Different solution than the one posted

GRE Quant Problem Solving
1

Hi there, just a quick post to check if the solution posted is actually correct. The solution in the video is 4. I think it’s 3.

Here’s the problem.
“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?”

What it means is that the sum of numbers must be divisible by 3.
If the sum equals 1 - the added number should be 2 to make it divisible by 3
if the sum equals 2 - we need to add 1
if the sum equals 3 - we need to add 0
if the sum equals 4 - we need to add 2
if the sum equals 5 - we need to add 1
if the sum equals 6 - we need to add 0
if the sum equals 7 - we need to add 2
if the sum equals 8 - we need to add 1
if the sum equals 9 - we need to add 0
for 10,11,17,2276 the logic is the same. Which means that the new number can only be 0, 1, 2.

What do you guys think?

2

No, and you’re not on the right track. For example, in the “if the sum equals 9” case, why can I not add 3 instead?

3

You’re absolutely right. We can add 3. But we can also add 6 or 9. In that instance the answer would be of a total of 6.

4

It isn’t “6” - in that case, there are 4 ways (0, 3, 6, 9) you can make the number divisible by 10. The question is asking for the maximum number of ways - which is 4.

5

I get what you mean. Though I really think the question is phrased ambiguously and can be understood in 2 different ways. ETS definitely knows how to keep one on their toes. Thanks for the quick replies!

6

I’m not quite sure how that’s ambigous.