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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 - 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
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
Solution Verification: How many positive integers less than $1000$ have . . . A positive integer less than $1000$ has a unique representation as a $3$-digit number padded with leading zeros, if needed To avoid a digit of $9$, you have $9$ choices for each of the $3$ digits, but you don't want all zeros, so the excluded set has count $9^3 - 1 = 728$
How many numbers from $1$ to $1000$ can be written as the sum of $4$s . . . I like the numbers $4$ and $5$ I also like any number that can be added together using $4$ s and $5$ s Eg, $$9 = 4+5 \qquad 40 = 5 + 5 + 5 + 5 + 5 + 5 + 5 +5$$ How many number have this property from 1 to 1000? Multiples of $4$ s and $5$ s are easy, but how do I calculate the number of numbers from different combinations of adding $4$ and $5$? (And which ones are different from multiples of