Difference between revisions of "2014 AMC 10B Problems/Problem 17"
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<math>\textbf{(A) } 2^{1002} \qquad\textbf{(B) } 2^{1003} \qquad\textbf{(C) } 2^{1004} \qquad\textbf{(D) } 2^{1005} \qquad\textbf{(E) }2^{1006}</math> | <math>\textbf{(A) } 2^{1002} \qquad\textbf{(B) } 2^{1003} \qquad\textbf{(C) } 2^{1004} \qquad\textbf{(D) } 2^{1005} \qquad\textbf{(E) }2^{1006}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | ||
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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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | First, we can write the expression in a more primitive form which will allow us to start factoring. | ||
+ | <cmath>10^{1002} - 4^{501} = 2^{1002} \cdot 5^{1002} - 2^{1002}</cmath> | ||
+ | Now, we can factor out <math>2^{1002}</math>. This leaves us with <math>5^{1002} - 1</math>. Call this number <math>N</math> Thus, our final answer will be <math>2^{1002+k}</math>, where <math>k</math> is the largest power of <math>2</math> that divides <math>N</math>. Now we can consider <math>N \pmod{16}</math>, since <math>k \le 4</math> by the answer choices. | ||
+ | |||
+ | Note that | ||
+ | <cmath> | ||
==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 02:06, 23 February 2014
Contents
[hide]Problem 17
What is the greatest power of that is a factor of ?
Solution 1
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 .
Solution 2
First, we can write the expression in a more primitive form which will allow us to start factoring. Now, we can factor out . This leaves us with . Call this number Thus, our final answer will be , where is the largest power of that divides . Now we can consider , since by the answer choices.
Note that
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.