Difference between revisions of "2014 AMC 10A Problems/Problem 25"
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{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}} | {{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}} | ||
− | |||
+ | == Problem == | ||
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>. How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath> | The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>. How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath> | ||
+ | |||
<math>\textbf{(A) }278\qquad | <math>\textbf{(A) }278\qquad | ||
\textbf{(B) }279\qquad | \textbf{(B) }279\qquad | ||
Line 9: | Line 10: | ||
\textbf{(E) }282\qquad</math> | \textbf{(E) }282\qquad</math> | ||
− | ==Solution== | + | == Solution 1 == |
+ | Between any two consecutive powers of <math>5</math> there are either <math>2</math> or <math>3</math> powers of <math>2</math> (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 <math>3</math> powers of <math>2</math>. | ||
− | + | 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=867</cmath><cmath>2x+3y=2013</cmath> | ||
+ | from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
− | + | ==Solution 2== | |
− | < | + | |
− | + | 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 | |
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
− | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | |
− | == | + | == Video Solution by Richard Rusczyk == |
+ | https://www.youtube.com/watch?v=DRJvUMsZtl4&t=4s | ||
+ | == See Also == | ||
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}} | {{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}} | ||
− | {{AMC12 box|year=2014|ab=A|num- | + | {{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:51, 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.