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

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== Problem ==
 
== Problem ==
 
Find the positive integer <math>n\,</math> for which
 
Find the positive integer <math>n\,</math> for which
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<center><math>
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<cmath>
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\lfloor \log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994</math></center>.
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\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994
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</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>)
  
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== Solution ==
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== Solution 1 ==
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{{solution}}
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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>.
 +
 
 +
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>.
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 +
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>.
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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>.
 +
 
 +
Alternatively, one could notice this is an [[arithmetico-geometric series]] and avoid a lot of computation.
 +
 
 +
== Solution 2 ==
 +
For this solution, we notice that values between ranges of <math>2^n</math> are the same. Let's look at these ranges and their total value. It is trivial to conclude <math>\log_2{1} = 0</math> so we will not write that.
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 +
\begin{array}{c|c}
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\text{Range of } k & \text{Expression (Total)} \
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\hline
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2 \leq k < 2^2 & 1 \cdot (2^2 - 2) = 2 \
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2^2 \leq k < 2^3 & 2 \cdot (2^3 - 2^2) = 8 \
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2^3 \leq k < 2^4 & 3 \cdot (2^4 - 2^3) = 24 \
 +
2^4 \leq k < 2^5 & 4 \cdot (2^5 - 2^4) = 64 \
 +
2^5 \leq k < 2^6 & 5 \cdot (2^6 - 2^5) = 160 \
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2^6 \leq k < 2^7 & 6 \cdot (2^7 - 2^6) = 384 \
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2^7 \leq k < 2^8 & 7 \cdot (2^8 - 2^7) = 896 \
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\end{array}
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 +
Sum of the values up to <math>1</math> less than <math>2^k</math> is <math>1538</math>. This seems close enough to our desired value of <math>1994</math> as any additional groups would exceed <math>1994</math>. The difference is <math>1996 - 1538 = 456</math>. Since the next numbers of the sequence will always be 8 since <math>\lfloor \log_2{2^8 + n < 2^9} \rfloor = 8</math>, we can just find the number of '8's, which is <math>\frac{456}{8} = 57</math>. So there are exactly 57 integers in that range. The largest integer, which is <math>a</math>, is equal to <math>2^8 + 56 = \boxed{312}</math>. ~Totient4Breakfast
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1994|num-b=3|num-a=5}}
 
{{AIME box|year=1994|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 07:15, 1 April 2025

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 1

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

Alternatively, one could notice this is an arithmetico-geometric series and avoid a lot of computation.

Solution 2

For this solution, we notice that values between ranges of $2^n$ are the same. Let's look at these ranges and their total value. It is trivial to conclude $\log_2{1} = 0$ so we will not write that.

Range of kExpression (Total)2≤k<221⋅(22−2)=222≤k<232⋅(23−22)=823≤k<243⋅(24−23)=2424≤k<254⋅(25−24)=6425≤k<265⋅(26−25)=16026≤k<276⋅(27−26)=38427≤k<287⋅(28−27)=896

Sum of the values up to $1$ less than $2^k$ is $1538$. This seems close enough to our desired value of $1994$ as any additional groups would exceed $1994$. The difference is $1996 - 1538 = 456$. Since the next numbers of the sequence will always be 8 since $\lfloor \log_2{2^8 + n < 2^9} \rfloor = 8$, we can just find the number of '8's, which is $\frac{456}{8} = 57$. So there are exactly 57 integers in that range. The largest integer, which is $a$, is equal to $2^8 + 56 = \boxed{312}$. ~Totient4Breakfast

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
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Problem 5
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