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- Finding all triangular numbers less than 1000 that are palindromic
I'm working on a Koshy's elementary number theory exercise and need help finding all triangular numbers less than 1000 that are palindromic The problem says as follows: Find all triangular numbers
- arithmetic - 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?
- combinatorics - How many numbers are there between 99 and 1000, having . . .
First of all, from 99 to 1000, we have 100 to 999, meaning $9*10*10$ since 1 to 9 is 9 numbers We have 900 numbers Then, to get all numbers with at least one $7$ in their digits, we can do: All
- Why is kg m³ to g cm³1 to 1000? - Mathematics Stack Exchange
I understand that changing the divisor multiplies the result by that, but why doesn't changing the numerator cancel that out? I found out somewhere else since posting, is there a way to delete this?
- 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
- 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
- combinatorics - The number of bacteria in a culture is 1000 and this . . .
The number of bacteria in a culture is 1000 and this number increases by 250% every two hours How many bacteria is present after 24 hours?
- $1000$ small cubes are assembled into a larger cube. If one layer of . . .
$1000$ is the number of small cubes in the original cube Each face of the original cube contains $10\times10=100$ small cubes, so the effect of removing the small cubes on all six faces, before allowing for double or triple counting, is a deduction of $600$
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