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Old 10-12-2012, 12:54 AM   #16
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Default Re: A riddle....

17 added
210 multiplied

numbers 7,5,3,2. That would be my guess...
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Old 10-12-2012, 04:56 AM   #17
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Default Re: A riddle....

Well the only way the number is not even when multiplied is if it is only multiplied by odd numbers and then will finish by 5 if 5 was used and by 9 if the 5 wasn't used.

So She would know the number he knows is either a 9 or a 5? I havent really thought much about what the next step would be....

Because if all the numbers were odd, all she had to do was add them up until it made sense and she would know all the numbers right away, so I guess discount that

Last edited by macmac : 10-12-2012 at 05:02 AM.
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Old 10-12-2012, 06:49 AM   #18
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Default Re: A riddle....

15. He knows it's a 7 because of the prime thing already
Mentioned. She knows it's a 1 because she knows how he got 7 and in order toget to 15 the remaining 3 numbers would have to be either 1,3,4 or 1,2,5. He now knows that there is a 1 and knows that she has to know there's a 1 based on the same principles she did. Knowing both of those if the number given to him is 84 he can figure it out by reducing it out same for 70. This is as close as I can get but it's wrong because he'd know a 7 and a 1 the first time
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Old 10-12-2012, 06:54 AM   #19
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Default Re: A riddle....

Quote:
Originally Posted by brwnman
17 added
210 multiplied

numbers 7,5,3,2. That would be my guess...
I thought it was this but it can't be. She'd know all 4 numbers the second time. She answers

Last edited by raiderfan19 : 10-12-2012 at 07:04 AM.
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Old 10-12-2012, 09:42 AM   #20
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Default Re: A riddle....

Quote:
Originally Posted by raiderfan19
I thought it was this but it can't be. She'd know all 4 numbers the second time. She answers

She wouldn't.

if it's 17 added. If Ann knows no number, Bob would know that one number is 7.

Since Bob now knows that one number (7), there are two other alternatives. Either 7,5,3,2 or 7,6,3,1. Ann would now know only 7 & 3. With that, Bob would know all 4 numbers.

Edit - Wait, she might, I am not really taking the multiple into account...

Last edited by brwnman : 10-12-2012 at 09:47 AM.
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Old 10-12-2012, 02:01 PM   #21
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Default Re: A riddle....

Quote:
Originally Posted by brwnman
She wouldn't.

if it's 17 added. If Ann knows no number, Bob would know that one number is 7.

Since Bob now knows that one number (7), there are two other alternatives. Either 7,5,3,2 or 7,6,3,1. Ann would now know only 7 & 3. With that, Bob would know all 4 numbers.

Edit - Wait, she might, I am not really taking the multiple into account...
If it was 7-6-3-1 he would know two numbers (7-1) at the beginning since 126 can only be gotten to by 7-6-3-1 or 7-9-2-1. She would know that and therefore know all 4 of the numbers the second time
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Old 10-12-2012, 02:15 PM   #22
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Default Re: A riddle....

The numbers add up to 18. 7-6-3-2. He can only factor out the 7 originally as either 7 9-4-1 or 7-6-3-2 would come out to the same multiplication answer. However there are simply too many ways to get to 21 for her to know the second number by taking out a 7. It could be 9-4-1 or 8-5-1 or 9-3-2 or 6-5-3 etc. with the knowledge that the second number has to be a 2(to get to 18 she has to have either 6-3-2 or 5-4-2 or 8-1-2. It becomes easy to factor out the other 2 numbers thus the number he was told is 252 and she was told 18 and our original numbers were 7-6-3-2.
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Old 10-12-2012, 02:42 PM   #23
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Default Re: A riddle....

On second thought, I don't think 18 is a solution. I think the only two solutions are sums of 15 and 31.

Last edited by REACTION : 10-12-2012 at 03:59 PM.
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Old 10-12-2012, 03:39 PM   #24
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Default Re: A riddle....

Here are the possible sums and their combinations...

Code:
13 [(1, 2, 3, 7), (1, 2, 4, 6), (1, 3, 4, 5)] 14 [(1, 2, 3, 8), (1, 2, 4, 7), (1, 2, 5, 6), (1, 3, 4, 6), (2, 3, 4, 5)] 15 [(1, 2, 3, 9), (1, 2, 4, 8), (1, 2, 5, 7), (1, 3, 4, 7), (1, 3, 5, 6), (2, 3, 4, 6)] 16 [(1, 2, 3, 10), (1, 2, 4, 9), (1, 2, 5, 8), (1, 2, 6, 7), (1, 3, 4, 8), (1, 3, 5, 7), (1, 4, 5, 6), (2, 3, 4, 7), (2, 3, 5, 6)] 17 [(1, 2, 4, 10), (1, 2, 5, 9), (1, 2, 6, 8), (1, 3, 4, 9), (1, 3, 5, 8), (1, 3, 6, 7), (1, 4, 5, 7), (2, 3, 4, 8), (2, 3, 5, 7), (2, 4, 5, 6)] 18 [(1, 2, 5, 10), (1, 2, 6, 9), (1, 2, 7, 8), (1, 3, 4, 10), (1, 3, 5, 9), (1, 3, 6, 8), (1, 4, 5, 8), (1, 4, 6, 7), (2, 3, 4, 9), (2, 3, 5, 8), (2, 3, 6, 7), (2, 4, 5, 7), (3, 4, 5, 6)] 19 [(1, 2, 6, 10), (1, 2, 7, 9), (1, 3, 5, 10), (1, 3, 6, 9), (1, 3, 7, 8), (1, 4, 5, 9), (1, 4, 6, 8), (1, 5, 6, 7), (2, 3, 4, 10), (2, 3, 5, 9), (2, 3, 6, 8), (2, 4, 5, 8), (2, 4, 6, 7), (3, 4, 5, 7)] 20 [(1, 2, 7, 10), (1, 2, 8, 9), (1, 3, 6, 10), (1, 3, 7, 9), (1, 4, 5, 10), (1, 4, 6, 9), (1, 4, 7, 8), (1, 5, 6, 8), (2, 3, 5, 10), (2, 3, 6, 9), (2, 3, 7, 8), (2, 4, 5, 9), (2, 4, 6, 8), (2, 5, 6, 7), (3, 4, 5, 8), (3, 4, 6, 7)] 21 [(1, 2, 8, 10), (1, 3, 7, 10), (1, 3, 8, 9), (1, 4, 6, 10), (1, 4, 7, 9), (1, 5, 6, 9), (1, 5, 7, 8), (2, 3, 6, 10), (2, 3, 7, 9), (2, 4, 5, 10), (2, 4, 6, 9), (2, 4, 7, 8), (2, 5, 6, 8), (3, 4, 5, 9), (3, 4, 6, 8), (3, 5, 6, 7)] 22 [(1, 2, 9, 10), (1, 3, 8, 10), (1, 4, 7, 10), (1, 4, 8, 9), (1, 5, 6, 10), (1, 5, 7, 9), (1, 6, 7, 8), (2, 3, 7, 10), (2, 3, 8, 9), (2, 4, 6, 10), (2, 4, 7, 9), (2, 5, 6, 9), (2, 5, 7, 8), (3, 4, 5, 10), (3, 4, 6, 9), (3, 4, 7, 8), (3, 5, 6, 8), (4, 5, 6, 7)] 23 [(1, 3, 9, 10), (1, 4, 8, 10), (1, 5, 7, 10), (1, 5, 8, 9), (1, 6, 7, 9), (2, 3, 8, 10), (2, 4, 7, 10), (2, 4, 8, 9), (2, 5, 6, 10), (2, 5, 7, 9), (2, 6, 7, 8), (3, 4, 6, 10), (3, 4, 7, 9), (3, 5, 6, 9), (3, 5, 7, 8), (4, 5, 6, 8)] 24 [(1, 4, 9, 10), (1, 5, 8, 10), (1, 6, 7, 10), (1, 6, 8, 9), (2, 3, 9, 10), (2, 4, 8, 10), (2, 5, 7, 10), (2, 5, 8, 9), (2, 6, 7, 9), (3, 4, 7, 10), (3, 4, 8, 9), (3, 5, 6, 10), (3, 5, 7, 9), (3, 6, 7, 8), (4, 5, 6, 9), (4, 5, 7, 8)] 25 [(1, 5, 9, 10), (1, 6, 8, 10), (1, 7, 8, 9), (2, 4, 9, 10), (2, 5, 8, 10), (2, 6, 7, 10), (2, 6, 8, 9), (3, 4, 8, 10), (3, 5, 7, 10), (3, 5, 8, 9), (3, 6, 7, 9), (4, 5, 6, 10), (4, 5, 7, 9), (4, 6, 7, 8)] 26 [(1, 6, 9, 10), (1, 7, 8, 10), (2, 5, 9, 10), (2, 6, 8, 10), (2, 7, 8, 9), (3, 4, 9, 10), (3, 5, 8, 10), (3, 6, 7, 10), (3, 6, 8, 9), (4, 5, 7, 10), (4, 5, 8, 9), (4, 6, 7, 9), (5, 6, 7, 8)] 27 [(1, 7, 9, 10), (2, 6, 9, 10), (2, 7, 8, 10), (3, 5, 9, 10), (3, 6, 8, 10), (3, 7, 8, 9), (4, 5, 8, 10), (4, 6, 7, 10), (4, 6, 8, 9), (5, 6, 7, 9)] 28 [(1, 8, 9, 10), (2, 7, 9, 10), (3, 6, 9, 10), (3, 7, 8, 10), (4, 5, 9, 10), (4, 6, 8, 10), (4, 7, 8, 9), (5, 6, 7, 10), (5, 6, 8, 9)] 29 [(2, 8, 9, 10), (3, 7, 9, 10), (4, 6, 9, 10), (4, 7, 8, 10), (5, 6, 8, 10), (5, 7, 8, 9)] 30 [(3, 8, 9, 10), (4, 7, 9, 10), (5, 6, 9, 10), (5, 7, 8, 10), (6, 7, 8, 9)] 31 [(4, 8, 9, 10), (5, 7, 9, 10), (6, 7, 8, 10)] 32 [(5, 8, 9, 10), (6, 7, 9, 10)] 34 [(7, 8, 9, 10)]

For a sum of 18, there are multiple combinations that involve 7. Ann can't determine a second number with certainty just by knowing the sum and that one number is 7. And remember, it can't be a unique combination otherwise she would say that she knew all the numbers.

Last edited by REACTION : 10-12-2012 at 04:03 PM.
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Old 10-12-2012, 08:02 PM   #25
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Default Re: A riddle....

Not true. You forget, the 7 isn't the only piece of info she has. She knows that he also can't factor out another number for sure which reduces the possibilities and why the route of 36 is important. It's one of the only numbers that can have 2 completely different factors(2-3-6,9-4-1) another addition way that would get to 18 would be 7-6-4-1. Except that she knows it's not possible because he could have extracted out a 4 and he would have know 2 numbers at the beginning.

It's time for the op to answer though
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Old 10-12-2012, 08:14 PM   #26
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Default Re: A riddle....

How could Bob know two numbers at the beginning? 7 is the only unique prime factor for the bounds of 1 to 10 with no repeats.
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Old 10-12-2012, 08:20 PM   #27
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Default Re: A riddle....

Quote:
Originally Posted by REACTION
How could Bob know two numbers at the beginning? 7 is the only unique prime factor for the bounds of 1 to 10 with no repeats.
Given 7/6/4/1 as the numbers, they they multiply out to 168. We already know he factored out a 7. That leaves 3 numbers to get to 24. There are only 2 3 number combinations that get to 24(assuming we can't reuse numbers) 6-4-1 and 2-3-4. 4 is involved in both of those combinations and he would therefore know a 7 and a 4. This is just an example
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Old 10-12-2012, 08:32 PM   #28
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Default Re: A riddle....

Quote:
Originally Posted by raiderfan19
There are only 2 3 number combinations that get to 24(assuming we can't reuse numbers) 6-4-1 and 2-3-4.

What about 8-3-1?
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Old 10-12-2012, 08:36 PM   #29
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Default Re: A riddle....

Quote:
Originally Posted by REACTION
What about 8-3-1?
Honestly I smooth forgot that. Back to the drawing board. But the factoring part is going to be an important part of the answer
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Old 10-12-2012, 08:37 PM   #30
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Default Re: A riddle....

Yeah, this is a tough (but good) problem. More than a riddle. I'm going to try to program something to help me compute all possible factorizations.
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