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?

Jun 11, 2023 · First of all, from 99 to 1000, we have 100 to 999, meaning $9*10*10$ since 1 to 9 is 9 numbers. We have 900 numbers. Then, to get all numbers with at least one $7$ in their digits, we …

Oct 28, 2017 · I found this question asking to find the last two digits of $3^{1000}$ in my professors old notes and review guides. What material must I know to solve problems like this with remainders.I …

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 ...

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 suffices to prove …

What do you call numbers such as $100, 200, 500, 1000, 10000, 50000$ as opposed to $370, 14, 4500, 59000$ Ask Question Asked 14 years, 1 month ago Modified 9 years, 9 months 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 …

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 …

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 …