Difference between revisions of "2014 AMC 10A Problems/Problem 25"
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<cmath>x+y=867</cmath><cmath>2x+3y=2013</cmath> | <cmath>x+y=867</cmath><cmath>2x+3y=2013</cmath> | ||
from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>. | from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
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+ | ==Solution 2== | ||
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+ | The problem is asking for between how many consecutive powers of <math>5</math> are there <math>3</math> power of <math>2</math>s | ||
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+ | There can be either <math>2</math> or <math>3</math> powers of <math>2</math> between any two consecutive powers of <math>5</math>, <math>5^n</math> and <math>5^{n+1}</math>. | ||
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+ | The first power of <math>2</math> is between <math>5^n</math> and <math>2 \cdot 5^n</math>. | ||
+ | |||
+ | The second power of <math>2</math> is between <math>2 \cdot 5^n</math> and <math>4 \cdot 5^n</math>. | ||
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+ | The third power of <math>2</math> is between <math>4 \cdot 5^n</math> and <math>8 \cdot 5^n</math>, meaning that it can be between <math>5^n</math> and <math>5^{n+1}</math> or not. | ||
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+ | If there are only <math>2</math> power of <math>2</math>s between every consecutive powers of <math>5</math> up to <math>5^{867}</math>, there would be <math>867\cdot 2 = 1734</math> power of <math>2</math>s. However, there are <math>2013</math> powers of <math>2</math> before <math>5^{867}</math>, meaning the answer is <math>2013 - 1734 = \boxed{\textbf{(B)}279}</math>. | ||
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+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == | ||
− | https:// | + | https://www.youtube.com/watch?v=DRJvUMsZtl4&t=4s |
== See Also == | == See Also == |
Latest revision as of 14:51, 15 August 2023
- The following problem is from both the 2014 AMC 12A #22 and 2014 AMC 10A #25, so both problems redirect to this page.
Problem
The number is between and . How many pairs of integers are there such that and
Solution 1
Between any two consecutive powers of there are either or powers of (because ). Consider the intervals . We want the number of intervals with powers of .
From the given that , we know that these intervals together have powers of . Let of them have powers of and of them have powers of . Thus we have the system from which we get , so the answer is .
Solution 2
The problem is asking for between how many consecutive powers of are there power of s
There can be either or powers of between any two consecutive powers of , and .
The first power of is between and .
The second power of is between and .
The third power of is between and , meaning that it can be between and or not.
If there are only power of s between every consecutive powers of up to , there would be power of s. However, there are powers of before , meaning the answer is .
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=DRJvUMsZtl4&t=4s
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.