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.
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) 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 The way you're getting your bounds isn't a useful way to do things
On the other end of the binomial expansion, you have terms like $999^ {1000}$, which swamp your bound by about 3000 orders of magnitude. 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?
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.
