Difference between revisions of "2014 AMC 10B Problems/Problem 17"
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==Solution 3== | ==Solution 3== | ||
Convert <math>4^{501}=2^{1002}</math>. We can factor out <math>2^{1002}</math> to get that <math>\nu_2(10^{1002}-2^{1002})=1002+\nu_2(5^{1002}-1)</math>. Using the adjusted Lifting The Exponent lemma (<math>\nu_2(a^n-b^n)=\nu_2(n)+\nu_2(a^2-b^2)-1</math> for all even <math>n</math> and odd <math>a,b</math>), we get that the answer is <math>2^{1002+\nu_2(1002)+\nu_2(24)-1}=2^{1002+1+3-1}=\boxed{\textbf{(D)}2^{1005}}</math> | Convert <math>4^{501}=2^{1002}</math>. We can factor out <math>2^{1002}</math> to get that <math>\nu_2(10^{1002}-2^{1002})=1002+\nu_2(5^{1002}-1)</math>. Using the adjusted Lifting The Exponent lemma (<math>\nu_2(a^n-b^n)=\nu_2(n)+\nu_2(a^2-b^2)-1</math> for all even <math>n</math> and odd <math>a,b</math>), we get that the answer is <math>2^{1002+\nu_2(1002)+\nu_2(24)-1}=2^{1002+1+3-1}=\boxed{\textbf{(D)}2^{1005}}</math> | ||
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+ | ==Solution 4== | ||
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+ | Factor out <math>2^{1002}</math> to get <math>2^{1002}(5^{1002} - 1)</math>. Since <math>5^{1002}-1\equiv 3^{1002}-1\equiv (9)^{501}-1\equiv 1^{501}-1\equiv 0\pmod{8}</math>, but <math>5^{1002}-1\equiv 11^{1002}-1\equiv 121\cdot (11^4)^{250} - 1\equiv 121 - 1\equiv 8 \pmod{16}</math>, <math>5^{1002}-1</math> has 3 factors of 2. Hence <math>2^{1002 + 3} =\boxed{2^{1005}}</math> is the largest power of two which divides the given number | ||
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/aCtvD8nitgg | https://youtu.be/aCtvD8nitgg |
Revision as of 21:18, 3 January 2022
Problem
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 more than two end in .... Also, all odd powers of five more than 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 (Notice that the other number is . The powers of end in , so the two powers of will end with . Adding will make it end in . Thus, this is an odd number). ends in ..., contributing two powers of two to the final result.
Or we can see that ends in , and is divisible by only. Still that's powers of .
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 The powers of cycle in with a period of . Thus, This means that is divisible by but not , so and our answer is .
Solution 3
Convert . We can factor out to get that . Using the adjusted Lifting The Exponent lemma ( for all even and odd ), we get that the answer is
Solution 4
Factor out to get . Since , but , has 3 factors of 2. Hence is the largest power of two which divides the given number
Video Solution
~savannahsolver
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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