How to represent XOR of two decimal Numbers with Arithmetic Operators Is there any way to represent XOR of two decimal Numbers using Arithmetic Operators (+,-,*,/,%).

I think what Sanisetty Pavan means is that he has two non-negative integers $a$ and $b$ which we assume to be in the range $0 \leq a, b < 2^{n+1}$ and thus representable as $(n+1)$-bit vectors $(a_n, \cdots, a_0)$ and $(b_n, \cdots, b_0)$
where 
$$
a = \sum_{i=0}^n a_i 2^i, ~~ b = \sum_{i=0}^n b_i 2^i.
$$
He wants an an arithmetic expression for the integer $c$ where
$$c = \sum_{i=0}^n (a_i \oplus b_i) 2^i 
= \sum_{i=0}^n (a_i +  b_i -2 a_ib_i) 2^i = a + b - 2 \sum_{i=0}^n a_ib_i 2^i$$
in terms of $a$ and $b$ and the arithmetic operators $+, -, *, /, \%$. Presumably integer constants are allowed in the expression.  The expression 
for $c$ above shows a little progress but I don't
think it is much easier to express $\sum_{i=0}^n a_ib_i 2^i$ than it is to
express $\sum_{i=0}^n (a_i \oplus b_i) 2^i$ in terms of $a$ and $b$, but perhaps Listing's gigantic formula might be a tad easier to write out, though Henning Makholm's objections will still apply.
Added note:  For fixed $n$, we can express $c$ as
$c = a + b - 2f(a,b)$ where $f(a, b)$ is specified recursively as
$$f(a, b) = (a\%2)*(b\%2) + 2f(a/2, b/2)$$
with $a\%2$ meaning the remainder when integer $a$ is divided by $2$
(that is, $a \bmod 2$) and $a/2$ meaning "integer division" which 
gives the integer quotient (that is, $a/2 = (a - (a\%2))/2$).
Working out the recursion gives a formula with $n+1$ terms for
$f(a, b)$.
