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
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?