- 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
- How much zeros has the number $1000!$ at the end?
1 the number of factor 2's between 1-1000 is more than 5's so u must count the number of 5's that exist between 1-1000 can u continue?
- probability - 1 1000 chance of a reaction. If you do the action 1000 . . .
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
- terminology - What do you call numbers such as $100, 200, 500, 1000 . . .
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, 6 months ago
- definition - What is the smallest binary number of $4$ bit? Is it . . .
My approach: Today, my teacher asked me that and I replied $ (1000)_2$ but my teacher said that it will be $ (0000)_2$ If I ask someone what is the smallest decimal value of $2$ digits, everyone will say $10$
- 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 - Probability of winning a prize in a raffle . . .
You'll be surprised The correct probability of winning at least one ticket is around $0 2242$ Assuming exactly one prize is given, your answer of $\frac {1} {160}$ is the probability of winning is correct That is, you go home empty-handed with probability $\frac {159} {160}$ However, $40$ tickets are chosen for prizes, not just one So even if you miss out on a prize the first time, you
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