The square of a number has four digits. One of its digits is 0 and the other three digits are even. If the square of the number is less than 7000, find the number.

Steps to Solve

1. Determine the range for the numbers To find numbers whose squares yield four-digit results, we need to find $1000 \leq n^2 < 10000$.

Taking square roots, we find: $$ \sqrt{1000} \approx 31.62 \quad \text{and} \quad \sqrt{10000} = 100 $$

Thus, $n$ must be in the range of $32 \leq n < 100$.

2. Refine the range based on the given condition We also need $n^2 < 7000$.

Taking the square root of 7000 gives us: $$ \sqrt{7000} \approx 83.67 $$

Thus, $n$ must also satisfy $32 \leq n < 84$.

3. Check each number for conditions We will check each integer from 32 to 83 to see if their square:

• Is a four-digit number.
• Contains the digit '0'.
• Contains three even digits.

4. Count the even digits and check for '0' For each candidate $n$, calculate $n^2$ and verify:

• Convert $n^2$ to string (or examine it).
• Count the even digits: $0, 2, 4, 6, 8$.
• Check if '0' is present in the digits.

5. List valid solutions Compile the numbers that meet all the conditions stated.