Jun 9, 2020 · Notes You can prove the claim by: multiplying throughout by $ x ^ \frac {2} {3}$, shifting terms, then cubing to get rid of cube roots. We started with $ x = 2$ because the $ x = 1 $ term had …

It doesnt let me make it look correct, it's supposed to be log 1000 of 1001 and log 999 of 1000

Feb 20, 2021 · Basically, what is the difference between $1000\\times1.03$ and $1000/.97$? For some reason I feel like both should result in the same number. I only ask because I'm working a problem …

Essentially, $1000/1024$ is the average number (or "expected" number) of coins that will have come up all heads, but that includes the cases where more than one coin comes up heads all the time, so it …

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 …

For the congruence modulo $4$ you don't even need to invoke Euler's Theorem; you can just note that since $2^2\equiv 0\pmod {4}$, then $2^ {1000}\equiv 0 \pmod {4}$.

Jan 1, 2014 · If we want the last two digits, we note that $\phi (1000)=400$. So $$ 9999 = 9600 + 399$$ So $$ 7^ {9999} \equiv 7^ {399} \mod 1000 $$ Since $399$ is 1 less than $400$ we can calculate the …

Mar 7, 2014 · 3 I'm being asked to compute $7^ {1000} \mod 24$. I have Fermat's Little Theorem and Euler's Theorem. How do I use these to compute $7^ {1000} \mod 24$? I'm stuck because $24$ is …

Oct 10, 2019 · Often in calculating probabilities, it is sometimes easier to calculate the probability of the 'opposite', the technical term being the complement. Because if something happens with probability p …

well, do you know how to compare $n^ {-10}$ and $ (n+1)^ {-10}$? can you see how small $2^ {-10}$ is compared to $10^ {-3}$?