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 ...
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)...
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 of being ...
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 consider $2$ numbers. 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). Can't we say $010 ...
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 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?
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
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.
