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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
Exactly $1000$ perfect squares between two consecutive cubes Since $1000$ is $1$ mod $3$, we can indeed write it in this form, and indeed $m=667$ works Therefore there are exactly $1000$ squares between the successive cubes $ (667^2)^3$ and $ (667^2+1)^3$, or between $444889^3$ and $444890^3$
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
What is mathematical basis for the percent symbol (%)? Percent means 1 part of 100 or 1 100 and is indicated with % Per mille means 1 part of 1000 or 1 1000 and is indicated with ‰, so it seems that these symbols indicate the mathematical operations