What combinations of numbers are possible?
You can use the formula
a * b * k = c where a, b, and c are the three single digit numbers and k is another possible factor of c. In this case, c will be larger than a and b so it cannot be 1 or 2. Let's look at factors of 3-9:
3: 1,3 - impossible because there are only 2 unique numbers
4: 1,2,4 - could be 1,2,4 (1 * 2 * 2 = 4 , so a and b could be 1 and 2 and c could be 4
5: 1,5 - impossible because there are only 2 unique numbers
6: 1,2,3,6 - could be 1,2,6 or 1,3,6 or 2,3,6
7: 1,7 - impossible because there are only 2 unique numbers
8: 1,2,4,8 - could be 1,2,8 or 1,4,8 or 2,4,8
9: 1,3,9 - could be 1,3,9
From this list of possibilities, we need to find one where one person can know what their number is by seeing two of the other numbers. Here are the possible numbers a person could see others having (lowest number first), and the possible numbers that the person could be.
1,2 - could be 4 or 8
1,3 - could be 6 or 9
1,4 - could be 2 or 8
1,6 - could be 2 or 3
1,8 - could be 2 or 4
1,9 - has to be 3 (1,3,9)
2,3 - has to be 6 (1,2,3)
2,4 - could be 1 or 8
2,6 - could be 1 or 3
2,8 - could be 1 or 4
3,6 - could be 1 or 2
3,9 - has to be 1 (1,3,9)
4,8 - could be 1 or 2
Now we have 3 possibilities for two single digit numbers that would allow someone to deduce their own number.
If I see 1 and 9, I know I am a 3. If I see a 3 and 9, I know I am a 1. If I see a 2 and a 3, I know I am a 6.
But two of those possibilities are for the same set of three numbers. That cannot be the case here because two of the people could not deduce what their numbers were.
If Mr. Hehe saw a 1 and 9, he would know he is a 3. The problem is one of the other two would see a 3 and a 9 and could deduce their were a 1. So that cannot be the case.
So now they all know that Mr. Hehe was able to deduce his number. That lets the first people know the three digits, otherwise Mr. Hehe could not have deduced his number. Since they can see the other two digits, they know what their own digit must be.
The other two are interchangeable for this problem, so let's assume Ms. Haha is 2 and Ms. Lola is 3:
Ms. Haha: I don't know my number. (sees 3 and 6, she could be a 1 or a 2)
Ms. Lola: I don't know my number. (sees 2 and 6, she could be a 1 or a 3)
Mr. Hehe: I knew my number before either of you spoke. (the only possible condition is if he saw a 2 and 3 and deduced he was a 6)
Ms. Haha and Ms. Lola: Now we both know our digits. (They can deduce that the combination is 2,3,6 because if either were a 1, Mr. Hehe wouldn't know for certain what his number was before anyone spoke. And since they can see 6 and the other person's number, they know they are the remaining number of the trio)