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How much zeros has the number $1000!$ at the end? yes it depends on $2$ and $5$ Note that there are plenty of even numbers Also note that $25\times 4 = 100$ which gives two zeros Also note that there $125\times 8 = 1000$ gives three zeroes and $5^4 \times 2^4 = 10^4$ Each power of $5$ add one extra zero So, count the multiple of $5$ and it's power less than $1000$
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 $9^3 - 1 = 728$ Hence the count you want is $999 - 728 = 271$
$1000$ 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 $600$
algebra precalculus - Partitions using only powers of two on $1000 . . . How many ways are there to write $1000$ as a sum of powers of $2,$ ($2^0$ counts), where each power of two can be used a maximum of $3$ times Furthermore, $1+2+4+4$ is the same as $4+2+4+1$ These count as one arrangement, not two separate ones
What does it mean when something says (in thousands) I'm doing a research report, and I need to determine a companies assets So I found their annual report online, and for the assets, it says (in thousands) One of the rows is: Net sales $ 26,234