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 …

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 …

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 …

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 …

May 28, 2017 · 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 $1000$ …

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 …

Oct 22, 2015 · If a coin is flipped 1000 times, 600 are heads, would you say it's fair? My first thought was to calculate the p-value. Assume it's fair, the probability of getting 600 or more …

Mar 2, 2012 · 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. …

Feb 2, 2016 · A number is chosen at random from the first 1,000 positive integers. What is the probability that it's divisible by 3,5, or 7? So I started off by breaking the problem up and …

Oct 11, 2015 · 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 …