copy and paste this google map to your website or blog!
Press copy button and paste into your blog or website.
(Please switch to 'HTML' mode when posting into your blog. Examples: WordPress Example, Blogger Example)
Input Form:Deal with this potential dealer,buyer,seller,supplier,manufacturer,exporter,importer
(Any information to deal,buy, sell, quote for products or service)
Exactly $1000$ perfect squares between two consecutive cubes 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 17 hours ago community wiki 5 revs R P A reflection on
algebra precalculus - Which is greater: $1000^ {1000}$ or $1001^ {999 . . . 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
algebra precalculus - Multiple-choice: sum of primes below $1000 . . . 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
combinatorics - Determine the number of odd binomial coefficients in . . . 4 Determine the number of odd binomial coefficients in the expansion of $ (x+y)^ {1000}$ Hint: The number of odd coefficients in any finite binomial expansion is a power of $2$ Is there a way to prove this without using something like Lucas's theorem or any other non-trivial result?