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# January 2015 - Alice and Bob (Solution)

The answer is 72. Make a table of possible combinations twice. Group one based on the sum and the other based on the difference. Now, work through the possibilities. This is Alice's table:

 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 1,10 2,10 3,10 4,10 5,10 6,10 7,10 8,10 9,10 N/A N/A 2,3 2,4 2,5 2,6 2,7 2,8 2,9 3,9 4,9 5,9 6,9 7,9 8,9 N/A N/A N/A N/A N/A N/A 3,4 3,5 3,6 3,7 3,8 4,8 5,8 6,8 7,8 N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A 4,5 4,6 4,7 5,7 6,7 N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A 5,6 N/A N/A N/A N/A N/A N/A N/A N/A

Alice only knows that the values of a and b are in a specific column. This is Bob's table:

 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 1,10 2,3 2,4 2,5 2,6 2,7 2,8 2,9 2,10 N/A 3,4 3,5 3,6 3,7 3,8 3,9 3,10 N/A N/A 4,5 4,6 4,7 4,8 4,9 4,10 N/A N/A N/A 5,6 5,7 5,8 5,9 5,10 N/A N/A N/A N/A 6,7 6,8 6,9 6,10 N/A N/A N/A N/A N/A 7,8 7,9 7,10 N/A N/A N/A N/A N/A N/A 8,9 8,10 N/A N/A N/A N/A N/A N/A N/A 9,10 N/A N/A N/A N/A N/A N/A N/A N/A

Bob only knows that the values of a and b are in a specific column. Now, consider the dialogue.

1: Alice: I don't know the numbers.

If the sum were 3, 4, 18, or 19, there would only be one possible pair of values. Hence Alice could not have received those numbers as the sum. We can rule out these pairs from both tables: 1,2; 1,3; 8,10; 9,10.

2: Bob: Neither do I.

If the difference were 9, there would only be one possible pair of values. Hence Bob could not have received 9 as the difference. We can rule out 1,10 from both tables.

3: Alice: I knew that.

If Alice had received 11 as the sum, she would not have known that the pair definitely was not 1,10 before Bob announced that. However, she did know. Therefore Alice did not receive 11 as the sum. We can rule out 2,9; 3,8; 4,7; 5,6 from both tables.

4: Bob: Even if you told me one of the numbers, I'm not sure I could figure out the other one.

Bob would know the difference and one of the numbers. However, he is not sure he could figure out the other one. The only way this could happen is if he did not know which number was greater. However, if the difference was 5, 6, 7, or 8, this could not happen as each number appears at most once. Thus if Bob was told a number, he would know which pair the number was in. Hence the difference was not 5, 6, 7, or 8. We can rule out 1,6; 2,7; 4,9; 5,10; 1,7; 2,8; 3,9; 4,10; 1,8; 3,10; 1,9; 2,10 from both tables.

5: Alice: I knew that.

If Alice had received any sum between 7 and 15, she would not have known that Bob may not have known the other number given one of them. Hence this could not have happened and Alice did not receive a sum between 7 and 15. It is easier to rebuild the table now. This is Alice's table:

 1,4 1,5 6,10 7,10 2,3 2,4 7,9 8,9

This is Bob's table:

 2,3 2,4 1,4 1,5 8,9 7,9 7,10 6,10

6: Bob: Now I know that 10 is not one of the numbers.

If Bob received a difference of 3 or 4, he would not be able to know for sure that 10 is not one of the numbers. Thus he received a difference of 1 or 2. We can rule out 1,4; 7,10; 1,5; 6,10 from both tables.

7: Alice: Now I know that 3 is not one of the numbers.

If Alice had received a sum of 5, she would not be able to know that 3 is not one of the numbers. (In fact, she would know that the pair is 2,3 so 3 is definitely one of the numbers) Thus she did not receive a sum of 5. We can rule out 2,3 from both tables.

8: Bob: Now I know the numbers.

If Bob had received a difference of 2, he would not be able to tell whether the pair was 2,4 or 7,9. Thus he received a difference of 1 and the pair is 8,9.

9: Alice: So do I.

Alice knows that if Bob knows the numbers, the pair is 8,9.

Hence the pair 8,9 is consistent with all clues. Thus the product is 72.

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