Difference between revisions of "2014 AMC 10B Problems/Problem 17"
(→Solution) |
(→Solution: minor latex $) |
||
Line 12: | Line 12: | ||
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 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 | + | Adding these extra <math>3</math> powers of two to the original <math>1002</math> factored out, we obtain the final answer of <math>\textbf{(D) } 2^{1005}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:02, 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 powers of two to the original
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.