Find all $n$ natural numbers such that $10\mid n^{10}+1$ Determine all natural numbers $n$ such that : 
$10$ divisor of $n^{10}+1$ 
My attempt : 
Let $n=r(\mod{10})$ so $n^{10}+1=(r^{10}+1)(\mod{10})$ 
This mean : $r^{10}+1=0(\mod{10})$ 
Now $r\in {0,1,2,3,4,5,6,7,8,9}$ after try I get $r=3,7$ 
So : $n=10k+3,10k+7$ 
Is my work correct ? 
Please I need other simple method to computing 
 A: You are correct but a few things to make you calculations much fewer.
1) if $n$ and $10$ are not relatively prime then any common divisor $a;a\ne 1$ will not divide $n^{10}+1$ so $10\not  \mid n^{10}+1$.
So we need only test $1,3,7,9$
2) $(10-i)^{10}\equiv i^{10}\pmod{10}$ so we only check $1,3$.
3) Eulers Thereom says as $\phi(10) =4$ that for $n$ relatively prime to $10$ that $n^4\equiv 1 \pmod {10}$ so $n^{10}\equiv n^2 \pmod {10}$
So we just need to check whether $k^2 + 1\equiv 0 \pmod {10}$ for $k= 1,3$.
We get $k=3$.
and so our solutions are $n\equiv \pm 3 \pmod {10}$ 
or we could say $n\equiv 3,7 \pmod {10}$
or we could say $n = 10k +3$ or $n=10k +7$ for some integer $7$
or we could say "$n$ is any number with the last digit of $3$ or $7$".
A: By Fermat's theorem, $n^5 \equiv n \bmod 5$. Therefore, $n^{10} \equiv n^2 \bmod 5$ and so $n^{10} +1 \equiv n^2 +1 \bmod 5$. Thus, if $10\mid n^{10}+1$, then $n^2 \equiv 4 \bmod 5$ and so $n \equiv \pm 2 \equiv 2,3 \bmod 5$. Thus, $n \equiv 2,3,7,8 \bmod 10$. Since $n$ is odd, we're left with $n \equiv 3,7 \bmod 10$.
Alternatively,
By Euler's theorem, $n^4 \equiv 1 \bmod 10$, and so $n^{10} \equiv n^2 
\bmod 10$. Thus, if $10\mid n^{10}+1$, then $n^2 \equiv 9 \bmod 10$. Therefore, $n \equiv 3,7 \bmod 10$.
A: Yes, your thoughts are correct.
Maybe you need to be more precice here and there. For example you do not tell us what $k$ is. Also you should not say "Let $n=r\mod 10$". $n$ is an arbitrary natural number, that we try to find. 
Basically you need to find the numbers which have the final digit $9$ when they are raised to the power of 10. 
These numbers are of the form you suggested and it is enough to test the numbers $0,\dotso, 9$ as you did.
A: The only digits such that one power $x$ are such that $m^x+1$ is divisible by $10$ are easily deduced to be the $3^{4m+2},\quad 7^{4m+2},\quad 9^{4m+1},\quad 9^{4m+3}$ and because of the exponent is fixed equal to $10$ the only possibilities are these you have correctly given.
A: Clearly $(n,10)=1$
$\implies n\equiv\pm1\pmod{10},n^r\equiv1\not\equiv-1$ for any integer $r$
Or
$n\pm3,n^{2m}\equiv(\pm3)^{2m}\equiv9^m\equiv(-1)^m(1-10)^m\equiv(-1)^m\pmod{10}$
which will be $\equiv-1$ as required if $m$ is odd
Here $m=5$
