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how many 0,s are there in 100 ! ? how can i solve it??

Note by Aritra Nandy
3 years, 5 months ago

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The last \(24\) digits are \(0\), since the exponent of \(5\) in \(100!\) is \(24\). There are \(30\) occurrences of \(0\) altogether. I found the other \(6\) using Mathematica. Mark Hennings · 3 years, 5 months ago

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@Mark Hennings i shall be highly obliged if u plzz show the process i m not getting it! Aritra Nandy · 3 years, 5 months ago

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There are \(20\) numbers between \(1\) and \(100\) which are multiples of \(5\) and \(4\) numbers between \(1\) and \(100\) which are multiples of \(5^2\). There are no multiples of \(5^3\) between \(1\) and \(100\). Thus, when we calculate \(100!\), it will have \(20+4=24\) factors of \(5\). We say that the exponent of \(5\) in \(100!\) is \(24\). Factors of \(2\) are much more common, so that \(100!\) has at least \(24\) factors of \(2\). The exponent of \(2\) in \(100!\) is at least \(24\). In fact, it is \(97\). The number of zeros at the end of \(100!\) is the number of times that \(100!\) can be divided by \(10\). This is the smaller of the exponents of \(2\) and \(5\) in \(100!\), which is \(24\). Thus \(100!\) ends in \(24\) zeros.

Mathematica tells me that \(100!\) is equal to \[ \begin{array}{l} 9332621544394415268169923885626670049071596826438162146859296389521759999322991 \\ 5608941463976156518286253697920827223758251185210916864000000000000000000000000 \end{array} \] which contains \(30\) zeros, including the \(24\) ones at the end. Mark Hennings · 3 years, 4 months ago

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@Mark Hennings thanx a lot Mr. Mark..!! Aritra Nandy · 3 years, 4 months ago

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What base are you working on? In base \(100! - 1\), we have \((100_{10})! = 11_{(100_{10})! - 1}\), and so no zero. In base \(2\), by using Wolfram|Alpha I get this:

11011001100001001011001001110110000111001010111011
10000100100000001101001010100101000110101010100101
11011110110100100000011010001011011101001011001101
11011111001101001110000111010110010000110110101101
10010100101000011101000110010000111001101111100010
00000111001000101110100010101010111000011001100101
01001010000100000110001101110110010110011101101110
01011101101001011101110100010110000001011101010001
00111001101011100011000011010000000000000000000000
00000000000000000000000000000000000000000000000000
0000000000000000000000000

And because I'm too lazy counting the zeroes, plugging another query to Wolfram|Alpha gives that there are 318 zeroes in \((100_{10})!_2\).

So, yeah, your question is incomplete.

In general, counting the number of zeroes (or any digit) of a factorial in any sufficiently small base is difficult; use some computer to do that. However, counting the number of zeroes at the end of the number of a factorial is easy. Mark H. has shown the method to do it; just generalize to other factorials and bases.

EDIT: Not to mention that your \(100\) is of unknown base either. Just realized this. Ivan Koswara · 3 years, 4 months ago

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@Ivan Koswara no base was mentioned in the paper from where i took up this sum! Aritra Nandy · 3 years, 4 months ago

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