Difference between revisions of "2005 AIME II Problems/Problem 7"
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== Solution 2 (Bashing) == | == Solution 2 (Bashing) == | ||
− | Let <math>y=\sqrt[16]{5}</math>, then | + | Let <math>y=\sqrt[16]{5}</math>, then expanding the denominator results in: |
<cmath>(y^8+1)(y^4+1) =(y^{12}+y^8+y^4+1)</cmath> | <cmath>(y^8+1)(y^4+1) =(y^{12}+y^8+y^4+1)</cmath> | ||
<cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath> | <cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath> | ||
Line 26: | Line 26: | ||
Therefore: | Therefore: | ||
− | <cmath>\frac{4}{y^{16}-1/y-1} = \frac{4(y-1)}{y^{16}-1}=\frac{4(y-1)}{4} = y-1 </cmath> | + | <cmath>\frac{4}{(y^{16}-1)/(y-1)} = \frac{4(y-1)}{y^{16}-1}=\frac{4(y-1)}{4} = y-1 </cmath> |
+ | |||
+ | It is evident that <math> x+1 = (y-1)+1 = \sqrt[16]5</math> as Solution 1 states. | ||
+ | |||
+ | == Solution 3 == | ||
+ | Like Solution <math>2</math>, let <math>z=\sqrt[16]{5}</math> | ||
+ | Then, the expression becomes | ||
+ | |||
+ | <math>x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}</math> | ||
+ | Now, multiplying by the conjugate of each binomial in the denominator, we obtain... | ||
+ | |||
+ | <math>x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}</math> | ||
+ | Plugging back in, | ||
+ | |||
+ | <math>x=\frac{4({\sqrt[16]{5}-1)}}{5-1}\implies x=\sqrt[16]{5}-1</math> | ||
+ | |||
+ | Hence, after some basic exponent rules, we find the answer is <math>\boxed{125}</math> | ||
− | |||
== See Also == | == See Also == | ||
{{AIME box|year=2005|n=II|num-b=6|num-a=8}} | {{AIME box|year=2005|n=II|num-b=6|num-a=8}} |
Latest revision as of 11:33, 27 October 2019
Problem
Let Find .
Solution 1
We note that in general,
.
It now becomes apparent that if we multiply the numerator and denominator of by , the denominator will telescope to , so
.
It follows that .
Solution 2 (Bashing)
Let , then expanding the denominator results in:
Therefore:
It is evident that as Solution 1 states.
Solution 3
Like Solution , let Then, the expression becomes
Now, multiplying by the conjugate of each binomial in the denominator, we obtain...
Plugging back in,
Hence, after some basic exponent rules, we find the answer is
See Also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.