May 13, 2014 · 1 If a number ends with n n zeros than it is divisible by 10n 10 n, that is 2n5n 2 n 5 n. A factorial clearly has more 2 2 s than 5 5 s in its factorization so you only need to count how many 5 5 …

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 …

The way you're getting your bounds isn't a useful way to do things. You've picked the two very smallest terms of the expression to add together; on the other end of the binomial expansion, you have terms …

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 ...

Mar 7, 2015 · 0 Can anyone explain why $1\ \mathrm {m}^3$ is $1000$ liters? I just don't get it. 1 cubic meter is $1\times 1\times1$ meter. A cube. It has units $\mathrm {m}^3$. A liter is liquid amount …

What do you call numbers such as $100, 200, 500, 1000, 10000, 50000$ as opposed to $370, 14, 4500, 59000$ Ask Question Asked 13 years, 11 months ago Modified 9 years, 7 months ago

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 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 …

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 …

Aug 25, 2023 · Keep rolling two dice until the cumulative sum hits 1000 Ask Question Asked 2 years, 3 months ago Modified 2 years, 3 months ago