Difference between revisions of "2014 AMC 10B Problems/Problem 17"
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We begin by factoring the <math>2^{1002}</math> out. This leaves us with <math>5^{1002} - 1</math>. | We begin by factoring the <math>2^{1002}</math> out. This leaves us with <math>5^{1002} - 1</math>. | ||
| − | We factor the difference of squares, leaving us with <math>(5^{501} - 1)(5^{501} + 1)</math>. We note that for all powers of 5 more than two, it ends in ...<math>25</math> Thus, <math>(5^{501} + 1)</math> would end in ...<math>26</math> and thus would contribute one power of two to the answer, but not more. | + | |
| − | We can continue to factor <math>(5^{501} - 1)</math> as a difference of cubes, leaving us with <math>(5^{167} - 1)</math> times an odd number. <math>(5^{167} - 1)</math> ends in ...<math>24</math>, contributing two powers of two to the final result. | + | We factor the difference of squares, leaving us with <math>(5^{501} - 1)(5^{501} + 1)</math>. We note that for all powers of 5 more than two, it ends in ...<math>25</math>. Thus, <math>(5^{501} + 1)</math> would end in ...<math>26</math> and thus would contribute one power of two to the answer, but not more. |
| + | |||
| + | We can continue to factor <math>(5^{501} - 1)</math> as a difference of cubes, leaving us with <math>(5^{167} - 1)</math> times an odd number. <math>(5^{167} - 1)</math> ends in ...<math>24</math>, contributing two powers of two to the final result. | ||
| + | |||
Adding these extra 3 power of two to the original 1002 factored out, we obtain the final answer of <math>\textbf{(D) } 2^{1005}</math>. | Adding these extra 3 power of two to the original 1002 factored out, we obtain the final answer of <math>\textbf{(D) } 2^{1005}</math>. | ||
Revision as of 16:01, 20 February 2014
Problem 17
What is the greatest power of 
that is a factor of 
?
Solution
We begin by factoring the 
out. This leaves us with 
.
We factor the difference of squares, leaving us with 
. We note that for all powers of 5 more than two, it ends in ...
. Thus, 
would end in ...
and thus would contribute one power of two to the answer, but not more.
We can continue to factor 
as a difference of cubes, leaving us with 
times an odd number. ends in ...
, contributing two powers of two to the final result.
Adding these extra 3 power of two to the original 1002 factored out, we obtain the final answer of 
.
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
