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1 if a number ends with n n zeros than it is divisible by 10n 10 n, that is 2n5n 2 n 5 n What material must i know to solve problems like this with remainders.i know w. 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 s are there in the factorization of 1000

You have a 1/1000 chance of being hit by a bus when crossing the street I found this question asking to find the last two digits of $3^{1000}$ in my professors old notes and review guides However, if you perform the action of crossing the street 1000 times, then your chance of being.

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). In pure math, the correct answer is $ (1000)_2$ Firstly, we have to understand that the leading zeros at any number system has no value likewise decimal One is $ (010)_2$ and another one is $ (010)_ {10}$

Let's work with the $2$ nd number $ (010)_ {10}= (10)_ {10}$ we all agree that the smallest $2$ digit number is $10$ (decimal) 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 like $999^ {1000}$, which swamp your bound by about 3000 orders of magnitude.

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 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

It has units $\mathrm {m}^3$ A liter is liquid amount measurement 1 liter of milk, 1 liter of water, etc Does that mean if i pump $1000$ liters of water they would take exactly $1$ cubic meter of space?

Keep rolling two dice until the cumulative sum hits 1000 ask question asked 2 years, 3 months ago modified 2 years, 3 months ago

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 $1000$ there are $167$ odd primes and 1 even prime (2), so the sum has to be odd, leaving only the first two numbers.

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