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 | + | We factor the difference of squares, leaving us with <math>(5^{501} - 1)(5^{501} + 1)</math>. We note that all even powers of 5 more than two end in ...<math>625</math>. Also, all odd powers of five more than 2 end in ...<math>125</math>. Thus, <math>(5^{501} + 1)</math> would end in ...<math>126</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>124</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>124</math>, contributing two powers of two to the final result. |
Revision as of 18:59, 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 all even powers of 5 more than two end in ...
. Also, all odd powers of five more than 2 end 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 |
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