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Find the last digit of this number:

$$({}_{4n+1} C_0 )^{4m+1} + ({}_{4n+1} C_1 )^{4m+1} +({}_{4n+1} C_2 )^{4m+1} + \cdots + ({}_{4n+1} C_{4n+1} )^{4m+1}\;,$$

where $n$, $m$ belong to the holy set of natural numbers.

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    The question as it stands is a bit hard to fix. Please write the equation in $\TeX$ properly. (You seem to know $\TeX$ and you can best fix this!)2012-02-13
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    For binomial coefficients, you can use `\binom{m}{n}` which looks like $\binom{m}{n}$.2012-02-13
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    "God created the integers". They sure are holy :-)2012-02-13
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    You can also use `{n \choose k}` for ${n \choose k}$.2012-02-13
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    @kanappan samnath i dont know tex, you are free to edit the question the way you like it to be,2012-02-13
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    For those who don't know, I created this question on our own, its one of my most celebrated creations,,,,,2012-02-13
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    @Aryabhata: funny, that's what I was telling him in the chatroom...2012-02-13
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    @J.M. I did exactly as you told me to do, and it worked too, it, the expression looks too fine to me, so whats the fuss all about2012-02-13
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    I did say that most of us here really prefer $\binom{n}{k}$ (`\binom{n}{k}`) as notation, Stom...2012-02-13
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    @Stom: What you call "the fuss" was people trying to help you make a very badly formatted question readable so that people might be able to answer it, which I presume was your intention in asking it. It seems inappropriate to criticize them for this attempt at assistance. I've now cleaned up the formatting in the question. In case you intend to ask more questions here in the future, it might be a good idea to take a look at the edits (by clicking on the "edited ... ago" link under the question) so you can do it yourself next time.2012-02-13
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    @joriki thanks for the edit.2012-02-13
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    @lhf if i tell what i have worked then the question is solved, i look for alternate solutions2012-02-13
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    @Stom, that's not how this site is supposed to operate.2012-02-13
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    But I am looking for alternate solutions, is that a bad thing?2012-02-14
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    @Stom, no, looking for alternative solutions is fine. Just be open about it. And disclosing your own proof will avoid duplication. It's perfectly ok to just add you own proof as an answer.2012-02-14
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    One more question : if i get the question deleted would get my points back which i lost in the process downvoting of this question?2012-02-20

2 Answers 2

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Here is my approach,

It is a fact that for any natural number n, n^4k+1 , has the same unit digit as n, itself, (for any natural number k)

so

in the series we just vanish all the powers of each terms, as we only have to find the unit digit,

and doing so gives us just the series representing sum of binomial coefficients of the series (1+x)^4n+1

whose sum is 2^4n+1

and last digit of 2^4n+1 is 2 itself, (declared earlier)

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    But i firmly believe that a more better solution , from number theory exists for this problem,2012-02-18
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    Did you understand joriki's answer?2012-02-18
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    I will one day.2012-03-18
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The ingredients you need to solve this are Euler's theorem (along with the value of Euler's totient function for the base of our decimal system) and the binomial theorem (applied to a power of $1+1$), or alternatively the fact that the total number of subsets of a $k$-element set is $2^k$.

$$$$

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    OK so i know now the ingredients to solve this, may i also have the recipe?2012-02-13
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    @Stom: Questions in the imperative without any indication why you're asking or what you've tried often don't get fully worked out answers here. In the present case, lhf asked you what you've tried and you haven't responded (yet).2012-02-13
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    I already know the answer, and am looking for alternate solutions, in fact i created the problem,myself2012-02-13
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    I guarantee that the question is legitimate, and the solution is as relevant as the answer is elegant.2012-02-13
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    @Stom: I find it rather bad style to pose a problem without mentioning the fact that you already know the answer. I for one am not going to put any more time into this question.2012-02-13