Prove that the probability that the sum of the digits used leaves the remainder 2 when divided by $4$ is $\frac{1}{4}.$ A number of 5 digits is written down at random.Prove that the probability that the sum of the digits used leaves the remainder 2 when divided by $4$ is $\frac{1}{4}.$
The sum of digits must be like 6,10,14 etc.But i am not sure how many digits we have to consider for summing.Please help me in solving this problem. 

While waiting for some smarter solution (if any), I'll give a direct count of how many 5-digit numbers are congruent to $2$ (mod $4$). 
Let's start with 1-digit numbers (from $0$ to $9$): we have $3$ of them congruent to $0$ (they are $0$, $4$ and $8$), $3$ congruent to $1$ ($1$, $5$ and $9$), $2$ congruent to $2$ ($2$ and $6$) and $2$ congruent to $3$ ($3$ and $7$).
Consider now a 2-digit number $n$ (between $00$ and $99$), composed of two single digits $x$ and $y$. We have $n\equiv0$ if:
$$
\begin{align}
&x\equiv0,\ y\equiv0\quad (3\times3=9\ \hbox{cases})\cr
&x\equiv1,\ y\equiv3\quad (3\times2=6\ \hbox{cases})\cr
&x\equiv2,\ y\equiv2\quad (2\times2=4\ \hbox{cases})\cr
&x\equiv3,\ y\equiv1\quad (2\times3=6\ \hbox{cases})\cr
\end{align}
$$
for an overall total of $9+6+4+6=25$ cases. In the same way one can count $n\equiv1$ in $26$ cases, $n\equiv2$ in $25$ cases and $n\equiv3$ in $24$ cases.
We can repeat the same procedure for a 4-digit number $n$ (between $0000$ and $9999$), which is composed of two 2-digit numbers (we'll denote the sums of their digits by $x$ and $y$). We have $n\equiv 0$ if: 
$$
\begin{align}
&x\equiv0,\ y\equiv0\quad (25\times25=625\ \hbox{cases})\cr
&x\equiv1,\ y\equiv3\quad (26\times24=624\ \hbox{cases})\cr
&x\equiv2,\ y\equiv2\quad (25\times25=625\ \hbox{cases})\cr
&x\equiv3,\ y\equiv1\quad (24\times26=624\ \hbox{cases})\cr
\end{align}
$$
so that we have a total of $2498$ cases with $n\equiv 0$. In a similar way we can compute a total of $2500$ cases with $n\equiv 1$, then $2502$ cases with $n\equiv 2$ and $2500$ cases with $n\equiv 3$.
The same process can be repeated for a 5-digit number: we must add the leftmost digit to a 4-digit number, taking into account that it cannot be zero. Between 1 and 9 there are 2 digits $\equiv0$, 3 digits $\equiv1$, 2 digits $\equiv2$ and 2 digits $\equiv3$. We can then form a 5-digit number congruent to $2$ in $2\times2502+3\times2500+2\times2498+2\times2500=22500$ different ways. So that the probability to pick one of them is $22500/90000=1/4$.
Notice that if we had allowed for a zero leftmost digit that probability would have been different.
