Difference between revisions of "2014 AMC 10A Problems/Problem 25"
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==Solution== | ==Solution== | ||
− | Between any two powers of 5 there are either 2 or 3 powers of 2 (because <math>2^2<5^1<2^3</math>). Consider the intervals <math>(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})</math>. We want the number of intervals with 3 powers of 2. | + | Between any two consecutive powers of 5 there are either 2 or 3 powers of 2 (because <math>2^2<5^1<2^3</math>). Consider the intervals <math>(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})</math>. We want the number of intervals with 3 powers of 2. |
From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these 867 intervals together have 2013 powers of 2. Let <math>x</math> of them have 2 powers of 2 and <math>y</math> of them have 3 powers of 2. Thus we have the system | From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these 867 intervals together have 2013 powers of 2. Let <math>x</math> of them have 2 powers of 2 and <math>y</math> of them have 3 powers of 2. Thus we have the system |
Revision as of 16:31, 14 April 2014
- 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 5 there are either 2 or 3 powers of 2 (because ). Consider the intervals . We want the number of intervals with 3 powers of 2.
From the given that , we know that these 867 intervals together have 2013 powers of 2. Let of them have 2 powers of 2 and of them have 3 powers of 2. Thus we have the system
\[x+y&=867\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[2x+3y&=2013\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
from which we get , so the answer is .
(Solution by superpi83)
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.