Suppose that $n$ is positive integer and that $a$ is the integer equal to $\frac{10^{2 n}-1}{3\left(10^{n}+1\right)}$. If the sum of the digits of $a$ is 567, what is the value of $n$ ?
Since $x^{2}-1=(x+1)(x-1)$, then setting $x=10^{n}$ gives $10^{2 n}-1=\left(10^{n}+1\right)\left(10^{n}-1\right)$. Therefore,

$$
a=\frac{10^{2 n}-1}{3\left(10^{n}+1\right)}=\frac{\left(10^{n}+1\right)\left(10^{n}-1\right)}{3\left(10^{n}+1\right)}=\frac{10^{n}-1}{3}
$$

because $10^{n}+1 \neq 0$.

Since $10^{n}$ is the integer consisting of a 1 followed by $n 0 \mathrm{~s}$, then $10^{n}-1$ is the integer consisting of $n 9 \mathrm{~s}$.

Thus, $a=\frac{10^{n}-1}{3}$ is the integer consisting of $n$ s. This is because each digit 9 in $10^{n}-1$ can be divided by 3 to obtain the new digit 3 without any re-grouping (borrowing).

Therefore, the sum of the digits of the integer $a$ is $3 n$.

Since the sum of the digits of $a$ is given to be 567 , then $3 n=567$ or $n=189$.

ANSWER: $n=189$