You multiply by 9 by adding a 0

As the problem image shows, there are three cases:

• Case 1: We strike a zero in the hundreds place.

But this is not even a valid case, as a three-digit number cannot begin with a zero.

• Case 2: We strike a zero in the units place.

This is the same as dividing by $10$ , so the number we get will be $\frac{N}{10}$ . If $\frac{N}{10} = \frac{N}{9}$ , we need to have $9N = 10N$ , or $N = 0$ , not a three-digit number.

• Case 3: We strike a zero in the tens place.

Let the number be $\overline{a0b}$ , so after removing the zero we get $\overline{ab}$ . Thus:

$\overline{a0b} = 9 \cdot \overline{ab}$

$100a + b = 9 (10a + b)$

$100a + b = 90a + 9b$

$10a = 8b$

$5a = 4b$

If $b = 0$ , then $a = 0$ too, which gives the number $\overline{a0b} = \overline{000} = 0$ , not a three-digit number too. So $b > 0$ . But since $b$ is a decimal digit, it must be less than $10$ .

Finally, as $a,b$ are integers, we need $5$ to divide $4b$ . Since $4$ is not divisible by $5$ , we need to have $5$ to divide $b$ . Thus $b$ is an integer divisible by $5$ that lies in the range $[1,9]$ ; the only possibility is $b = 5$ , which causes $a = 4$ and $N = \overline{a0b} = \boxed{405}$ . It can be easily verified that the number works.

Another way is by programming, although clearly it's not an intended solution for a math problem and is rather hard with various conversions between integers and strings. Here's a Python solution:

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# iterate over all three-digit numbers
for number in range(100, 1000):

    # cast to string, as deleting a character is a string operation
    number = str(number)

    # iterate over each character
    for i in range(len(number)):

        # if the character is not a zero, skip
        if number[i] != "0": continue

        # try removing the character
        removed = number[0:i] + number[i+1:]

        # check whether the result is the number divided by 9
        if int(removed) * 9 == int(number):

            # print the number; this one is valid
            print(number)

# prints a single number '405'