fact

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

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Comments

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 - 7 years, 9 months ago

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

Aritra Nandy - 7 years, 9 months ago

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 - 7 years, 9 months ago

thanx a lot Mr. Mark..!!

Aritra Nandy - 7 years, 9 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:

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 - 7 years, 9 months ago

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

Aritra Nandy - 7 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$