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How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7? Further, $991$ and $997$ are below $1000$ so shouldn't have been removed either This gives $224+2+2=228$ numbers relatively prime to $210$, so $1000-228=772$ numbers are divisible by $2$, $3$, $5$, or $7$
Differential equations : A tank contains 1000 L of brine 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
Why is kg m³ to g cm³1 to 1000? - Mathematics Stack Exchange 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 delete this?
algebra precalculus - Multiple-choice: sum of primes below $1000 . . . 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 $1000$ there are $167$ odd primes and 1 even prime (2), so the sum has to be odd, leaving only the first two numbers
$1000$ small cubes are assembled into a larger cube. If one layer of . . . $1000$ is the number of small cubes in the original cube Each face of the original cube contains $10\times10=100$ small cubes, so the effect of removing the small cubes on all six faces, before allowing for double or triple counting, is a deduction of $600$
How many digits does $2^ {1000}$ contain? - Mathematics Stack Exchange 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 $2^ {1000}$ should have close to, but perhaps a few more, than 300 digits