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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 If i flip it $1000$ times and it lands heads up for each flip, what is the probability that the coin is unfair, and how do we quantify that if it is unfair 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
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 using the command line utility bc: $ bc sqrt((667^2)^3) 296740963 sqrt((667^2+1)^3-1) 296741963 Cite edited Nov 27 at 22:11 community wiki 5 revs R.P. A reflection. How many integers between $1000$ and $9999$ inclusive consist of (a) distinct odd digits, (b) distinct digits, (c) from the number of integers obtained in (b), how many are odd integers? 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.
If you are taking $0$ as a natural number, then add 3 to the count Since the problem asks for the numbers less than $1000$, subtract off $2$ from the count since $1000$ is divisible by $2$ and $5$ but not $3$. 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?
If a coin is flipped 1000 times, 600 are heads, would you say it's fair Assume it's fair, the probability of getting 600 or more heads will be.5^1. It means 26 million thousands Essentially just take all those values and multiply them by $1000$
So roughly $\$26$ billion in sales. 49 how to solve this problem, i can not figure it out If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9 The sum of these multiples is 23
Find the sum of all the multiples of 3 or 5 below 1000.
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