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> | + | 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>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> | + | 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. |
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>. | 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>. |
Revision as of 17:58, 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 |
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