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?
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 $9000 - 1000 + 1 = \boxed{8001}$.
Let's break down the problem step by step:
Step 1: Consider a four-digit number N.
Step 2: Remove the leftmost digit from N to obtain a three-digit number.
Step 3: The three-digit number obtained is one-ninth of N.
To solve this problem, we can express the four-digit number N as:
N = 1000A + 100B + 10C + D,
where A, B, C, and D represent the digits of N.
According to step 3, the three-digit number obtained by removing the leftmost digit is one-ninth of N. Mathematically, this can be written as:
100A + 10B + C = (1/9) * (1000A + 100B + 10C + D).
Simplifying the equation, we get:
900A + 90B + 9C = 1000A + 100B + 10C + D.
Rearranging the terms and canceling out similar terms, we arrive at:
100A + 10B - 80C - D = 0.
From this equation, we can deduce that D = 100A + 10B - 80C.
To find the number of four-digit numbers N that satisfy the given condition, we need to consider the possible values of A, B, and C.
Since N is a four-digit number, A cannot be zero. Therefore, A can take values from 1 to 9.
Similarly, since N is a four-digit number, B, C, and D can take values from 0 to 9.
Now, let's go through all possible values of A, B, and C and count the number of solutions that satisfy the equation.
Possible values of A: 1, 2, 3, 4, 5, 6, 7, 8, 9 (9 possibilities)
Possible values of B: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 possibilities)
Possible values of C: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 possibilities)
Multiplying the number of possibilities for A, B, and C together, we get:
9 * 10 * 10 = 900.
Therefore, there are 900 four-digit numbers N that have the property that the three-digit number obtained by removing the leftmost digit is one-ninth of N.