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We make a 10 by having a 2 and a 5 in the prime factorization. Since we obviously have more 2s than 5s in the prime factorization of 100!, we just need to find to find the number of 5s in the factorization of 100!.
Of the first 100 integers 100/5=20 of then contain a 5 and 100/25 contain a second 5.
Thus there are a total of 20+4=24 0s at the end of 100!.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
We make a 10 by having a 2 and a 5 in the prime factorization. Since we obviously have more 2s than 5s in the prime factorization of 100!, we just need to find to find the number of 5s in the factorization of 100!. Of the first 100 integers 100/5=20 of then contain a 5 and 100/25 contain a second 5. Thus there are a total of 20+4=24 0s at the end of 100!.
24 zeroes as [ 100/5 ] + [ 100/5^2 ] = 24
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using...gif....rite...?basic p and c
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what do you mean?
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Numbers of zeros in factorial is equal to \sum_{i=1}^k \lfloor \frac{n}{5^i} \rfloor where 5^k \leq n
WE HAVE TO FIND THE NUMBER OF 5'S GREATEST INTEGER FUNCTION(100/5) + GREATEST INTEGER FUNCTION(100/5^2 ) SO THER WILL BE 24 ZEROS
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I suggest turning off the "Caps Lock" button. It would make all of us a little happier.
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He initially typed in "29 ZEROS",and thought that others were wrong and he was right. So that is why the use of caps. Modified his answer later.