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| − | [https://artofproblemsolving.com/community/my-aops] | + | [https://artofproblemsolving.com/community/my-aops OliverA's profile] |
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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} = 1199</math>. There are <math>44</math> numbers between <math>2004</math> and <math>2047</math> inclusive that satisfy it. So the answer is <math>1199-44=1\boxed{155}</math>
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