How many four digit numbers N have the property that the three digit number obtained by removing the leftmost digit is one ninth of N?

Answer 1

Let $x$ be the three digit number obtained by removing the leftmost digit of $N$. Then, we can write \[N = 1000x + x,\]which simplifies to $N = 1001x$. Since $N$ has four digits while $x$ has three, we know that \[1000 \le 1001x < 10000.\]Dividing both sides of the inequality by 1001, we have \[\frac{1000}{1001} \le x < \frac{10000}{1001}.\]Then, \[\frac{1000}{1001} \le x < \frac{99}{100} = \frac{99 \cdot 100}{100^2} = \frac{9900}{100^2},\]which implies \[1000 \le 1001x < \frac{9900}{10^2} \times 1000 = 9900.\]We count the number of integers $n$ satisfying \[1000 \le n < 9900.\]The smallest multiple of 1000 in this range is 1000 while the largest multiple of 1000 in this range is 9000. Thus, the number of solutions is $9000 - 1000 + 1 = \boxed{8001}$.