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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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    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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    I will one day.2012-03-18
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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    @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