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/sci/ - Science & Math

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5146098 No.5146098[DELETED]  [Reply] [Original]

We pick four numbers from 1 - 10.

Then we call 2 perfect logicians called Ann and Bob. They know we have four numbers but neither of them know the numbers.

We add those numbers and tell only Ann the result.

We multiply those numbers and tell only Bob the result.

Then we put them in a room together and ask them how many numbers do they know from the addition and multiplication of the numbers:

Ann: I don't know any number
Bob: I only know one number
Ann: I now know two numbers
Bob: I now know all the numbers!

What are the numbers?

>> No.5146105

Whichever four we picked to begin with.

>> No.5146115

This riddle is impossible to solve.

How can Bob determine a number from just the product? Say A * B * C * D = 1492. Now tell me one of those variables.... you can't. Even if primes are involved.

>> No.5146131

>>5146115
Well, if we're given the bounds of 1 to 10 with no repeats, wouldn't there be a number where if it's large enough and a multiple of 10, then 10 would have to be one of the four numbers?

Let's say the numbers are the highest possible set: 7, 8, 9, 10. That would give Bob a product of 5040 and he'd know 10 is one necessary number, right?

>> No.5146127

>>5146098
This looks like fun :)

So first, how can Ann not know any numbers?
It must be 14 or higher, or she would know that a 1 must be present, and it must be 30 or lower, or she would know that a 10 is present.

That is all I have so far, but I am thinking, and I will beat this

>> No.5146142

>>5146115
If it's 1-10, then a factor of 7 can only be due to 7.

>>5146127
yep

>>5146131
no, it could be 2*5
see above

>> No.5146149 [DELETED] 

>>5146127
Me again.
I am wondering if Bob knows 1 of the numbers right from the start, or if it is dependent on the information of the fact that Ann does not know any numbers?

For him to definitely know a number between 14 and 30, I suppose it would have to be a 7.
A value of 14, 21, or 28 would mean that a 7 must be present, whereas all other numbers have multiple combinations, I think, and one could not be certain.

>> No.5146153

>>5146127
Me again.
I am wondering if Bob knows 1 of the numbers right from the start, or if it is dependent on the information of the fact that Ann does not know any numbers?

>> No.5146159

>>5146131
No.
It could be a 5 and a 2, and two other numbers

>> No.5146167

>>5146142
I agree, one of the numbers must be a 7.

>> No.5146168

Shit.. my brain hurts

getting flashbacks of word problems

>> No.5146169

>>5146098
can any two numbers be the same?

>> No.5146171

>>5146149
It can't be 14, 21, or 28 as it would have to be 7 * 1 * CD, where combinations of the other numbers wouldn't make a product of 2, 3, or 4.

>> No.5146174

>>5146167
So if they determine 7 is one number, how does Ann find two others from that?

>> No.5146175

Looks like a system of equations. You'd have to rewrite the fuck out of them though.

>> No.5146178

Nice problem OP.

>> No.5146179

>>5146167
Based on this, the new lower bound is 2,3,4,7 for there to be definitely no 1's, which is 16.

The upper bound is still the same.

For her to now know 2 numbers, she must know that one of the numbers is a 1, and the sum is either 14 or 15.
It can not be 16.

She knows 2 numbers
They are 1, and 7

Out of what is left, for Bob to know them all, it must be a unique combination

I will find it.

>> No.5146186

I think the sum is 15. He knows there's a 7 because of the prime thing already mentioned. She knows there'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... I think I'm stuck

>> No.5146194

Wait, are they sharing their findings with each other or working independently?

>> No.5146199

>>5146179
Me again.

Combinations are 1,7,2,4, if it is 14

1,7,2,5 or 1,7,3,4 If they add up to 15.

These all have unique products, and if he is a perfect logician, he will know this,
So it could be any of these 3, and he knows all of the numbers because he knows the product?

Now I am stuck
:(

Is there a unique solution, OP?

>> No.5146206

7,
4,
1,
3.
This would be my guess

>> No.5146207

>>5146194
They do not tell each other either the sum or the product, the only information that they transfer is how many numbers they know.

We see the entire transcript of their conversation in the OP's post

>> No.5146208

>>5146206
Do not guess.
Prove.

>> No.5146218

>>5146207
Okay, so they only tell each other how many numbers they know, but not what the numbers are?

>> No.5146225

>>5146218
Yes.

>> No.5146228

>>5146218
Correct.

I have narrowed it down to 3 possible combinations, unless there is some problem with my logic.
See >>5146199

>> No.5146235

I see how Bob would find 7, but how would Ann get two numbers just from Bob telling her that he found a number? She doesn't know it's 7, unless she happens to use the same thought process as Bob to arrive at 7. But even then... she doesn't know the product. Only the sum.

>> No.5146267

If Ann knows no numbers, then the sum cannot be 10, 11, 33, or 34, which correspond to
1, 2, 3, 4
1, 2, 3, 5
6, 8, 9, 10
7, 8, 9, 10

Bob can only know one number if 7 is a factor. Ann can deduce this too when Bob states that he knows one number.

Ann then states that she now knows 2 numbers. The sum must therefore be 31, which can only be achieved with:
4, 8, 9, 10,
5, 7, 9, 10,
6, 7, 8, 10

Because Ann knows 7, she can exclude the first choice. She thus knows 10 as well, which Bob can also deduce.

Bob can then determine whether it is
5, 7, 9, 10
or
6, 7, 8, 10

>> No.5146262

>>5146235
>She doesn't know it's 7, unless she happens to use the same thought process as Bob to arrive at 7.

She is a perfect logician
She knows that for him to be certain that he knows one number, one of the numbers is a 7
This narrows down the possibilities, and she knows from this, and from knowing the sum, another number
This can only means she knows the number 1, as only the lower bound shifts due to a 7 being a certainty.

3 combinations remain, and he knows for certain which one, because the products are all different.

7,1,2,4,
7,1,2,5,
7,1,3,4

But from here, where do we go?

>> No.5146264

>>5146235
Wait, nevermind... I guess she could conclude that 7 is the only possible number that he found (without knowing the product) because it's the only number whose factors necessitates its inclusion in the set.

Right?

>> No.5146273

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.

>> No.5146278

>>5146273
I think it's been solved.

>> No.5146279

>>5146267
Just adding that the excluded possibilities in the first step are actually irrelevant to the rest of the logic.

>> No.5146288

>>5146279
one more thing

When Bob states that he knows all the numbers, it is the only time that Ann cannot figure them out herself just from Bob's certainty.

We can thus conclude that Bob is a prick.

>> No.5146290

>>5146267
>Ann then states that she now knows 2 numbers. The sum must therefore be 31
Why?
Why can the sum not be 14 or 15?
The only information she has gained, is that a 7 is one of the numbers

The range of acceptable sums for this is still 13 - 34
You do not know for certain that the extra number that she knows is a 10

She could also know a 1.

>> No.5146296

>>5146208
will it be a proof enough if i show that this can be a valid case ?
If so , then suppose the numbers are 7,1,3,4;
The sum person doesn't know anything.
The product guy knows there's a 7 in them, so he knows one number.
The total sum is 15 , therefore the sum lady now knows that 7 is a number ( since he said he knew one on his first turn ) , so she now knows that others should add up to 8;
Now , he was unsure of the other numbers , this implies that the product when divided by 7 , gave a number in which at least one prime number had its degree more than 1;
therefore she can say that the other numbers can be (3,4,1), (2,6,1) (5,10,1) (8,2,4) ( 8,4,1) or others like that.
She knows their sum so she can fairly say that all the numbers would be less than 7.
That would be the case only with (2,6,1) or ( 3,4,1)
out of these only the latter adds up to 8 , which adds with 7 to give 15.

What are all the fallacies in the above claim ?

>> No.5146300

>>5146273
Oh, this works.
Well done.

>> No.5146310

>>5146290
Nope.

There are a limited number of possible combinations of numbers. She knows the sum. When she learns that 7 is a number, she states that she then knows two numbers. Of all the possible sums, there is only one for which the additional information of 7 would allow her to deduce a second number.

>> No.5146314

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)]

>> No.5146312

>>5146296
That also seems to work, and was one of my 3 combinations.

It seems that there are multiple solutions to this.

>> No.5146317

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)]

>> No.5146321

>>5146310
If the sum is 14 or 15 she would know that one of the numbers has to be 1.

At least 4 possible solutions exist.

>> No.5146327

>>5146296
That doesn't allow Ann to determine a second number. She isn't guessing. She knows for certain.

See
>>5146267
and
>>5146314
>>5146317

Only a sum of 31 allows her to know a second number.

>> No.5146342

>>5146327
>Only a sum of 31 allows her to know a second number.

What about 18? >>5146273

>> No.5146346

>>5146327
Fuck, I take that back. 15 works as well. Bob then knows that it's one of the following:
5, 7, 9, 10
6, 7, 8, 10
1, 2, 5, 7
1, 3, 4, 7
and can determine which one from the product.

>> No.5146359

>>5146327
>Only a sum of 31 allows her to know a second number.
Incorrect

If the sum is 14, and she knows that there is a 7, is there any legitimate combination without a 1?
If the sum is 18, and she knows that there is a 7, is there any legitimate combination without a 2?
If the sum is 15 and she knows that there is a 7, is there any legitimate combination without a 1?

There are multiple values of the sum, that means that she certainly does know the next number
So you are wrong.

>> No.5146363

>>5146346
As does 14, and 18

As I say, multiple combinations of options that force the same dialogue between them.

No unique solution, and we can not be certain of the numbers.
Only Bob can.
:)

>> No.5146375

Just want to say...

Bob > Ann

Women shouldn't be involved in math/hard science.

>> No.5146379

>>5146359
>>5146363
You're both wrong.

15 and 31 are the only ones that work.

For 14:
[(1, 2, 3, 8), (1, 2, 4, 7), (1, 2, 5, 6), (1, 3, 4, 6), (2, 3, 4, 5)]

There is only one combination with 7 that gives a sum of 14. Ann would know all numbers, not 2.

For 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)]

There are multiple combinations that involve 7. She cannot determine a second number with certaintly just by knowing the sum and that one number is 7.

15 and 31 are the only possibilities. I've checked it programmatically.

Remember, it can't be a unique combination otherwise she would say that she knew all the numbers.

>> No.5146391 [DELETED] 

>>5146379
2,3,4,5 is not valid because it does not contain a 7
All other combinations have a 1 in them.


You are still wrong

There are still multiple solutions here.

>> No.5146395

A point of grammar.
>Ann: I now know two numbers
This is also true, in particular, if she knows all the numbers.

>> No.5146397

This is completely reliant on whether or not we can use two of the same numbers from 1-10. (ie. 1 1 2 3)

>> No.5146401

>>5146379
I agreed, the sum must be 31 or 15.

>> No.5146404

>>5146391
Ok, we both agree that Bob determines that 7 is one of the numbers.

Ann knows the sum. If there is a unique combination that involves 7 for that sum, then she would know all 4 numbers.

All sums that involve 7 are listed in >>5146314 and >>5146317

Show me something other than 15 or 31 that allows Ann to determine a second number and no more.

>> No.5146409

>>5146395
Ha, she would not say that though. She would say exactly how many numbers she knows; she would say the maximum.

If she knew all of the numbers, then she would tell Bob this.
She would not troll him.

>> No.5146411

>>5146404
I realised my error. You are correct, and I deleted my post.

>> No.5146416

>>5146397
You can't. No repeats.

>> No.5146417

>>5146411
I figured that after I saw that the post was gone.

An argument on 4chan that didn't involve random ad hominems and general nastiness...

Well played, gents.

>> No.5146419

>>5146416
Alright.

>> No.5146427

>>5146417
That is why I like /sci/ and not /b/
:)

I am glad we agree.
We are done, I think.

>> No.5146438

So the two possible answers are sums of 15 and 31?

What are the possible number combinations then?

>> No.5146452

>>5146438
see >>5146346

>> No.5146457

>>5146452
But the problem seems to be asking for a specific set of numbers.

Are you saying there's like 8+ total possible combinations?

>> No.5146460

>>5146438
31 is not in the bracket where ann wouldn't know 10 was one of the numbers in the first step. so it can only be 15.

>> No.5146471

>>5146460
so A=1, B=7, C and D can be determined from the multiplication that Bob was told about.

>> No.5146472

>>5146460
Oh wow, yes, absolutely correct.

I did not see that.

So the numbers are 1,2,5,7 or 1,3,4,7.

Any way to narrow it down 1 last time?
Is there any reason that either of these would not work with the dialogue that they exchanged?

>> No.5146476

>>5146471
Yes, but the question "What are the numbers?" implies a unique solution.

Perhaps there is a reason why one of these 2 combinations would not lead to the conversation in the OP?

>> No.5146479

So these are all the possible answers?

1, 2, 5, 7 (Sum: 15, Product: 70)
1, 3, 4, 7 (Sum: 15, Product: 84)
5, 7, 9, 10 (Sum: 31, Product: 3150)
6, 7, 8, 10 (Sum: 31, Product: 3360)

>> No.5146488

>>5146472
>Any way to narrow it down 1 last time?

There has to be. I'm not going to start my afternoon fap until I narrow this down.

>> No.5146494

>>5146098
answer is 5,3,4,7

>> No.5146496

>>5146472
Yatta! I have it!

The numbers must be 1,3,4,7
:D


If the numbers were 1,2,5,7, then as soon as Bob is told the product is 70, then he would know that a 7 must be present, and then the last 3 numbers have a product of 10
This is only possible with 1,2,and 5.

Bob does not know all the numbers at this stage, not just 1
So this can not be true.

He is told that the product is 84.
He knows that there must be a 7, and he knows that the last 3 numbers have a product of 12.

This is possible in 2 different ways.
7,1,3,4
Or 7,1,2,6

...Oh
So he knows that there must be a 1?

Should he not know 2 numbers at this stage?

Oh darn, I thought I had it.
:(

>> No.5146500

>>5146496
*Just 1.

>> No.5146519

>>5146496
maybe he wasn't such a perfect logician afterall

>> No.5146521

>>5146496
lol thats what i was just thinking

>> No.5146522

the solution is 10,7,5,9

>> No.5146527

Ok, last piece of the puzzle. We've narrowed it down to:
1, 2, 5, 7 (Sum: 15, Product: 70)
1, 3, 4, 7 (Sum: 15, Product: 84)

1, 2, 5, and 7 are all prime numbers. Bob would have been able to determine them right away as there is only one combination of 4 numbers that gives a product of 70.

84, however, could be
1, 2, 6, 7
or
1, 3, 4, 7
but then Bob would know 2 numbers, not one.

>> No.5146529

>>5146479
1,4,3,7 cannot be the answer

if you multiply them the answer is 84

divide that by 7 which you assume it's the number bob discovered and you get 12 now are there 3 such unique intergers that multiply to 12 that do not involve 1? In other words can you find integers to satisfy this where none of them is equal 1 x*y*z=12

12=2*2*3

here we see that we could only multilply by two unique intergers and therefore Bob would seen that and also would have known that 1 was a number. They would have said he knew two numbers. 7 and 1 but he didn't therfore 1 is not a number

>> No.5146537

>>5146522
that would mean the sum is 31 which would mean Ann would have figured out 1 of the numbers

>> No.5146588

The only way for Bob to know exactly one of the
numbers, that I could think of is if the product
is 252. It could then be either 2,3,6 & 7 or 1,4,7 & 9.

This would mean that Ann's sum would be either 18 or 21.

However, Ann should know that those are the only solutions and by knowing the sum, she'd know all four numbers right away... Why can't I solve this?

>> No.5146622

Is OP and epic troll?

Well fucking played if he is. 10/10

>> No.5146624

>>5146622
No.

>> No.5146645

Guys I think we're wrong in the initial assumption that if Bob knows one number it must be 7.

If the product of 4 numbers is 540 that means either:

(1,6,9,10)
(2,3,9,10)
(2,6,9,5)
(3,4,9,5)

so he would know that 1 number is 9.

>> No.5146641

>>5146622
why? is it unsolvable?

>> No.5146648

>>5146645
>Guys I think we're wrong in the initial assumption that if Bob knows one number it must be 7

NO.... WE'VE COME TOO FAR TO TURN BACK NOW

IT HAS TO BE 7

>> No.5146655

Is this a Turing Award problem? Cuz this shit is hard even for me, and I've had calculus 2 (including integrals and even Taylor series)

>> No.5146660

http://en.wikipedia.org/wiki/Divisibility_rule
Obviously these are in play.

>> No.5146662

>>5146648
But we followed the reasoning that one number is 7 and it doesn't work

>> No.5146671

This seems like it should be basic fucking algebra but I'm not seeing a solution.

>> No.5146669

>>5146648
we did

I'm working it out right now.
Bob can determine 1, 7, 8 or 9 depending on the product.

I'll post a full solution as soon as I have it.

Awesome question, OP.

>> No.5146676

>>5146314
>>5146317

Out of curiosity, what did you use to calculate those combinations?

>> No.5146679

>>5146671
I think this is more advanced than just algebra. Seems like Algebraic Combinatorics or something.

>> No.5146687

I've arrived at 1 and 7. The other two should be below 7.

>> No.5146695

combinations: 210

Ann eliminates (1, 2, 3, 4)
Ann eliminates (1, 2, 3, 5)
Ann eliminates (6, 8, 9, 10)
Ann eliminates (7, 8, 9, 10)
remaining combinations after Ann's first statement: 206

>> No.5146696

Bob could know 9 if the product is 540 [(1, 6, 9, 10), (2, 3, 9, 10), (2, 5, 6, 9), (3, 4, 5, 9)]
Bob could know 9 if the product is 1080 [(2, 6, 9, 10), (3, 4, 9, 10), (3, 5, 8, 9), (4, 5, 6, 9)]
Bob could know 1 if the product is 60 [(1, 2, 3, 10), (1, 2, 5, 6), (1, 3, 4, 5)]
Bob could know 7 if the product is 630 [(1, 7, 9, 10), (2, 5, 7, 9), (3, 5, 6, 7)]
Bob could know 8 if the product is 960 [(2, 6, 8, 10), (3, 4, 8, 10), (4, 5, 6, 8)]
Bob could know 9 if the product is 2160 [(3, 8, 9, 10), (4, 6, 9, 10), (5, 6, 8, 9)]
Bob could know 7 if the product is 168 [(1, 3, 7, 8), (1, 4, 6, 7), (2, 3, 4, 7)]
Bob could know 7 if the product is 210 [(1, 3, 7, 10), (1, 5, 6, 7), (2, 3, 5, 7)]
Bob could know 9 if the product is 216 [(1, 3, 8, 9), (1, 4, 6, 9), (2, 3, 4, 9)]
Bob could know 7 if the product is 1260 [(2, 7, 9, 10), (3, 6, 7, 10), (4, 5, 7, 9)]
Bob could know 7 if the product is 252 [(1, 4, 7, 9), (2, 3, 6, 7)]
Bob could know 9 if the product is 270 [(1, 3, 9, 10), (1, 5, 6, 9), (2, 3, 5, 9)]
Bob could know 7 if the product is 280 [(1, 4, 7, 10), (1, 5, 7, 8), (2, 4, 5, 7)]
Bob could know 8 if the product is 288 [(1, 4, 8, 9), (2, 3, 6, 8)]
Bob could know 7 if the product is 560 [(1, 7, 8, 10), (2, 4, 7, 10), (2, 5, 7, 8)]
Bob could know 7 if the product is 840 [(2, 6, 7, 10), (3, 4, 7, 10), (3, 5, 7, 8), (4, 5, 6, 7)]
Bob could know 7 if the product is 336 [(1, 6, 7, 8), (2, 3, 7, 8), (2, 4, 6, 7)]
Bob could know 7 if the product is 1680 [(3, 7, 8, 10), (4, 6, 7, 10), (5, 6, 7, 8)]
Bob could know 8 if the product is 1440 [(2, 8, 9, 10), (3, 6, 8, 10), (4, 5, 8, 9)]
Bob could know 7 if the product is 420 [(1, 6, 7, 10), (2, 3, 7, 10), (2, 5, 6, 7), (3, 4, 5, 7)]
Bob could know 9 if the product is 432 [(1, 6, 8, 9), (2, 3, 8, 9), (2, 4, 6, 9)]
Bob could know 7 if the product is 504 [(1, 7, 8, 9), (2, 4, 7, 9), (3, 4, 6, 7)]
remaining combinations after Bob's first statement: 68

>> No.5146701

Ann could know [2, 3] if the sum is 16 [(1, 2, 3, 10), (2, 3, 4, 7)]
Ann could know [9, 10] if the sum is 29 [(2, 8, 9, 10), (4, 6, 9, 10)]
remaining combinations after Ann's second statement: 4

product 60 with combination (1, 2, 3, 10)
product 2160 with combination (4, 6, 9, 10)
product 168 with combination (2, 3, 4, 7)
product 1440 with combination (2, 8, 9, 10)
Bob can then determine the combination because he knows the product.

>> No.5146707

>>5146676
Python.

Same thing that I used for
>>5146695
>>5146696
>>5146701


Not sure if I'm still missing something.

>> No.5146716

>>5146707
Can I see the python code to calculate sum combinations?

>> No.5146724

>>5146716
http://pastebin.com/b9Y2XZRe

>> No.5146740

I narowed it down to two posibilities:
2 8 9 10
4 6 9 10
the sum is definitely 29
i got this result by brute force, i wrote a short program that calculated all the posibilities, than i started to pick them out, these two remained

>> No.5146745

There are so many product combinations that have a factor of 7 though.

>> No.5146748

>>5146740
Compare it to my code above. I was not able to eliminate 16. Which one of us made the mistake?

You may as well paste-bin your code too.

>> No.5146749

>>5146701
by your logic this works better (7,5,4,3)

sum is 19

bob found 7 so sum is now 12

leaving three variables

the sums can only in combination of (2,6,4) or (3,4,5) because 9,7,8,1 cannot be use because bob found 7 which would lead to finding 1 and making 9 and 8 impossible to sum to 12 using two more intergers. So (2,4,6) and (3,4,5) here we sse that the for is in common and therefore is the second number found. This leaves (2,6) or (3,5) depending on the product that bob knows.

Prove me wrong

>> No.5146750

>>5146740
So which number are you saying Bob found first and how?

10? 9?

>> No.5146756

I really liked the problem and the discussion around it.

What stuff do I need to know to solve this type of problems?

Thank you.

>> No.5146760

>>5146749
ladies we have a winner

>> No.5146758

>prime factorization

>> No.5146762

>>5146756
Factors of sums.

There are combinatorics tricks which could probably solve general cases of this sort of problem, but in the specific it's just algebra and number theory.

>> No.5146767

>>5146748
>>5146750
my code just calculated the possibilities, then i started to handpick the wrong ones out like an autist (i see the potential of mistakes here). i cant say which one bob guessed at first, i left in all the possibilities when he could guess exactly one number, but there was not a specific one

>> No.5146764

>>5146749
Read through the thread. Assuming that Bob finds 7 (due to it being a unique prime factor) leads to a contradiction.

See the brute force methods above. They follow the process of elimination logically and handle all cases.

>> No.5146778

>>5146764
what contradiction is that? Prove where my answer is contracditive?

>> No.5146781

>>5146767
I think you made a mistake when manually selecting them.

>> No.5146789

I think the numbers are (1,2,5,8)

product is 80, :

lets say Bob can determine one number is 1

So if Ann can determine other number it is 8 from possibilities:

(1,2,5,8)
(1,3,4,8)

Since Bob chose from

1*2*5*8=80
1*3*4*8=96

it must be (1,2,5,8) since if it was 96 he could determine all 4 immediately

tell me if I'm wrong

>> No.5146794

>>5146778
It was already eliminated in one of the previous steps. Do you want me to add explicit output to the program so you can see exactly which step eliminates it?

>> No.5146802

I have a similar but easier problem:
A bartender chats with a guest. He says he has three kids and the product of their ages is 72. The guest says that's not enough information to tell their ages. The bartender tells him to go out and look at the house number of the pub, that is the sum of their ages. The guest complains he still cant tell how old are they. the bartender replies: the youngest one likes strawberry ice cream.
How old are the kids?

>> No.5146809

>>5146789
I think my 1st step is wrong if it was 80 he would be able to determine 2 numbers

>> No.5146828

>>5146794
>>5146794
I don't trust your program. 7 is the number that can be eliminated first. It's common sense. The answer (3,4,5,7) or (2,6,4,7)

>> No.5146832

>>5146828
Here, with verbose output.

http://pastebin.com/Ta6QUaCJ

The program is pasted above too. You can check the logic yourself.
>>5146724

>> No.5146835

>>5146701
the 1,2,3,10 is not good, because the combination 1,3,4,5 falls out at the beginning: the tum is 13 and Ann would know 1 number if the sum were 13. This leaves two combinations with the sum of 60: (1,2,5,6) and (1,2,3,10). if the product were 60 bob would know 2 numbers

i think 2,3,4,7 is also wrong but i dont know why yet

>> No.5146840

>>5146835
i ment to say product of 60

>> No.5146849

>>5146828
Relevant lines for (3,4,5,7) or (2,6,4,7)

For the sum 19, Ann considers [(1, 3, 7, 8), (1, 5, 6, 7), (2, 3, 5, 9), (2, 3, 6, 8), (2, 4, 6, 7), (3, 4, 5, 7)]
Ann could not determine two and only two numbers if the sum were 19
Ann eliminates [(1, 3, 7, 8), (1, 5, 6, 7), (2, 3, 5, 9), (2, 3, 6, 8), (2, 4, 6, 7), (3, 4, 5, 7)]

You are wrong. Look it over. The answer has been posted above.

>> No.5146857

>>5146835
Look at the verbose output.
>>5146832

All combinations are in numerical order (e.g. "2, 4, 6, 8"). You can search for them and see exactly where they are eliminated and why.

Feel free to inspect the logic.

>> No.5146884

>>5146857
Yes you forget that when the sum is 13 or 12, all the possible combinations include the number 1, therefore Ann would know that number when asked first time. Sums 31 and 32 should also be excluded for similar reasons.

>> No.5146897

Some statements were missing:
http://pastebin.com/mmdixFwf

>> No.5146902

>>5146884
nice catch

updating

>> No.5146924

New output
http://pastebin.com/v8Gw1NR9

Conclusion:
product 2160 with combination (4, 6, 9, 10)
product 1440 with combination (2, 8, 9, 10)
Bob can then determine the combination because he knows the product.

Is there any way to eliminate one of those possibilities?

>> No.5146937

>>5146924
thatts what i said an hour ago, and no i dont have any idea :(

>> No.5147068

>>5146645
you sir are one SICK FUCK

>> No.5147100

If anyone is still working on this, here's some nice tabulated output from the brute-force method.
http://pastebin.com/uK2Jfhu3

>> No.5147191

Someone tell me why 2, 3, 6, 7 isn't the answer.

>>5146273
>>5146273

>> No.5147209

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, 1-4-9) another addition way that would get to 18 would be 1-4-6-7. 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.

>> No.5147228 [DELETED] 

>>5147191
Forgot the possibility of 1, 4, 6, 7 which adds up to 18 and multiplies to 168 (7 x 24) So she never finds out that the second number is 2.
2, 4, 5, 7
2, 3, 6, 7
2, 8, 1, 2

>> No.5147255

>>5147191
See the second table for Ann in >>5147100

>> No.5147339

>>5147255

So it has two solutions? Can't be narrowed down?

Product 2160 with combination (4, 6, 9, 10)
Product 1440 with combination (2, 8, 9, 10)

>> No.5147346

>>5147100
That table only works if Bob and Ann share their combinations. They don't.

>> No.5147374

>>5147339
There seem to be two solutions in the absence of more information.

>>5147346
Actually, they do. They're perfect logicians. When Bob says that he knows 1 number, Ann can eliminate all combinations for which Bob could not know exactly 1 number. She can do this because she knows that Bob is a perfect logician and he knows the product.

Ann is left with a subset of combinations. Bob can also determine which subsets Ann could be left with. Ann then says that she knows 2 numbers. Bob can determine which of the remaining combinations would give her that information for a given sum.

Etc.

Either you have misunderstood the tables, or you have misunderstood the problem.

>> No.5147387

>>5147100
That's a sexy tabulated output anon.

>> No.5147395

>>5147374
But the process of eliminating sets of numbers is different for the product and sum. Ann only knows the sum, and Bob only knows the product.

I can see Ann finding the 7, but after that she'd have to know the product to find the same sets that Bob finds, no?

>> No.5147409

>>5147395
That's not how it works. When Bob says that he knows 1 number, Ann doesn't know which one. All she knows is that there are different products that would allow Bob to know 1 number. Look at the second table.

Ann knows that if the product is 168 then Bob would know that 7 is one of the numbers. The same for 210. She knows if the product is 216, then Bob would conclude that 9 is one of the numbers, etc. She doesn't know what product it is or what number Bob knows, but she can narrow down the number of combinations to the products that would allow Bob to know 1 number.

That leaves here with a limited set of combinations. She knows the sum, so with that sum she then says that she knows 2 numbers. Bob can consider all possible sums of the remaining combinations and figure which sums would allow Ann to know exactly 2 numbers, etc.

Just read through the output from the top. It doesn't require knowledge of the sum or the product that either Ann or Bob know. Each step is deduced entirely from the statements they make and the possible combinations that could correspond to those statements.