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Difference between revisions of "2003 AIME I Problems/Problem 13"
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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>.
+
In order for there to be more <math>1</math>'s than <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>.
  
 
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>.

Revision as of 16:32, 18 October 2015

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

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 $k+1 > \frac{d+1}{2} \Longrightarrow k > \frac{d-1}{2} \Longrightarrow k \ge \frac{d}{2}$. 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 $\boxed{155}$.

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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