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Difference between revisions of "2003 AIME I Problems/Problem 13"
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Let <math> N </math> be the number of positive integers that are less than or equal to <math>2003</math> and whose base-<math>2</math> representation has more <math>1</math>'s than <math>0</math>'s. Find the [[remainder]] when <math> N </math> is divided by <math>1000</math>.
 
Let <math> N </math> be the number of positive integers that are less than or equal to <math>2003</math> and whose base-<math>2</math> representation has more <math>1</math>'s than <math>0</math>'s. Find the [[remainder]] when <math> N </math> is divided by <math>1000</math>.
  
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== Solution ==
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== Solution 1 ==
 
In base-<math>2</math> representation, all positive numbers have a leftmost digit of <math>1</math>. Thus there are <math>{n \choose k}</math> numbers that have <math>n+1</math> digits in base <math>2</math> notation, with <math>k+1</math> of the digits being <math>1</math>'s.  
 
In base-<math>2</math> representation, all positive numbers have a leftmost digit of <math>1</math>. Thus there are <math>{n \choose k}</math> numbers that have <math>n+1</math> digits in base <math>2</math> notation, with <math>k+1</math> of the digits being <math>1</math>'s.  
  
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In order for there to be more <math>1</math>'s then <math>0</math>'s, we must have <math>k+1 > \frac{d+1}{2} \Longrightarrow k > \frac{d-1}{2} \Longrightarrow k \ge \frac{d}{2}</math>. Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in [[Pascal's Triangle]], from rows <math>0</math> to <math>10</math> (as <math>2003 < 2^{11}-1</math>). Since the sum of the elements of the <math>r</math>th row is <math>2^r</math>, it follows that the sum of all elements in rows <math>0</math> through <math>10</math> is <math>2^0 + 2^1 + \cdots + 2^{10} = 2^{11}-1 = 2047</math>. The center elements are in the form <math>{2i \choose i}</math>, so the sum of these elements is <math>\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351</math>.
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In order for there to be more <math>1</math>'s than <math>0</math>'s, we must have <math>k+1 > \frac{n+1}{2} \implies k > \frac{n-1}{2} \implies k \ge \frac{n}{2}</math>. Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in [[Pascal's Triangle]], from rows <math>0</math> to <math>10</math> (as <math>2003 < 2^{11}-1</math>). Since the sum of the elements of the <math>r</math>th row is <math>2^r</math>, it follows that the sum of all elements in rows <math>0</math> through <math>10</math> is <math>2^0 + 2^1 + \cdots + 2^{10} = 2^{11}-1 = 2047</math>. The center elements are in the form <math>{2i \choose i}</math>, so the sum of these elements is <math>\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351</math>.
  
 
The sum of the elements on or to the right of the line of symmetry is thus <math>\frac{2047 + 351}{2} = 1199</math>. However, we also counted the <math>44</math> numbers from <math>2004</math> to <math>2^{11}-1 = 2047</math>. Indeed, all of these numbers have at least <math>6</math> <math>1</math>'s in their base-<math>2</math> representation, as all of them are greater than <math>1984 = 11111000000_2</math>, which has <math>5</math> <math>1</math>'s. Therefore, our answer is <math>1199 - 44 = 1155</math>, and the remainder is <math>\boxed{155}</math>.
 
The sum of the elements on or to the right of the line of symmetry is thus <math>\frac{2047 + 351}{2} = 1199</math>. However, we also counted the <math>44</math> numbers from <math>2004</math> to <math>2^{11}-1 = 2047</math>. Indeed, all of these numbers have at least <math>6</math> <math>1</math>'s in their base-<math>2</math> representation, as all of them are greater than <math>1984 = 11111000000_2</math>, which has <math>5</math> <math>1</math>'s. Therefore, our answer is <math>1199 - 44 = 1155</math>, and the remainder is <math>\boxed{155}</math>.
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 +
==Solution 2==
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We seek the number of allowed numbers which have <math> k</math> 1's, not including the leading 1, for <math> k=0, 1, 2, \ldots , 10</math>.
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For <math> k=0,\ldots , 4</math>, this number is
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<math> \binom{k}{k}+\binom{k+1}{k}+\cdots+\binom{2k}{k}</math>.
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By the Hockey Stick Identity, this is equal to <math> \binom{2k+1}{k+1}</math>. So we get
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<math> \binom{1}{1}+\binom{3}{2}+\binom{5}{3}+\binom{7}{4}+\binom{9}{5}=175</math>.
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For <math> k=5,\ldots , 10</math>, we end on <math> \binom{10}{k}</math> - we don't want to consider numbers with more than 11 digits. So for each <math> k</math> we get
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<math> \binom{k}{k}+\binom{k+1}{k}+\ldots+\binom{10}{k}=\binom{11}{k+1}</math>
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 +
again by the Hockey Stick Identity. So we get
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<math> \binom{11}{6}+\binom{11}{7}+\binom{11}{8}+\binom{11}{9}+\binom{11}{10}+\binom{11}{11}=\frac{2^{11}}{2}=2^{10}=1024</math>.
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The total is <math> 1024+175=1199</math>. Subtracting out the <math> 44</math> numbers between <math> 2003</math> and <math> 2048</math> gives <math> 1155</math>. Thus the answer is <math> 155</math>.
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==Solution 3==
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We will count the number of it <math>< 2^{11}=2048</math> instead of <math>2003</math> (In other words, the length of the base-2 representation is at most <math>11</math>. If there are even digits, <math>2n</math>, then the leftmost digit is <math>1</math>, the rest, <math>2n-1</math>, has odd number of digits. In order for the base-2 representation to have more <math>1</math>'s, we will need more <math>1</math> in the remaining <math>2n-1</math> than <math>0</math>'s. Using symmetry, this is equal to
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<math>\frac{2^9+2^7+..+2^1}{2}</math>
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Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of <math>1</math>'s at least as the number of <math>0</math>'s. So it's equal to
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<math>\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}</math>
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Summing both cases, we have <math>\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 199</math>. There are <math>44</math> numbers between <math>2004</math> and <math>2047</math> inclusive that satisfy it. So the answer is <math>199-44=\boxed{155}</math>
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~minor edits by minediamonds
  
 
== See also ==
 
== See also ==

Latest revision as of 18:42, 7 October 2023

Problem

Let $N$ be the number of positive integers that are less than or equal to $2003$ and whose base-$2$ representation has more $1$'s than $0$'s. Find the remainder when $N$ is divided by $1000$.

Solution 1

In base-$2$ representation, all positive numbers have a leftmost digit of $1$. Thus there are ${n \choose k}$ numbers that have $n+1$ digits in base $2$ notation, with $k+1$ of the digits being $1$'s.

In order for there to be more $1$'s than $0$'s, we must have $k+1 > \frac{n+1}{2} \implies k > \frac{n-1}{2} \implies k \ge \frac{n}{2}$. Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in Pascal's Triangle, from rows $0$ to $10$ (as $2003 < 2^{11}-1$). Since the sum of the elements of the $r$th row is $2^r$, it follows that the sum of all elements in rows $0$ through $10$ is $2^0 + 2^1 + \cdots + 2^{10} = 2^{11}-1 = 2047$. The center elements are in the form ${2i \choose i}$, so the sum of these elements is $\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351$.

The sum of the elements on or to the right of the line of symmetry is thus $\frac{2047 + 351}{2} = 1199$. However, we also counted the $44$ numbers from $2004$ to $2^{11}-1 = 2047$. Indeed, all of these numbers have at least $6$ $1$'s in their base-$2$ representation, as all of them are greater than $1984 = 11111000000_2$, which has $5$ $1$'s. Therefore, our answer is $1199 - 44 = 1155$, and the remainder is $\boxed{155}$.

Solution 2

We seek the number of allowed numbers which have $k$ 1's, not including the leading 1, for $k=0, 1, 2, \ldots , 10$.

For $k=0,\ldots , 4$, this number is

$\binom{k}{k}+\binom{k+1}{k}+\cdots+\binom{2k}{k}$.

By the Hockey Stick Identity, this is equal to $\binom{2k+1}{k+1}$. So we get

$\binom{1}{1}+\binom{3}{2}+\binom{5}{3}+\binom{7}{4}+\binom{9}{5}=175$.

For $k=5,\ldots , 10$, we end on $\binom{10}{k}$ - we don't want to consider numbers with more than 11 digits. So for each $k$ we get

$\binom{k}{k}+\binom{k+1}{k}+\ldots+\binom{10}{k}=\binom{11}{k+1}$

again by the Hockey Stick Identity. So we get

$\binom{11}{6}+\binom{11}{7}+\binom{11}{8}+\binom{11}{9}+\binom{11}{10}+\binom{11}{11}=\frac{2^{11}}{2}=2^{10}=1024$.

The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.

Solution 3

We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $199-44=\boxed{155}$

~minor edits by minediamonds

See also

2003 AIME I (Problems • Answer Key • Resources)
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Problem 14
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