M1. The positive integer $N$ has five digits.
The six-digit integer $P$ is formed by appending the digit 2 to the front of $N$. The six-digit integer $Q$ is formed by appending the digit 2 to the end of $N$. Given that $Q=3 P$, what values of $N$ are possible?
Solution
We have $P=N+200000$ and $Q=10 N+2$. Hence
$$
\begin{aligned}
10 N+2 & =3 \times(N+200000) \\
& =3 N+600000
\end{aligned}
$$

Therefore, subtracting $3 N+2$ from each side, we get
$$
7 N=599998
$$
and so, dividing each side by 7 , we obtain
$$
N=85714 .
$$

Hence the only possible value of $N$ is 85714 .