Difference between revisions of "OliverA"

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<center><u><h1 style="color:black;font-size:200px;font style="font-family:Brush Script MT,courier,arial,helvetica;">[https://artofproblemsolving.com/community/user/350017 '''OliverA's profile''']</h1></u></center>
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
(I invented this first, OK? :P)
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
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>

Latest revision as of 23:45, 7 October 2020

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(I invented this first, OK? :P)

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