Making a century

The solution is $\large 3 \frac{69258}{714}$ i.e. $\boxed{369258714}$

The solution is little lengthy but you have to take time to do the sum.

If the integer part is 1 .

The fraction becomes $\large 1 \frac{99 \times x}{x}$

The numerator of the fraction must be divisible by 9 as it will be a multiple of 99. That means the sum of digits in the numerator is a multiple of 9. The entire set of digits from $2,3,..,9$ add up to 1 less than a multiple of 9, so the denominator must be three digits that add up to 1 less than a multiple of 9. That way, any arrangement of the 5 remaining digits is divisible by 9.

The sums that are 1 less than a multiple of 9 are 8, 17, 26. The smallest possible 3-digit sum is $2+3+4=9$ and the largest is $9+8+7=24$ , so only 17 works. The sets of 3 digits from $2,3,...,9$ that sum to $17$ are: $(2, 6, 9), (2, 7, 8), (3, 5, 9), (3, 6, 8), (4, 5, 8), (4, 6, 7)$

That's 6 sets, and there are 6 ways to arrange each set, so you have 36 possible denominators to test. Multiply each denominator by 99 and see if the result is a arrangement of the remaining digits.

To avoid the trial and error , we see that ,
$99 \times \overline{ghi} \geq 99 \times \overline{g23} = 9900 \times g + 2277 > 9900 \times g + 100 \times g = 10000g$
This makes $g$ and $b$ the same number .

Hence we discard 1.

If the integer part is 2

The fraction becomes $\large 2 \frac{98 \times x}{x}$

Then the fraction is an improper form of 98 and the numerator is divisible by 98. If you add the denominator to the numerator, you get a multiple of 9 as the total sum will be a multiple of 99. So the sum of the 3 digits in the denominator + the sum of the 5 digits in the numerator must be a multiple of 9. That's impossible, since the sum of all the digits is $1 + 3 + 4 + .. + 9 = 43$ is not a multiple of 9. That's easy...no testing required for 2.

If the integer part is 3

The fraction becomes $\large 3 \frac{97 \times x}{x}$

Then the numerator plus 2 times the denominator is divisible by 9 as the number will be a multiple of 99. Since the sum of digits is $1 + 2 + 4 + ...+ 9 = 42$ is three less than a multiple of 9, then the denominator must have a sum that is 3 more than a multiple of 9. The sums go 3, 12, 21, 30.
The smallest possible 3-digit sum is $1+2+4=7$ and the largest is $9+8+7=24$ , so only 12 and 21 are feasible. (1, 2, 9), (1, 4, 7), (2, 4, 6), (4, 8, 9), (5, 7, 9), (6, 7, 8)

Again, 6 x 6 = 36 total denominators to test.
But 714 satisfies and hence the answer is $\boxed{369258714}$