Riddle Me This

I hate proffesor Sorites!! He always does this bull*****!!

 

My list of unique products is:
1 - 1,1 =2
2 - 1,2 =3
3 - 1,3 =4
5 - 1,5 =6
7 - 1,7 =8
10 - 2,5 =7
14 - 2,7 =9
15 - 3,5 =8
20 - 4,5 =9
25 - 5,5 =10
27 - 3,9 =12
30 - 5,6 =11
32 - 4,8 =12
35 - 5,7 =12
40 - 5,8 =13
42 - 6,7 =13
48 - 6,8 =14
49 - 7,7 =14
54 - 6,9 =15
56 - 7,8 =15
63 - 7,9 =16
72 - 8,9 =17
81 - 9,9 =18
with the number after the equal sign being the associated sum.
From that the unique sums are:
2 - 1,1 =0
3 - 1,2 =1
4 - 1,3 =2
6 - 1,5 =4
7 - 2,5 =3
10 - 5,5 =0
11 - 5,6 =1
16 - 7,9 =2
17 - 8,9 =1
18 - 9,9 =0
with the number after the equal sign being the associated difference.
From that set, the unique differences are only:
3 - 2,5
4 - 1,5

1 and 5 would return a value of 2 for D's clue.
2 and 5 would return a value of 1 for D's clue.
So it seems like either 1 and 5 or 2 and 5 could be a workable answer.
This all assumes that each successive person has knowledge of the form of the clue given to the previous person. Otherwise, 1 and 5 for example would have a non-unique sum of 6. 6 is a unique sum of all the possible pairs with a unique product though.

 

I read this:

I'm thinking of two digits; each is from 0 to 9 inclusive.

As suggesting the digits were different. If he's thinking of 2 & 2, for example, he's really just thinking of one digit twice. If that were the case, then that's how you eliminate the 9 & 9 solution you have. But Commodore said later that the digits can be the same. If that's the case, I agree - I don't see why 9&9 and 8&9 can't both be valid answers.

 

Cool puzzle.

For Annie, there are 27 different number combinations that provide a unique product. I'm throwing out duplicates. 6 and 9 are listed but not 9 and 6.

1 * 1 = 1
1 * 2 = 2
1 * 3 = 3
1 * 5 = 5
1 * 7 = 7
2 * 5 = 10
2 * 7 = 14
3 * 5 = 15
3 * 7 = 21
3 * 9 = 27
4 * 5 = 20
4 * 7 = 28
4 * 8 = 32
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
6 * 7 = 42
6 * 8 = 48
6 * 9 = 54
7 * 7 = 49
7 * 8 = 56
7 * 9 = 63
8 * 8 = 64
8 * 9 = 72
9 * 9 = 81

Bert knows his numbers have to be in this set of combinations. 20 of the above combinations would not produce a unique sum:

1 + 7 = 8 occurs 2 times
2 + 7 = 9 occurs 2 times
3 + 5 = 8 occurs 2 times
3 + 7 = 10 occurs 2 times
3 + 9 = 12 occurs 3 times
4 + 5 = 9 occurs 2 times
4 + 7 = 11 occurs 2 times
4 + 8 = 12 occurs 3 times
5 + 5 = 10 occurs 2 times
5 + 6 = 11 occurs 2 times
5 + 7 = 12 occurs 3 times
5 + 8 = 13 occurs 2 times
5 + 9 = 14 occurs 3 times
6 + 7 = 13 occurs 2 times
6 + 8 = 14 occurs 3 times
6 + 9 = 15 occurs 2 times
7 + 7 = 14 occurs 3 times
7 + 8 = 15 occurs 2 times
7 + 9 = 16 occurs 2 times
8 + 8 = 16 occurs 2 times

That leaves the following 7 options for Bert:
1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
1 + 5 = 6
2 + 5 = 7
8 + 9 = 17
9 + 9 = 18

Chester then would be able to eliminate 4 of the combinations, because they would not produce a unique difference:

1 - 1 = 0 occurs 2 times
2 - 1 = 1 occurs 2 times
9 - 8 = 1 occurs 2 times
9 - 9 = 0 occurs 2 times

That leaves the following three differences:

3 - 1 = 2
5 - 1 = 4
5 - 2 = 3

Dagmar now has the three combinations to plug into his formula (Bert + Chester) / Annie. Two result in the same number and can be eliminated:

1 & 3: [Bert(4) + Chester(2)] / Annie(3) = 2
1 & 5: [Bert(6) + Chester(4)] / Annie(5) = 2

Which leaves just one combination:
2 & 5: [Bert(7) + Chester(3)] / Annie(10) = 1

The final answer is 2 and 5.

 

Pretty impressive, ScriboErgoSum.

What threw me off is that I did not tackle it as systematically as you did and did not take into consideration that every other person's clue only looks at the subset left by the previous clue. E.g., I would have thought it cannot be 2 and 5 because I would have looked at Bert's clue in an isolated fashion and would have said 2 and 5 is not unique because there is also 3+4, 1+6, etc, but they were already eliminated at the previous stage.

 

Hmmm... I am leaning toward no solution for Dagmar


Let say the numbers are N1 and N2, N1>N2

Dagmar's clue is this:

(B+C)/A = [ (N1+N2)+(N1-N2) ] / (N1N2) = 2/N2

The answer has to be a whole number, therefore, N2 = 2 or 1. Since the rule was told to all parties, ABCD should all base their decisions on this rule.

Knowing this, A can solve 1,3,5,7,9,10,12,14,16,18
B can solve 4,6,7,9, base on the set [1,3],[1,5],[2,5],[2,7]
C can solve 2,3,4,5, using the same set
If any element in this set is eliminated, then D would have a unique answer. But as it is, D is screwed


TLDR: ScriboErgoSum's solution eliminates [2,7] due to the presence of [4,5]. However, we know right away that [4,5] will not work for D. Keeping [2,7] in the valid set leads to non-unique answer for D.

 

Damn, I was just about to post that, scribo. nice job.

 

D's constraint is what I considered the most important in my solution too.
I haven't understood your TLDR edit though...

 

It comes down to this: if A,B, and C don't care about D's clue, then ScriboErgoSum's solution fits well. The existence of D's clue can alter the decision made by ABC. If we take that into consideration, then D ends up with a non-unique solution.

In this case, B can't rule out [2,7], because he should already know that [4,5] is not a valid answer.

 

2 and 5.

Edit: Didn't read the thread before answering. Saw Sribo got it earlier.

I basically figured out all the numbers Annie could have come up with where she could know the answer immediately, with there being no possibility of another answer. Since Bert knows Annie has flawless logic, he'd know the answer would be restricted to these numbers. So I added the remaining number combinations together and threw out the duplicates. I then took the cominbations that were left and subtracted the smaller number from the larger number, again throwing out the duplicates. That only left 3 number cominations (1,3), (1,5), and (2,5). Plugging those numbers into the equation given to Dagmar, the first 2 combos came out to "2" and the last one came out to "1". Therefore, 2 and 5 were the only numbers that could have given unique numbers to all 4 students.