# Difference between revisions of "OliverA"

Line 1: | Line 1: | ||

[https://artofproblemsolving.com/community/my-aops] | [https://artofproblemsolving.com/community/my-aops] | ||

− | |||

− | |||

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 | 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 |

## Revision as of 11:21, 26 June 2020

We will count the number of it $\< 2^{11}=2048$ (Error compiling LaTeX. ! Undefined control sequence.) instead of (In other words, the length of the base-2 representation is at most . If there are even digits, , then the leftmost digit is , the rest, , has odd number of digits. In order for the base-2 representation to have more 's, we will need more in the remaining than 's. Using symmetry, this is equal to Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of 's at least as the number of 's. So it's equal to Summing both cases, we have . There are numbers between and inclusive that satisfy it. So the answer is