
− | &=(-x^5+x^4+x^3) | + | &=(-x^5+x^4+x^3) \\ |
− | &=-x^3(x^2-x-1) | + | &=-x^3(x^2-x-1) = \boxed{(\textbf{B}) 0} |
− | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
~Lcz | ~Lcz | ||
+ | |||
+ | ==Solution 3== | ||
+ | We can immediately note that the exponents of <math>x^{11}-7x^7+x^3</math> are an arithmetic sequence, so they are symmetric around the middle term. So, <math>x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})</math>. We can see that since <math>x+\frac{1}{x} = \sqrt{5}</math>, <math>x^2+2+\frac{1}{x^2} = 5</math> and therefore <math>x^2+\frac{1}{x^2} = 3</math>. Continuing from here, we get <math>x^4+2+\frac{1}{x^4} = 9</math>, so <math>x^4-7+\frac{1}{x^4} = 0</math>. We don't even need to find what <math>x^3</math> is! This is since <math>x^3\cdot0</math> is evidently <math>\boxed{(B) 0}</math>, which is our answer. | ||
+ | |||
+ | ~sosiaops | ||
+ | |||
+ | ==Solution 4== | ||
+ | We begin by multiplying <math>x+\frac{1}{x} = \sqrt{5}</math> by <math>x</math>, resulting in <math>x^2+1 = \sqrt{5}x</math>. Now we see this equation: <math>x^{11}-7x^{7}+x^3</math>. The terms all have <math>x^3</math> in common, so we can factor that out, and what we're looking for becomes <math>x^3(x^8-7x^4+1)</math>. Looking back to our original equation, we have <math>x^2+1 = \sqrt{5}x</math>, which is equal to <math>x^2 = \sqrt{5}x-1</math>. Using this, we can evaluate <math>x^4</math> to be <math>5x^2-2\sqrt{5}x+1</math>, and we see that there is another <math>x^2</math>, so we put substitute it in again, resulting in <math>3\sqrt{5}x-4</math>. Using the same way, we find that <math>x^8</math> is <math>11\sqrt{5}x-29</math>. We put this into <math>x^3(x^8-7x^4+1)</math>, resulting in <math>x^3(0)</math>, so the answer is <math>\boxed{(B)~0}</math>. | ||
+ | |||
+ | ~purplepenguin2 | ||
+ | |||
+ | == Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials) == | ||
+ | https://youtu.be/hzcSPVGFbC8 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution by Interstigation (Simple Silly Bashing) == | ||
+ | https://youtu.be/Hdk2SDOcw7c | ||
+ | |||
+ | ~ Interstigation | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | Not the most efficient method, but gets the job done. | ||
+ | |||
+ | https://youtu.be/L1iW94Ue3eI?t=1468 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Revision as of 08:52, 2 March 2021
Contents
Problem
The real number satisfies the equation . What is the value of
Solution 1
We square to get . We subtract 2 on both sides for and square again, and see that so . We can divide our original expression of by to get that it is equal to . Therefore because is 7, it is equal to .
Solution 2
Multiplying both sides by and using the quadratic formula, we get . We can assume that it is , and notice that this is also a solution the equation , i.e. we have . Repeatedly using this on the given (you can also just note Fibonacci numbers),
~Lcz
Solution 3
We can immediately note that the exponents of are an arithmetic sequence, so they are symmetric around the middle term. So, . We can see that since , and therefore . Continuing from here, we get , so . We don't even need to find what is! This is since is evidently , which is our answer.
~sosiaops
Solution 4
We begin by multiplying by , resulting in . Now we see this equation: . The terms all have in common, so we can factor that out, and what we're looking for becomes . Looking back to our original equation, we have , which is equal to . Using this, we can evaluate to be , and we see that there is another , so we put substitute it in again, resulting in . Using the same way, we find that is . We put this into , resulting in , so the answer is .
~purplepenguin2
Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials)
~ pi_is_3.14
Video Solution by Interstigation (Simple Silly Bashing)
~ Interstigation
Video Solution by TheBeautyofMath
Not the most efficient method, but gets the job done.
https://youtu.be/L1iW94Ue3eI?t=1468
~IceMatrix
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.