Difference between revisions of "2014 AMC 10A Problems/Problem 25"
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From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these <math>867</math> intervals together have <math>2013</math> powers of <math>2</math>. Let <math>x</math> of them have <math>2</math> powers of <math>2</math> and <math>y</math> of them have <math>3</math> powers of <math>2</math>. Thus we have the system | From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these <math>867</math> intervals together have <math>2013</math> powers of <math>2</math>. Let <math>x</math> of them have <math>2</math> powers of <math>2</math> and <math>y</math> of them have <math>3</math> powers of <math>2</math>. Thus we have the system | ||
− | <cmath>x+y | + | <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>. | ||
Revision as of 00:29, 16 March 2015
- 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
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 .
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.