Difference between revisions of "1994 AIME Problems/Problem 4"
m |
I like pie (talk | contribs) (Started solution) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
Find the positive integer <math>n\,</math> for which | Find the positive integer <math>n\,</math> for which | ||
| − | < | + | <cmath> |
| − | \lfloor \log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994</ | + | \lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994 |
| + | </cmath> | ||
(For real <math>x\,</math>, <math>\lfloor x\rfloor\,</math> is the greatest integer <math>\le x.\,</math>) | (For real <math>x\,</math>, <math>\lfloor x\rfloor\,</math> is the greatest integer <math>\le x.\,</math>) | ||
== Solution == | == Solution == | ||
| − | {{solution}} | + | Notice that <math>\lfloor\log_2{a}\rfloor</math> is equal to <math>\log_2{2^{x}}</math>, where <math>2^x</math> is the largest power of <math>2</math> less than or equal to <math>a</math>. Using this fact, the equation can be rewritten as |
| + | <cmath> | ||
| + | \log_2{1}+\log_2{2}+\log_2{2}+\log_2{4}+\cdots+\log_2{n}=1994 | ||
| + | </cmath> | ||
| + | Simplifying the [[logarithm]]s, we get | ||
| + | <cmath> | ||
| + | 0+1+1+2+2+2+2+\ldots+2^m\cdot m=1994 | ||
| + | </cmath> | ||
| + | {{incomplete|solution}} | ||
== See also == | == See also == | ||
{{AIME box|year=1994|num-b=3|num-a=5}} | {{AIME box|year=1994|num-b=3|num-a=5}} | ||
Revision as of 11:54, 26 April 2008
Problem
Find the positive integer 
for which
![\[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\]](http://latex.artofproblemsolving.com/f/d/6/fd6dfef2630f9d364c3d85c8652f289df5a017f7.png)
(For real 
, 
is the greatest integer 
)
Solution
Notice that 
is equal to 
, where 
is the largest power of 
less than or equal to 
. Using this fact, the equation can be rewritten as
![\[\log_2{1}+\log_2{2}+\log_2{2}+\log_2{4}+\cdots+\log_2{n}=1994\]](http://latex.artofproblemsolving.com/3/a/b/3ab032a183e17a601770c627537ced1bd6c5d254.png)
Simplifying the logarithms, we get
![\[0+1+1+2+2+2+2+\ldots+2^m\cdot m=1994\]](http://latex.artofproblemsolving.com/4/d/9/4d94db63b1cb64dca4c3252b2368d16c755096af.png)
Template:Incomplete
See also
| 1994 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
