What is $x$ if $x-\frac{13951396^{2016}}{90} $ is an integer? 
What is $x$ if $x-\frac{13951396^{2016}}{90} $ is an integer: $\frac{46}{90}$ or $\frac{47}{90}$or $\frac{35}{90}$

Is there any simple way to solve this?
 A: $\begin{align}{\rm Note}\ \ x \, =\ &\dfrac{a}{90} =\, n + \dfrac{(2j)^{\large k}}{90},\  \ k>0\\[0.3em]
  \Rightarrow\ a &\,=\, 90 n + (2j)^{\large k}\,\ \text{is even, so $\,a\,$ can only be $46$, not $47$ or $35$}\end{align}$
A: $$\dfrac{13951396^{2016}}{90}=\\
\dfrac{13951396^{2015}.13951396}{90}=\\
\dfrac{13951396^{2015}.6975698}{45}=\\$$ so you can check by mod 45 
$$13951396\equiv1 \space (mod \space 45)\\
13951396^{2015} \equiv 1 \space (mod \space 45)\\6975698 \equiv23 \space (mod \space 45)\\ \to 
13951396^{2015}.6975698\equiv1\times 23 \space (mod \space 45)\\
13951396^{2015}.6975698=45q+23$$ so 
$$\dfrac{13951396^{2015}.6975698}{45}=\dfrac{45q+23}{45}=\dfrac{90q+46}{90}$$ now $$x-\dfrac{90q+46}{90} $$ is integer $$x=\dfrac{46}{90}\\\dfrac{46}{90}-\dfrac{90q+46}{90} \in Z$$
A: We have, modulo 90:
$13951396\equiv(1+3+9+5+1+3+9)\times 10+6=316\equiv 46$
To get this we cast out nines from the tens and larger digits and then tack on the units digit.
Then
$46^2=2116\equiv 46$ mod $90$
so residue $46$ is idempotent, thus we are sure that the numerator in the original fraction has that residue.
