2

$$11-1=10 \\ 121-1=120 \\ 1331-1=1330$$

Now it can be seen that the tens digit increases by 1 at each increment of exponent. So, only in case of $11^{10}$ the tens digit is zero and the units digit will be $1$. So, total $2$ zeros at the end. After that at each power increment of $10$ the occurence of 0 will shift until $11^{20}$. So, number of zeroes should be $2$. Because $(5!)!$ is a multiple of 10.

On second thought: Is it like tens place 0 has a cyclicity of 10; hundreth place has 10^2; thousand place 10^3......Now 120! has 28 number of 10s. So, the tens and hundredth place will have zero. But the thousand place will not. So total 3 trailing zeros. Is my understanding ok?

But the answer given is $1$.

jimjim
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archangel89
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    Dig a bit deeper! You noticed that $11^{10}$ ends $..a01$ for some digit $a$. +1 for experimenting this yourself. But what about $11^{20}$, $11^{30}$ et cetera? What about $11^{100}$? – Jyrki Lahtonen Dec 31 '14 at 08:55
  • 11^11 will be similar to 11. And it will be a cycle, I think. – archangel89 Dec 31 '14 at 08:56
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    Yes, you are correct in that the tens digit becomes $0$ for every tenth power, and repeats in a cycle of length ten. Now turn your attention to the hundreds digit. To that end it is convenient that you only look at the powers of the form $11^{10n}$, where you already know that the tens digit is equal to zero. – Jyrki Lahtonen Dec 31 '14 at 08:57
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    See : http://math.stackexchange.com/questions/231350/number-of-consecutive-zeros-at-the-end-of-11100-1 – lab bhattacharjee Dec 31 '14 at 09:04
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    So $11^{10}=25937424601$ ending with $601$. Why is this suffcient information to decide the last three digits of $11^{20}$? – Jyrki Lahtonen Dec 31 '14 at 09:04
  • What is your answer? I have checked with 11^100. I think the remainder obtained when 11^100 is divided by 100 is 01. So, it matches the idea.@Jyrki Lahtonen yes you are right. It is not sufficient. @lab bhattacherjee its ok to use binomial but what is the answer to my question? – archangel89 Dec 31 '14 at 09:12
  • @archangel89, $1+$(due to $10$) the number of zeros in $120!$ – lab bhattacharjee Dec 31 '14 at 09:15
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    Is it like tens place 0 has a cyclicity of 10; hundreth place has 10^2; thousand place 10^3......Now 120! has 28 number of 10s. So, the tens and hundredth place will have zero. But the thousand place will not. So total 3 trailing zeros. Is my understanding ok? – archangel89 Dec 31 '14 at 09:24
  • @ user2345215 can you help me with this question? – archangel89 Dec 31 '14 at 09:29
  • @ lab bhattacharjee Are you sure you are talking about trailing zeroes? – archangel89 Dec 31 '14 at 09:30

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