Oct 19, 2025 · Therefore there are exactly $1000$ squares between the successive cubes $ (667^2)^3$ and $ (667^2+1)^3$, or between $444889^3$ and $444890^3$. Finally, we can verify all of this by …

Jul 17, 2019 · 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 …

May 13, 2014 · 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?

May 12, 2017 · Hint $\ $ Examining their factorizations for small $\rm\,N\,$ shows that the power of $3$ dividing the former exceeds that of the latter (by $2),$ so the former cannot divide the latter. It …

Jan 30, 2017 · 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 …

Aug 23, 2015 · I'm trying to solve the following exercise: $$2017^ {2016^ {2015}} \mod 1000,$$ here's what I've already come up with: Using Euler's conrgruence, one finds that $$2017^ {2016^ {2015}} …

Oct 31, 2017 · It means "26 million thousands". Essentially just take all those values and multiply them by $1000$. So roughly $\$26$ billion in sales.

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 ...

Jan 12, 2018 · If I pay you $1000$ now and $2000$ a year from now, then the total present value is $1000$ plus $2000/1.1$. Why would you then go back and add another $1000$ from nowhere?

Question: Express the function $\frac {n^3} {1000} - 100n^2 - 100n + 3$ in terms of the Θ notation and prove that your expression in fact fits into the Θ definition.