Difference between revisions of "2008 AMC 10A Problems/Problem 7"
Piemax2713 (talk | contribs) (→Solution) |
Piemax2713 (talk | contribs) (→Solution) |
||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | Simplifying, we get <cmath>\frac{3^4016-3^4012}{3^4014-3^4010}.</cmath> factoring out <math>3^4012</math> on the top and factoring out <math>3^4010</math> on the bottom gives us <cmath>\frac{(3^4-1)(3^4012)}{(3^4-1)(3^4010)}.</cmath> Canceling out <math>3^4-1</math> gives us <math>\frac{3^4012}{3^4010}=\frac{3^2}{3^0}=9.</math> | + | Simplifying, we get <cmath>\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.</cmath> factoring out <math>3^{4012}</math> on the top and factoring out <math>3^{4010}</math> on the bottom gives us <cmath>\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.</cmath> Canceling out <math>3^4-1</math> gives us <math>\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9.</math> |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2008|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:18, 30 November 2019
Problem
The fraction
simplifies to which of the following?
Solution
Simplifying, we get factoring out
on the top and factoring out
on the bottom gives us
Canceling out
gives us
See also
2008 AMC 10A (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 • 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.