Hi,

This is a question given under GRE Quant Problem Solving section,

One digit in a 10-digit number is missing. What is the maximum number of ways that the missing digit could be filled so that the final number is divisible by 3?

for which the answer is " 4 ".

But, the way I approached to the solution gives me only 4 possibilities i.e. 0,1,2,3,6,9

because for any no. to be divisible by 3 : the sum of the digits should be divisible by 3.

which leaves us with 3 cases :

- remainder is 0 which means add 0, 3, 6, 9 to the sum and the no is div by 3.
- remainder is 1 = add 2 to the sum and the no is div by 3
- remainder is 2 = add 1 to the sum the no is div by 3

looking at all these cases the only probable answer to the blank could be 0,1,2,3,6,9

but, according to the video (answer explanation by Greg) is 0,3,6,9.

which will say that the 10-digit number is already divisible by 3 and it certainly satisfies just 1 case.

Can anyone help.

I did this by filling the first 9 places with 3’s and then counting the number of ways in which I can fill the last slot.

\underline{3}-\underline{3}-\underline{3}-\underline{3}-\underline{3}-\underline{3}-\underline{3}-\underline{3}-\underline{3}-\underline{X}

Now, as divisibility by 3 says that the sum should be divisible by 3, I can replace the ‘X’ with 0,3,6,9. Also, as the question said could thus, I needed only a single test-case.

Okay, it can be true if you fill up all 9 places with same digit …

and the fact that consecutive 3 nos are definitely div by 3.

But , since its no where mentioned to constraint our imagination… if we take number like -

7 8 7 1 2 5 5 9 5 _

The sum of the digits till 9th place = 49

Adding a 0,3,6,9 - wont make this number divisible by 3.

Now , we have to add a 2.

Please guide if there is a condition given that I am missing out on.

Or 5 or 8. But they are asking for a maximum.

Maximum number of ways * to fill the black space.

Oh! No got it !!

At a time - it can be 3 or 4 so - yep !

Thank you !