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?

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, …

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 11, 2017 · Alternate Method: We want to count the number of times the digit $5$ appears in the list of positive integers from $1$ to $1000$.

Sep 29, 2024 · In pure math, the correct answer is $ (1000)_2$. Here's why. Firstly, we have to understand that the leading zeros at any number system has no value likewise decimal. Let's …

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

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 $2^ {1000}\equiv 0 \pmod {4}$.

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

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 …

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 …