HMMT November, 2017

Solution:

Let $d$ be the number of digits of $N$. Suppose that $N$ does not itself have leftmost digit $1$. Then the number of integers $1, 2, \ldots, N$ which have leftmost digit $1$ is
$$1 + 10 + 10^{2} + \ldots + 10^{d-1} = \frac{10^{d} - 1}{9}$$
so we must have $\frac{10^{d} - 1}{9} = \frac{2N}{5}$, or $5\left(10^{d} - 1\right) = 18N$. But the left-hand side is odd, so this is impossible.

Thus $N$ must have leftmost digit $1$. In this case, the number of integers $1, 2, \ldots, N$ which have leftmost digit $1$ is
$$\begin{aligned} & 1 + 10 + 10^{2} + \ldots + 10^{d-2} + \left(N - 10^{d-1} + 1\right) \\ = & \frac{10^{d-1} - 1}{9} + N - 10^{d-1} + 1 \\ = & N - 8\left(\frac{10^{d-1} - 1}{9}\right) . \end{aligned}$$
Therefore we need $N - 8\left(\frac{10^{d-1} - 1}{9}\right) = \frac{2}{5} N$, or $N = \frac{40}{3}\left(\frac{10^{d-1} - 1}{9}\right)$. Then, $\frac{10^{d-1} - 1}{9}$ must be divisible by $3$. The base-ten representation of $\frac{10^{d-1} - 1}{9}$ has $d-1$ ones, so $d-1$ must be divisible by $3$. Both $d=1$ and $d=4$ make $N$ less than $2017$, but $d=7$ gives the answer $N = \frac{40}{3}(111111) = 1481480$.