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Difference between revisions of "1994 AIME Problems/Problem 4"

(Started solution)
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== Solution ==
 
== Solution ==
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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
+
Note that if <math>2^x \le a<2^{x+1}</math> for some <math>x\in\mathbb{Z}</math>, then <math>\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x</math>.
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<cmath>
 
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\log_2{1}+\log_2{2}+\log_2{2}+\log_2{4}+\cdots+\log_2{n}=1994
 
−
</cmath>
 
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Simplifying the [[logarithm]]s, we get
 
−
<cmath>
 
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0+1+1+2+2+2+2+\ldots+2^m\cdot m=1994
 
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</cmath>
 
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{{incomplete|solution}}
 
  
 +
Thus, there are <math>2^{x+1}-2^{x}=2^{x}</math> integers <math>a</math> such that <math>\lfloor\log_2{a}\rfloor=x</math>. So the sum of <math>\lfloor\log_2{a}\rfloor</math> for all such <math>a</math> is <math>x\cdot2^x</math>.
 +
 +
Let <math>k</math> be the integer such that <math>2^k \le n<2^{k+1}</math>. So for each integer <math>j<k</math>, there are <math>2^j</math> integers <math>a\le n</math> such that <math>\lfloor\log_2{a}\rfloor=j</math>, and there are <math>n-2^k+1</math> such integers such that <math>\lfloor\log_2{a}\rfloor=k</math>.
 +
 +
Therefore, <math>\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994</math>.
 +
 +
Through computation: <math>\sum_{j=0}^{7}(j\cdot2^j)=1538<1994</math> and <math>\sum_{j=0}^{8}(j\cdot2^j)=3586>1994</math>. Thus, <math>k=8</math>.
 +
 +
So, <math>\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}</math>.
 +
 
== 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 20:52, 22 July 2008

Problem

Find the positive integer $n\,$ for which \[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\] (For real $x\,$, $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$)

Solution

Note that if $2^x \le a<2^{x+1}$ for some $x\in\mathbb{Z}$, then $\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x$.

Thus, there are $2^{x+1}-2^{x}=2^{x}$ integers $a$ such that $\lfloor\log_2{a}\rfloor=x$. So the sum of $\lfloor\log_2{a}\rfloor$ for all such $a$ is $x\cdot2^x$.

Let $k$ be the integer such that $2^k \le n<2^{k+1}$. So for each integer $j<k$, there are $2^j$ integers $a\le n$ such that $\lfloor\log_2{a}\rfloor=j$, and there are $n-2^k+1$ such integers such that $\lfloor\log_2{a}\rfloor=k$.

Therefore, $\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994$.

Through computation: $\sum_{j=0}^{7}(j\cdot2^j)=1538<1994$ and $\sum_{j=0}^{8}(j\cdot2^j)=3586>1994$. Thus, $k=8$.

So, $\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}$.

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
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