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# Savoury

### #21

Posted 19 August 2008 - 04:24 PM

I think there is a leap of logic here, b/c there are exceptions.

For example, if Product Guy knows the Product is 16, the sum could be 8 (4+4) or 10 (8+2).

Let's say the sum is 10 and Sum Guy knows it. Sum Guy says "the sum is less than 14" - which is still true (true), but that doesn't help Product Guy solve anything.

We are relying on Product Guy saying "now I know the numbers" as meaning he DOES KNOW THE CORRECT NUMBERS... but he may not!

So, why can't the numbers be 8 and 2 ?

### #22

Posted 16 September 2008 - 11:30 PM

The first one says he doesnt know the sum, so he knows the product. Another thing we can derive from this that the product has more than 2 divisors, since if the number had only 2 divisors it would be pretty easy to make the product as follows:

x is the product and x = a*b is the only way to write x as a product of smaller numbers then the sum is a + b.

So x can be produced by at least 2 different product like x = a * b = c * d, where a is different from both c and d.

The second 1 says that the sum is smaller than 14.

From this the first 1 knows the numbers. So if x can be written as x = a1 * b1 = a2 * b2 = .... = an * bn, there can be only 1 i in {1,n} for which ai + bi < 14 since the first couldnt guess the numbers if there were more pairs like that.

But from the information that the "product guy" can guess the numbers if he knows that the sum is less than 14 he can guess the numbers too. This means that if his number(the sum) y can be written as y = c1 + d1 = ... = cm + dm there should be only 1 pair say cj,dj such that cj * dj = x is the only way to write x as a product such that the sum is smaller than 14. With such a pair both can decide as the conversation goes in the excercise.

Lets see the solution:

I try for y-s from 13

13 = 7+6 where with product reordering it can be written as 14 + 3 and 21 + 2 so its a good candidate.

13 = 8 + 5 with reordering 10 + 4 , 20 + 2 so it is a good candidate too

since there are at least 2 good candidates the "sum guy" cant decide which is good so the sum cant be 13

12 = 6 + 6 with reordering 12 + 3, 18 + 2 so it is a good candidate

12 = 7 + 5 which is not good becouse it is the only way we can write 35 as a product is 5 * 7 so the "product guy" would know the sum from start

12 = 8 + 4 with reordering 16 + 2 so it makes it a good candidate and

12 fails the same way as 13 with 2 candidates.

(*)

11 = 6 + 5 with reordering 10 + 3 which makes it a bad candidate since the product guy has 30 but he doesnt know if the numbers are 3,10 or 6,5 since they are both below 14.

11 = 7 + 4 with reordering 14 + 2 so its a good candidate

11 = 8 + 3 with reordering 6 + 4 so the problem is the same as with 6 + 5

11 = 9 + 2 with reordering 6 + 3 so the problem is the same as above.

11 can be a sum and 4,7 is a solution pair for the exercise since its a lonely candidate.

Lets see how the conversation goes as a proof of (4,7) solution:

-The first knows 42, but he doesnt know the sum which can be 4 + 7 = 11 or 14 + 2 = 16

-The second tells its below 14.

-Now the first knows the numbers.

-The second can guess the numbers using the discussion from paragraph (*)

There can be more solution pairs which have a sum of less than 11 I didnt try those since 1 solution was enough for me.

### #23

Posted 17 September 2008 - 06:08 PM

### #24

Posted 28 September 2008 - 12:43 AM

### #25

Posted 26 October 2008 - 02:47 PM

The sum is...............the parts of the giraffes left over after the crocs ate.

The product is..........what came out in the end.

**I'd give my right arm to be ambidextrous**.

Bows

### #26

Posted 21 February 2009 - 05:55 AM

You don't have to take all combinations to work this one out. It is based on the following:

If a number is the product of two primes, then you know their sum.

Clues:

A: I don't know the sum. So the number is not the product of two primes. One of the numbers must be 4, 6, 8, 9 or 10.

B: I knew that. This is the biggest clue! If the sum can also be a sum of two primes, then B would not be absolutely sure that A does not know the numbers eg if B's sum was 6, then it could be that A's product is 9=3x3, from which A could immediately deduce the two numbers. So the fact that B knew that A did not know the sum limits the sum to be 11. Every other number under 14 is the sum of two primes: 4=2+2, 5=2+3, 6=3+3, 7=3+4, 8=3+5, 9=2+7, 10=3+7, 12=5+7, 13=2+11.

So the possibilities are 2 9, 3 8, 4 7, or 5 6.

B: The sum is less than 14

A: I knew that. This implies that the product cannot be factored out into two numbers whose sum is larger or equal to 14. Now the largest sum that the factors can add to is when we take a large and a small factor. This eliminates 3x8=24=2x12, which allows 2+12=14;v or 4x7=28=2x14 with 2+14>14; or 5x6=30=2x15, 2+15>14; leaving only 2x9=18=3x6. So the numbers are 2 and 9.

hi dude ididnt get the solution can u pls eloborate it.

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