. If a number 418X367 is completely divisible by 9, then the smallest whole number in place of
X is:

For a number to be divisible by 9, the sum of its digits must also be divisible by 9. Let's apply this rule to solve the problem.

The number given is: **418X367**

### Step 1: Find the sum of the digits
The sum of the digits is:
\( 4 + 1 + 8 + X + 3 + 6 + 7 = 29 + X \)

### Step 2: Determine the smallest value of \( X \)
We need to find the smallest value of \( X \) such that the sum \( 29 + X \) is divisible by 9.

To find the smallest value of \( X \), we need to check the remainder when \( 29 \) is divided by 9:
\[
29 \div 9 = 3 \text{ remainder } 2
\]
This means that \( 29 \) leaves a remainder of 2 when divided by 9.

To make \( 29 + X \) divisible by 9, we need to add a value of \( X \) such that the remainder is zero. So, the smallest \( X \) that satisfies this is:
\[
X = 9 - 2 = 7
\]

### Conclusion
The smallest whole number that should replace \( X \) to make the number 418X367 divisible by 9 is:
\[
\boxed{7}
\]

Would you like more details or have any questions about the solution?

Here are 5 related questions to consider:
1. What is the divisibility rule for 3, and how does it compare to the rule for 9?
2. How would the approach change if the number was divisible by 11 instead of 9?
3. If another digit besides \( X \) was unknown, how would you solve for that digit using divisibility rules?
4. What are some real-world applications of divisibility rules in number theory?
5. How can you determine if a number is divisible by both 9 and 6 simultaneously?

**Tip:** The sum of the digits rule for divisibility by 9 is a great shortcut in arithmetic checks!

If a number 418X367 is completely divisible by 9, then the smallest whole number in place of X is:

Learn how to find the smallest whole number X to make the number 418X367 divisible by 9. This problem explores the divisibility rule for 9, involving basic arithmetic and digit sum calculations. Suitable for students in Grades 6-8.

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