[Guest]

There are two intelligent people (A and B) and a 5 digit palindromic number (a number that reads the same forwards and backwards). A is told the sum of the digits of the number. B is told the product of the digits of the number. Both A and B know all these details, but not the 5 digit palindrome or the other person's number. They go on to have the following conversation:

A to B: "I don't know what your number is."

B to A: "I don't know what your number is.

A to B: "I don't know what your number is."

B to A: "I know what your number is."

What was the original palindromic number?

[bonanova]

There are two intelligent people (A and B) and a 5 digit palindromic number (a number that reads the same forwards and backwards). A is told the sum of the digits of the number. B is told the product of the digits of the number. Both A and B know all these details, but not the 5 digit palindrome or the other person's number. They go on to have the following conversation:

A to B: "I don't know what your number is."

B to A: "I don't know what your number is.

A to B: "I don't know what your number is."

B to A: "I know what your number is."

What was the original palindromic number?

Are there three distinct digits in the original number?

Or do we not need to know that?

[Guest]

There are two intelligent people (A and B) and a 5 digit palindromic number (a number that reads the same forwards and backwards). A is told the sum of the digits of the number. B is told the product of the digits of the number. Both A and B know all these details, but not the 5 digit palindrome or the other person's number. They go on to have the following conversation:

A to B: "I don't know what your number is."

B to A: "I don't know what your number is.

A to B: "I don't know what your number is."

B to A: "I know what your number is."

What was the original palindromic number?

I don't think there is a unique number ...

[Guest]

Are there three distinct digits in the original number?

Or do we not need to know that?

You don't need to know that.

[Guest]

I don't think there is a unique number ...

You are partly right, in that there are two solutions to this answer. Namely, if the palindrome is abcba then bacab is also a solution. However, with this exception, there is a unique answer to this!

[Guest]

I think I have it. Please correct me if im wrong (which I probably am). Well I think that the number is 00000. Am I right?

[Guest]

it makes the most sense out of any of them

[Guest]

I think I have it. Please correct me if im wrong (which I probably am). Well I think that the number is 00000. Am I right?

Sorry, no. If that was the number then the sum would be 0 and the product would be 0. But that's the only combination that can give a sum of 0, so A would know straight away what B's number was...

Although not necessarily vice versa...

[Guest]

12321 or 12121 (ababa) or binary 10101

Sorry for guessing but struggling to see it!

[Guest]

Sorry for guessing but struggling to see it!

12321 or 12121 (ababa) or binary 10101

Bad luck. This is the first puzzle I've posted on here and it seems to have people stumped. I don't want to completely give the game away, but to help things along here's a pretty big clue:

Start by asking "What are the possible solutions to a revised version of the question where A's first comment is "I know your number" and there are no further comments? (The answer to this alone should prove that the 10101 guess is incorrect)

If you continue to consider revised versions of the question where "I know your number" is said earlier, you will find there are multiple possible solutions to each one. The original question as posed is the only version where there is a unique solution. ("Unique" meaning there is only one set of values for a, b, c where a, b, c make up the two palindromes abcba and bacab.)

[bonanova]

Great clues ... answer is imminent, I think.

[Guest]

So here's what I think...

The answer can NOT have a zero in it. If it has a zero, there is no way either person can guess - 23032 - product = 0, sum = 10. You can come up with other ways to get a sum of 10, so no way person can KNOW what other's number is.

IF person has product of 28,224, though, he can quickly deduce the only 5 numbers that can produce that product are 78987 (which adds up to be 39, the other person's number). So my answer here is 78987 or 87978.

11111. Sum = 5, Product = 1. What 5 numbers multiplied together = 1? The only reason I DON'T think it's this is because B would have IMMEDIATELY known what A's # was if he had a product of 1. He didn't know until 2nd round, so that's why I like first answer better. This was a GOOD one (and if my answer is wrong, I retract that thought

)

[bonanova]

Sum = 36 Product = 10368 -> Number is 98289 or 89298

In dealing with the number, since abcba and bacab have the same sum and product let's assume a>=b.

For a sum of 36, A knows there are 9 possible numbers and products:

98289 10368
97479 15876
96669 17496
95859 16200
88488 16384
87678 18816
86868 18432
77877 19208

99099	 0

So A has to say "I don't know your number" For these 9 product values, A knows that B can deduce these possible numbers and sums:

0 a0c0a - a whole slew of sums
10368 66866 - 32
10368 94849 - 34
10368 98289 - 36
15876 97479 - 36 B would say "I know your number"
17496 96669 - 36 B would say "I know your number"
16200 95859 - 36 B would say "I know your number"
16384 88488 - 36 B would say "I know your number"
18816 87678 - 36 B would say "I know your number"
18432 86868 - 36 B would say "I know your number"
19208 77877 - 36 B would say "I know your number"

0		 ab0ba - a whole slew of sums

Since B says "I don't know your number" A eliminates the last seven cases, leaving the first two: 0 or 10368.

So A still has to say "I don't know your number".

Crap, I can't read the notes I made - and it's late.

But, I believe B now knows A's number is 36.

[Guest]

Sorry if Someone has solve it. But my guess is

First of all, none of the digits can be zero. Because if either 1 digit is zero, the product will be zero as well and B will never know what number will A has after the first loop. Too many possibilities to get combination to become zero for the product.

So if this is the case. Then we can start will all 5 digits = 1. Forgive me as I am using the very slow step-by step manner to figure:

As you can see, the first combination that the conversation can happen as what neida describe will be when the number is

1 1 4 1 1

which mean A has the number of 8 and B has the number of 4

Correct?

Edited March 21, 2008 by woon

[Guest]

Sorry guys. Everyone's starting to think along the right lines, but aren't quite there yet. Let me know if you need more clues or want me to explain why any answers so far are wrong and I'll put it in a spoiler.

This is a great one - I heard it many years ago and it took me a while to get it, but have always remembered it as one of my favorites! Glad to see it's got people puzzled here too!

[Guest]

Yes, please explain why the recently submitted answers are wrong - I'd love to figure this one out so I can sleep tonight

[Guest]

OK - here's why answers so far aren't right, by person so you can only look at your own if you wish...

Proposed answer: 78987 - Sum: 39, Product: 28,224

If A has a sum of 39, that can be made up in quite a few different ways, so A can't say what B's number is.

If B has a product of 28,224, that can only be made with 78987 (or 87978), so B would know that A's number is 39. Hence B would be able to say the answer straight away, which he does not in the puzzle so this can't be the answer.

Proposed answer: 11111 - Sum: 5, Product: 1

Similar to the first - if A has a sum of 5, he can't say what B's number is. But if B has a product of 1, he knows the palindrome must be 11111 so knows that A's number is 5. So he wouldn't say that he didn't know what A's number was.

Proposed answer: 98289 - Sum: 36, Product: 10,368

This is possibly the closest so far, but just falls down because you missed one thing in your reasoning. Your step 1 reasoning is spot on, but when you get on to step 2 if the product was 15,876, B would only know that A's number was either 36 (97479) (as you said) or possibly 35 (67976) (which you missed), so he wouldn't be able to say he knew A's number. It's only a small miss, but it makes all the difference!

Proposed answer: 11411 - Sum: 8, Product: 4

If A's sum is 8, then he knows that B's number must be 0, 4 or 8 (I won't bother listing out all the ways here) so he can only say that he doesn't know B's number.

If B's product is 4, then he knows that A's number must be 7 or 8, so he can only say that he doesn't know A's number.

Looking at what they can each deduce from each others statement:

From B's perspective: A could be 7 or 8. If A's number was 7 he would only know that Bs number would be 0, 3 or 4. If A's number was 8 he would only know that Bs number would be 0, 4 or 8. So, regardless of whether A's number is 7 or 8 he would say "I don't know what your number is", so B can't deduce anything.

From A's perspective: B could be 0, 4 or 8. If B's number was 0 he would have a huge list of possibilities for A. If B's number was 4 he would know A's number is 7 or 8. If B's number was 8 he would know A's number was 8 or 12. So regardless of what B's number was, he would still say "I don't know what your number is", so A can't deduce anything.

We haven't deduced anything so far, and if we carry on we'll just go through this loop indefinitely, so neither could actually figure out what the other persons number was.

I think your logic was falling down at saying "you can't be 7, because if you were you'd know I'd be 4", whereas as you can see above, all it would actually tell A is that B was either 0, 3 or 4, and there is no reason to think that it shouldn't be any of these possibilities.

Hope that all helps!

Really looking forward to seeing who gets this one!

[Guest]

OK, this seems to have got everyone stuck. On request from grey cells (on a different thread), here's the answer for anyone keen to know:

Just before giving the answer, I'll explain the reasoning. Some of you may then wish to find the answer and check back before looking at the actual answer.

This one is far easier to solve by looking for all possible solutions, rather than looking at an individual answer and trying to prove it (disproving the previous guesses was far harder work than solving the puzzle!!) I also find it easier to do this with the help of a spreadsheet to keep track of the numbers.

Start by writing down all the possible palindromes - there are 1,000 of them. An easy way to do this is to write the numbers 0 to 999 and use the hundreds, tens and units figures as your a, b and c for the palindrome abcba. Against each palindrome, put in the sum and product. (You can see why a spreadsheet is useful.)

(A to B: I don't know what your number is): Now, if A initially knew B's number, that would mean he would have a sum which only had one corresponding product. E.g. if his sum was 0, B's number could only possibly be 0, so A would know it. The key here is to identify all the sums that only have one corresponding product (either by manually sorting and checking, or by using pivottables, or whatever way you find easiest). You should find that 0, 1, 2, 3, 4, 44 and 45 are all sums that only have 1 product associated with them. As A doesn't know B's number, we know that the sum can not be one of these values, so we can delete all lines from the spreadsheet that have a sum equal to one of these values.

(B to A: I don't know what your number is): Now, we have a reduced list of possibilities. If B knew A's number then he must have a product that has only one corresponding sum in the reduced list. Using a similar method to before we want to find all such products. There are over 150 of them at this stage and, because B doesn't know A's number, we know that the product can not be one of these values, so we can delete all lines from the spreadsheet that have a product equal to one of these values.

(A to B: I don't know what your number is): We now go through this again from A's perspective (looking for sums that only have one corresponding product), but with the new reduced list of possibilities. You will find that we can now knock another 5 possibilities off what the sum must be and reduce the list further still.

(B to A: I know what your number is): Finally, B says that he knows A's number. Again find any products that only have one corresponding sum. At this point you will find there is only one, and so you have your answer.

Hopefully all this makes sense, but let me know otherwise.

Anyone wanting to give this one last try should now do so before looking at the actual solution.

Sum: 37

Product: 20,736

Palindrome: 68986 (or 86968)

Edited March 23, 2008 by neida

[Guest]

neida,

I have follow your instruction.

It is true that your method really works!!!

Thanks for sharing this puzzle.

[grey cells]

OK, this seems to have got everyone stuck. On request from grey cells (on a different thread), here's the answer for anyone keen to know:

Just before giving the answer, I'll explain the reasoning. Some of you may then wish to find the answer and check back before looking at the actual answer.

This one is far easier to solve by looking for all possible solutions, rather than looking at an individual answer and trying to prove it (disproving the previous guesses was far harder work than solving the puzzle!!) I also find it easier to do this with the help of a spreadsheet to keep track of the numbers.

Start by writing down all the possible palindromes - there are 1,000 of them. An easy way to do this is to write the numbers 0 to 999 and use the hundreds, tens and units figures as your a, b and c for the palindrome abcba. Against each palindrome, put in the sum and product. (You can see why a spreadsheet is useful.)

(A to B: I don't know what your number is): Now, if A initially knew B's number, that would mean he would have a sum which only had one corresponding product. E.g. if his sum was 0, B's number could only possibly be 0, so A would know it. The key here is to identify all the sums that only have one corresponding product (either by manually sorting and checking, or by using pivottables, or whatever way you find easiest). You should find that 0, 1, 2, 3, 4, 44 and 45 are all sums that only have 1 product associated with them. As A doesn't know B's number, we know that the sum can not be one of these values, so we can delete all lines from the spreadsheet that have a sum equal to one of these values.

(B to A: I don't know what your number is): Now, we have a reduced list of possibilities. If B knew A's number then he must have a product that has only one corresponding sum in the reduced list. Using a similar method to before we want to find all such products. There are over 150 of them at this stage and, because B doesn't know A's number, we know that the product can not be one of these values, so we can delete all lines from the spreadsheet that have a product equal to one of these values.

(A to B: I don't know what your number is): We now go through this again from A's perspective (looking for sums that only have one corresponding product), but with the new reduced list of possibilities. You will find that we can now knock another 5 possibilities off what the sum must be and reduce the list further still.

(B to A: I know what your number is): Finally, B says that he knows A's number. Again find any products that only have one corresponding sum. At this point you will find there is only one, and so you have your answer.

Hopefully all this makes sense, but let me know otherwise.

Anyone wanting to give this one last try should now do so before looking at the actual solution.

Sum: 37

Product: 20,736

Palindrome: 68986 (or 86968)

Thanks for posting the answer and method of solving it .But I am not able to make sense of it at present.Maybe I will get hold of it tomorrow.Have to sleep.And by the way I tried for 196 till 30 odd iterations and till now I have not got the answer . But i will keep trying . Maybe I may become a maths nobel laureate(not sure of spelling)

[Guest]

Thanks for posting the answer and method of solving it .But I am not able to make sense of it at present.Maybe I will get hold of it tomorrow.Have to sleep.And by the way I tried for 196 till 30 odd iterations and till now I have not got the answer . But i will keep trying . Maybe I may become a maths nobel laureate(not sure of spelling)

I hate to end your dream, but there is NO Nobel Prize in mathematics...

[Guest]

Crap - I have the correct number written next to my original solution, 78987. Guess if I weren't such an ADD person I may have pursued that. Nice problem, though!

[grey cells]

I hate to end your dream, but there is NO Nobel Prize in mathematics...

Hey. I was just joking . I know that there is no such thing as a mathematics nobel lauraete.What do you think the was for?

And anyway before posting anything here I would definitely cross-check it . The world's top brains are here(P.S.Not joking) And thanks for replying to my post.

Hey thinking again my discovery could force the noble prize organisers to rethink the categories.I could become the first Mathematics nobel lauraete.

[grey cells]

Thanks Neida . Finally I got hold of the reasoning.I Agree that with 196 You have the upper hand .