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, 2 months ago

Log in to reply

@Mark Hennings
–
i shall be highly obliged if u plzz show the process i m not getting it!
–
Aritra Nandy
·
3 years, 2 months ago

Log in to reply

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, 2 months ago

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:

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, 2 months ago

Log in to reply

@Ivan Koswara
–
no base was mentioned in the paper from where i took up this sum!
–
Aritra Nandy
·
3 years, 2 months ago

## Comments

Sort by:

TopNewestThe 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, 2 months ago

Log in to reply

– Aritra Nandy · 3 years, 2 months ago

i shall be highly obliged if u plzz show the process i m not getting it!Log in to reply

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, 2 months ago

Log in to reply

– Aritra Nandy · 3 years, 2 months ago

thanx a lot Mr. Mark..!!Log in to reply

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:

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 numberof 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, 2 months ago

Log in to reply

– Aritra Nandy · 3 years, 2 months ago

no base was mentioned in the paper from where i took up this sum!Log in to reply