A number is said to be a “digifac” if each of its digits is a factor

For the number to be divisible by 9, its sum should be a multiple of 9. This number will also be divisible by 3.
The sum of digits we have now is 9+5+3 = 17
For number to be divisible by 5, it's unit digit should be 0 or 5. But 0 cannot be a factor of any number as 0 multiplied with any number will be 0.
So, the unit's digit will be 5. The sum now becomes, 17+5 = 22
We need a multiple of 9 greater than 22 i.e 27.
so, the ten's digit will be 27-22 = 5.
and the sum of missing digits will be 5+5=10

Answer:- D

For the number to be divisible by 3 and 9, the sum of the digits should be divisible by 3 and 9. Only D=10 satisfies. The sum is 27
Answer D

VERITAS PREP OFFICIAL SOLUTION:

Here, the term “digifac” should look intimidating. You probably haven’t studied digifacs before, so how should you approach this problem? Well, keep in mind that digifacs aren’t being tested; in fact, the author of this question just made that term up, and then defined it for you. What makes this question hard is that the non-challenge-seeker (I think I just made that term up, too…) will see the unfamiliar term “digifac” and lose faith immediately. “I don’t know what that is!” She who finds the challenge in the GMAT fun, however, will read the definition and think “got it – I need to find the two digits that ensure that 9, 5, and 3 are both factors of the overall number, and that the remaining two digits are also factors”. And work from there. The number must be divisible by 5, so the only units digits that work are 0 or 5. And the number must be divisible by 9 (and also 3), so we need the sum of all digits to be a multiple of 9. 9 + 5 + 3 = 17, so our only options are to get the sum to 18 (by adding 1) or to 27 (by adding 10). A quick glance at the answer choices shows that 0 1 isn’t an option. Why not? That would require 0 to be one of the digits…and 0 isn’t a factor of anything. So the units digit must be 5, making the tens digit 5, and we have 95,355. That number is a multiple of 5, 3, and 9, so it works: the correct answer is D, and more importantly this fun challenge required no “trivial” information about digifacs…that term only existed to obscure the link between the given information and the path to the answer.

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Hmm good question, got me thinking a bit. It's just recognizing divisibility rules. If each digit is a factor, then the number must be divisible by 3, 5, 9 (all provided). Everything divisible by 9 is divisible by 3 so ignore 3. If we take 9 + 5 + 3, we get 17. The only way to get a number divisible by 9 and 5 is to have it be a multiple of 9 and end in 5 or 0. Only adding 10 (D) gives a choice that is divisible by 9 and ends in 0.

9+5+3 = 17

17 +x = 9k
x = 5y

y = 2 ---> x=10
17+10 = 27
27/9 = 3

9, 5, 3, 5, 5

Every number should be a factor , thus it should have 5 in the end 9 and 3 multiples have 5 but 5 only has 0 and 5 as multiple. Now 953x5 to make a number divisible by 9 the sum of number should be multiple of 9. thus if we put 5, the sum would be 27 also multiple of 3, thus number would be 95355

I think the choices given make it a whole lot easier to figure out.
Here the 3 digits given are 9, 3, and 5. For a number to be divisible by 9 or 3 the sum of digits should be a multiple of 9 or 3.
The only answer choice in which this is true is D. ( 9+3+5+10= 27 which is a multiple of both 9 and 3). All the other choices lead to a sum that are not multiples of 9 and 3.

EDIT - I read the question as "what are the remaining two digits" and not what is the sum of the remaining two digits. My approach is incorrect just thought I'd point it out rather than delete.

Answering a question 9 years after it was posted   feels like doing a treasure hunt.

So here's I approached it:

Since it's a 5 digit number we can ignore options A,B and C. Which leaves us with D & E.

It says the entire 5 digit number has each of its digit as its factors so we can simply test the options:

D : 10

9,5,3,1,0 ...... 1 Is a factor of any number so option D looks possible.

E : 14

9,5,3,1,4 ....... 1 is a factor so that checks out but the number is not divisible by 4 or 4 is not a factor so E is not possible. (divisibility rule of 4: for a number to be divisible by 4 the last two digits should also be divisible by 4. In this case 14 is not divisibe)

Answer D