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 …

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?

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 …

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 …

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 …

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^ …

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 …

Aug 5, 2019 · $1000$ 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 …

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 …

Jul 10, 2017 · Just modular exponentiation. Eventually you do have to do some arithmetic. I suspect there is some trick to this problem in $7^ {7^3} \equiv 7^3 \pmod {1000}$ that creates …