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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
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
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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49 How to solve this problem, I can not figure it out: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of
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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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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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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A tank contains 1000 L of brine (that is, salt water) with 15 kg of dissolved salt. Pure water enters the top of the tank at a constant rate of 10 L / min. The solution is thoroughly mixed and
A hypothetical example: You have a 1/1000 chance of being hit by a bus when crossing the street. However, if you perform the action of crossing the street 1000 times, then your chance of being
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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