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For a quick back-of-the-envelope computation, you can note that $2^ {10}$ is only a little larger than $10^3$, so $2^ {1000} = (2^ {10})^ {100}$ is larger than $10^ {300}$, though not by much; so $2^
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A big part of this problem is that the "1 in 1000" event can happen multiple times within our attempt. Compare this to if you have a special deck of playing cards with 1000 cards in it, exactly
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Given that there are $168$ primes below $1000$. Then the sum of all primes below 1000 is (a) $11555$ (b) $76127$ (c) $57298$ (d) $81722$ My attempt to solve it: We know that below
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What does the sum of all the numerals from the numbers from $100$ up to $1000$ equal to?
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First of all, from 99 to 1000, we have 100 to 999, meaning $9*10*10$ since 1 to 9 is 9 numbers. We have 900 numbers. Then, to get all numbers with at least one $7$ in their digits, we
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1 the number of factor 2''s between 1-1000 is more than 5''s.so u must count the number of 5''s that exist between 1-1000.can u continue?
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An integer is chosen at random from the first 1000 positive integers. Probability that is a multiple of both 6 and 8? Ask Question Asked 11 years, 7 months ago Modified 7 years, 7 months ago
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I understand that changing the divisor multiplies the result by that, but why doesn''t changing the numerator cancel that out? I found out somewhere else since posting, is there a way to
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The number of bacteria in a culture is 1000 and this number increases by 250% every two hours. How many bacteria is present after 24 hours?
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I would like to find all the expressions that can be created using nothing but arithmetic operators, exactly eight $8$''s, and parentheses. Here are the seven solutions I''ve found (on the Internet)...
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