modulo...

We have to work mod 7 , since the denominator is 7. Hence,

$23^{1029} + 13^{1029} + n ≡ 0$ (mod 7)

Now, $23^{1029} ≡ 2^{1029}$ (mod 7)

Also, $2^3 ≡ 1$ (mod 7) $\Rightarrow (2^3)^{343} ≡ 1^{343} ≡ 1$ (mod 7)

Along similar lines, $13^{1029} ≡ (-1)^{1029} ≡ -1$ (mod 7)

Hence, $23^{1029} + 13^{1029} + n ≡ 0 \Leftrightarrow 1 -1 +n ≡ 0$ (mod 7)

The lowest possible whole number is $\boxed{0}$

$23=7k+2$ This implies that $23 \times 23 \times 23=7y+1$ ( using modular arithmetic) This implies that $23^{1029}=7m+1$ Similarly, $13^{1029}=7n+6$ Therefore $23^{1029}+13^{1029}=7k$ Thus n=0 & n=7