If $a^{7!} +b^{8!} +c^{9!} +d^{10!} =x$ where a,b,c and d are natural numbers that are not multiples of 10, the..... If $a^{7!} +b^{8!} +c^{9!} +d^{10!} =x$ where a,b,c and d are natural numbers that are not multiples of 10, then how many distinct values of unit's digit of x are possible ? 
How to proceed in such question, I don't have any idea on this. .. please guide thanks a lot ...
 A: Notice $4$ divides all of $7!, 8!, 9!,10!$.
Euler's theorem says that if $\gcd(k,10) = 1$ then $k^4 \equiv 1 \pmod {10}$ so the last digit of $k^{v!}$ is $1$ if $k$ is an odd number that doesn't end with $5$.
If you don't know Euler's theorem or modular arithmetic notice that if $k = 10w + v$ where $v=\pm 1, \pm 3$ then $k^4 = 10^4w^4 + 4*10^3w^3v + 6*10^2w^2v^2 + 4*10w*v^3 + v^4$ so the last digit if $k^4$ is the same as the last digit of $v^4$ and $v^4 =1$ or $v^4 = 81$.  So the last digit is one.  So $k^{4m}$ will have the last digit of $1$ if $k$ ends with $1, 3, 10-3=7, $ or $10-1=9$.
The other things to worry about are if $k$ is even or $k$ ends in $5$.
If $k$ is ends with $5$ then $k^m$ ends with $5$ (that's obvious isn't it?)
And if $k$ is even... well by the chinese remainder theorem $k^w\equiv 0 \pmod 2$ and $k^4\equiv 1 \pmod 5$ but Euler Th so $k^4 \equiv 6\pmod {10}$.
If you don't know Euler's th or CRT... well,... if $k =2j$ is even then $k^4 = 2^4 j^4 = 16*j^4$.  If $j$ is odd and not a multiple of $5$ then $j^4$ ends with $1$ and $k^4$ ends with $6$.  If $j$ is even just repeat: $j = 2l$ and $k^4 = 16*16*l^4$ and that ends with $6$ if $l$ is odd and if $l$ is even repeat as often as necessary.
So you have the last digits are $1, 5$ or $6$.
So we can have $4$ ones and end with $4$.
We can have $3$ ones and a five and end with $8$
We can have $3$ ones and a six and end with $9$.
We can have $2$ ones and $2$ fives and end with $2$ and .... so on.
Edit:  an improved way of enumerating the possibile sums $\bmod 10$
Once you render the units digit of each term as $\in\{1,5,6\}$, you can consider the sum separately $\bmod 5$ and $\bmod 2$.
In $\bmod 5$ the residue matches the number of $1$s plus the number of $6$s in the sum.  This can be any of $0,1,2,3$ or $4$.
In $\bmod 2$ you can have either residue $0$ or $1$ by swapping a $1$ for a $6$ or vice versa, provided that at least one of these is included.  But that requires an overall residue of $1,2,3$, or $4\bmod 5$ from the above.  The combination with residue $0\bmod 5$, $5+5+5+5$, does not allow this swapping and thus forces a residue of $0\bmod 2$.
So the can all combinations of resudues $\bmod 2$ and $\bmod 5$ except $(1\bmod 2, 0\bmod 5)$ allowing to all units digits except $5$.
