# fact

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

Note by Aritra Nandy
4 years, 11 months ago

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## Comments

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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.

- 4 years, 11 months ago

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

- 4 years, 11 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.

- 4 years, 11 months ago

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thanx a lot Mr. Mark..!!

- 4 years, 11 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.

- 4 years, 11 months ago

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

- 4 years, 10 months ago

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