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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!) –  user21436 Feb 13 '12 at 14:29
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For binomial coefficients, you can use \binom{m}{n} which looks like $\binom{m}{n}$. –  Aryabhata Feb 13 '12 at 14:33
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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. –  joriki Feb 13 '12 at 15:33
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@Stom, that's not how this site is supposed to operate. –  lhf Feb 13 '12 at 19:39
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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. –  lhf Feb 14 '12 at 10:43

2 Answers 2

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? –  Tomarinator Feb 13 '12 at 16:35
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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). –  joriki Feb 13 '12 at 16:43
    
I already know the answer, and am looking for alternate solutions, in fact i created the problem,myself –  Tomarinator Feb 13 '12 at 17:21
    
I guarantee that the question is legitimate, and the solution is as relevant as the answer is elegant. –  Tomarinator Feb 13 '12 at 17:40
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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. –  joriki Feb 13 '12 at 23:49
up vote -1 down vote accepted

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, –  Tomarinator Feb 18 '12 at 14:44
    
Did you understand joriki's answer? –  Aryabhata Feb 18 '12 at 18:53
    
I will one day. –  Tomarinator Mar 18 '12 at 18:09

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