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?

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)...

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 …

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 …

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 ...

Oct 3, 2023 · 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?

Feb 5, 2017 · Compute 3^1000 (mod13) Ask Question Asked 9 years, 1 month ago Modified 9 years, 1 month ago

Feb 24, 2023 · Question Statement An investor has $\$20,000$ to be invested amongst $4$ possible investments. Each investment must be a unit of $\$1,000$. If all the money needs to be invested …

Jun 27, 2018 · 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 …

Oct 23, 2016 · The exponent of 13 on the factorisation of $1000!$ is $\lfloor\frac {1000} {13}\rfloor+\lfloor\frac {1000} {13^2}\rfloor$ do the same for $326!$ and $674!$ and you'll find that …