< b="" />Solution:< />

To find the smallest natural number that is made up of each of the digits 1 through 9 exactly once and is divisible by 99, we need to analyze the divisibility rules of 99.

< b="" />Divisibility Rule of 99:< />
A number is divisible by 99 if and only if it is divisible by both 9 and 11.

< b="" />Divisibility Rule of 9:< />
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

< b="" />Divisibility Rule of 11:< />
A number is divisible by 11 if and only if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is divisible by 11.

Now, let's analyze the given options.

< b="" />Option A: 123456789< />
The sum of the digits is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, which is divisible by 9. However, the difference between the sum of the digits at odd positions (1 + 3 + 5 + 7 + 9) and the sum of the digits at even positions (2 + 4 + 6 + 8) is not divisible by 11. Therefore, option A is not divisible by 99.

< b="" />Option B: 123475869< />
The sum of the digits is 1 + 2 + 3 + 4 + 7 + 5 + 8 + 6 + 9 = 45, which is divisible by 9. The difference between the sum of the digits at odd positions (1 + 3 + 7 + 8 + 9) and the sum of the digits at even positions (2 + 4 + 5 + 6) is 2, which is divisible by 11. Hence, option B is divisible by 99.

< b="" />Option C: 125364789< />
The sum of the digits is 1 + 2 + 5 + 3 + 6 + 4 + 7 + 8 + 9 = 45, which is divisible by 9. However, the difference between the sum of the digits at odd positions (1 + 5 + 6 + 7 + 9) and the sum of the digits at even positions (2 + 3 + 4 + 8) is not divisible by 11. Therefore, option C is not divisible by 99.

< b="" />Option D: 314256789< />
The sum of the digits is 3 + 1 + 4 + 2 + 5 + 6 + 7 + 8 + 9 = 45, which is divisible by 9. However, the difference between the sum of the digits at odd positions (3 + 4 + 5 + 7 + 9) and the sum of the digits at even positions (1 + 2 + 6 + 8) is not divisible by 11. Therefore, option D is not divisible by 99.

Hence, the smallest natural number that is made up of each of the digits 1 through 9 exactly

Question Analysis:
We are given a task to find the smallest natural number that is made up of each of the digits 1 through 9 exactly once and is divisible by 99. To solve this question, we need to understand the divisibility rule of 99 and find the number that satisfies this rule.

Divisibility Rule of 99:
To determine if a number is divisible by 99, we need to check if the sum of its digits is divisible by 9 and if the difference between the sum of its odd-placed digits and the sum of its even-placed digits is divisible by 11.

Solution:
To find the smallest number that satisfies the given condition, we can start by arranging the digits 1 through 9 in ascending order and check if the number is divisible by 99. If it is not divisible, we can try rearranging the digits until we find the smallest possible number that satisfies the condition.

Step 1:
Arrange the digits 1 through 9 in ascending order: 123456789.

Step 2:
Check if the sum of the digits is divisible by 9.
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
Since 45 is divisible by 9, we move to the next step.

Step 3:
Check if the difference between the sum of the odd-placed digits and the sum of the even-placed digits is divisible by 11.
Odd-placed digits: 1, 3, 5, 7, 9
Even-placed digits: 2, 4, 6, 8
Sum of odd-placed digits: 1 + 3 + 5 + 7 + 9 = 25
Sum of even-placed digits: 2 + 4 + 6 + 8 = 20
Difference: 25 - 20 = 5
Since 5 is not divisible by 11, we need to rearrange the digits.

Step 4:
Rearrange the digits until we find the smallest possible number that satisfies the condition.
We can try different rearrangements and check if the number is divisible by 99.

Let's try rearranging the digits:
123475869

Step 5:
Check if the sum of the digits is divisible by 9.
1 + 2 + 3 + 4 + 7 + 5 + 8 + 6 + 9 = 45.
Since 45 is divisible by 9, we move to the next step.

Step 6:
Check if the difference between the sum of the odd-placed digits and the sum of the even-placed digits is divisible by 11.
Odd-placed digits: 1, 3, 5, 7, 9
Even-placed digits: 2, 4, 6, 8
Sum of odd-placed digits: 1 + 3 + 5 + 7 + 9 = 25
Sum of even-placed digits: 2 + 4 + 6 + 8 = 20
Difference: 25 - 20 = 5
Since 5 is not divisible by