BRAIN TEASERS THREAD

[alphatmw]

[ QUOTE ]
One of my favorites from long ago, I hope I get it right. I have purposefully changed the usual wording to prevent easy googling:

A & B are integers greater than 1 and less than 100.

Mr. Add knows (A+B)
Mr. Multiply knows (A*B)

Mr. Add says to Mr. Multiply, "You don't know A and B."
Mr. Multiply says, "Now I do know A and B."
Mr. Add says, "Now I do too."

What are A & B?

~MagicMan

[/ QUOTE ] A*B can't be prime, or else mr. multiply would know. thats where my ideas stopped. i also suspect that A*B must be less than 1600, as 1600 might be the largest number represented by two products of numbers between 1-100 (16*100 and 40*40).

[Deorum]

[ QUOTE ]
A*B can't be prime, or else mr. multiply would know.

[/ QUOTE ]

Also, A*B cannot be prime because the problem states that A and B are both greater than 1. Through my various attempts to solve this, I always get stuck between the first two statements: the first seems to imply that all possible sets of numbers derived from the sum that could be used as A and B have products that have at least five factors (including 1, which leaves us with 4 useful factors). For example, the sum cannot be 9, as the sets for A and B would then be {(2,7),(3,6),(4,5)} and the product of 2*7 is 14, which only has four factors. However, if all sets for A and B given by the sum of A and B have more than four factors, which I think is the implication of the first statement, "You do not know A and B", then he would need to be able to rule out all but one set of factors. I cannot figure out how he would do that.

[alphatmw]

if the numbers were (3,4), then mr. add would see 7 and know that mr. multiply is seeing either 10 or 12. if he were seeing 10, he would know the numbers are 5,2 so he must be seeing 12. given that its 12, mr. multiply doens't know if its 6,2 or 3,4.

mr. multiply is seeing 12. he knows that it's either 6,2 or 3,4. if it were 6,2, mr add would be seeing 8 and would think its 2,6...3,5...or 4,4.

...(insert more complicated thoughts?!)...

therefore, mr. multiply knows it is not 6,2. thus he knows the numbers are 3,4. and once he figures that out, mr. add knows it's 3,4 also.

lol i thought i had it but the he thinks-i think-he thinks progession got too complicated. i think i might have the right answer though. i think A and B have to be between 1 and 10, and that OP said 1-100 to throw us off. i don't think multiplying numbers greater than 10 allow for this type of situation to be feasible.

then again, this is all speculation because i'm not smart enough to figure it out.

[Magic_Man]

[ QUOTE ]
[ QUOTE ]
One of my favorites from long ago, I hope I get it right. I have purposefully changed the usual wording to prevent easy googling:

A & B are integers greater than 1 and less than 100.

Mr. Add knows (A+B)
Mr. Multiply knows (A*B)

Mr. Add says to Mr. Multiply, "You don't know A and B."
Mr. Multiply says, "Now I do know A and B."
Mr. Add says, "Now I do too."

What are A & B?

~MagicMan

[/ QUOTE ]

I would really appreciate it if someone would please post the answer to this with the reasoning.

[/ QUOTE ]

I have to admit that when I first saw this problem in high school, I mostly approached it with trial and error. That said, there are a few eliminations we can make:

1) Mr. Add knows that Mr. Multiply doesn't know A&B. This means that A*B is not the product of two primes, since it would then have a unique factorization.

2) Mr. Add must somehow then know that there is no combination of two primes A & B that make up his quantity (A+B). For example, A+B cannot be 12, because one possibility is A = 7, B = 5, both prime numbers. It just so happens that for "relatively small" numbers (certainly less than 100), any even number can be expressed as the sum of two primes (try it). Thus, A+B is odd. Notice that this means that A*B is even, since A & B must be a pair of odd & even numbers.

3) Also note that in order for Mr. Add's first statement to be true, A*B cannot be the cube of a prime. The only possible factors A & B would then be (prime1 * prime1^2). This, Mr. Add must see a number (A+B) that cannot be expressed as (prime1 + prime1^2).

4) We can now make a list of all possible numbers that Mr. Add sees. Start with the list 4, 5, 6 ... 198 and eliminate all numbers that can be expressed as the sum of two primes, or the sum of a prime and its square. Part of this list is:

(A+B) = 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97, ...

5) Now Mr. Product knows the numbers. This means that for some factors A*B, the quantity (A+B) is in the list above. If this were not true, then Mr. Sum could not have made his first statement. For example, if A*B = 16, the only choices are 4*4, and 2*8. 4+4 = 8 and 2+8 = 10, neither of which are in the list above. (They are not in that list because if (A+B) = 8, then there is a possibility A = 3, B = 5, in which case A*B = 15 and Mr. Product would know the two numbers immediately. Similar reasoning for A+B = 10. Again, A+B cannot be even.)

We can make a list of all the possibilities for Mr. Product, given this criterion. The fact that he actually found them means that for all the possible factors (A*B) that Mr. Product found, there is a unique number (A+B) that only occured once. Knowing this, Mr. Sum can find the two numbers, and they are:

Answer in white: <font color="white"> A = 4, B = 13 </font>


For a thorough explanation, see http://mathforum.org/library/drmath/view/55655.html

Or search google for "Mr. Sum and Mr. Product," since that is the traditional name of this puzzle.
~MagicMan

[Magic_Man]

[ QUOTE ]
It takes 10 men 8 days to dig 13 holes. How long would it take one man to dig half a hole?

~MagicMan

[/ QUOTE ]

Obviously, this is a silly trick question. There's no such thing as half a hole; that would, in fact, be itself a hole.

~MagicMan

[bigpooch]

Now that baseball season is over, maybe it's time for a
"baseball brain-teaser".

You are the team manager of a baseball team. You have nine
players on the roster that have unusual batting averages.
Call these players A, B, C, ..., I. It turns out that for
EACH of the last nine seasons, the batting averages of A,
B, C, ..., I are such that season after season,

the batting average of A was better than that of B,
the batting average of B was better than that of C,
...,
the batting average of H was better than that of I.

[By "better", I mean strictly better.]

Well, there's nothing usual about that so far. On the other
hand, each of these nine players had only played these nine
seasons, so it wasn't hard for you to compute their league
batting averages over all nine of these seasons. To your
surprise, the overall batting averages were such that

The league batting average for I was better than that for H,
the league batting average for H was better than that for G,
...
the league batting average for B was better than that for A.

How is this even possible?

[Sephus]

players with higher career averages have more at bats in years where everyones averages were higher and/or fewer when their averages were lower?

[bigpooch]

Could you yourself find an example using specific numbers,
or are you simply guessing?

[Sephus]

if players can have any number of at bats in a given year then yes i believe that i myself could come up with an example using specific numbers, but i won't because i expect it to be too tedious.

[Magic_Man]

Let's try it with three players:
Season 1:
A: AB: 10, H: 3, Avg = .300
B: AB: 8, H: 2, Avg = .250
C: AB: 6, H: 1, Avg = .167

Season 2:
A: AB: 10, H: 5, Avg = .500
B: AB: 50, H: 24, Avg = .480
C: AB: 100, H: 47, Avg = .470

Season 3:
A: AB: 10, H: 6, Avg = .600
B: AB: 100, H: 59, Avg = .590
C: AB: 200, H: 117, Avg = .585

League:
A: AB: 30, H: 14, Avg = .467
B: AB: 158, H: 85, Avg = .538
C: AB: 306, H: 165, Avg = .539

Extrapolate to 9 players.

~MagicMan