# number riddle

nescio | 00:45 Mon 14th May 2007 | Riddles
This seems to make no sense whatever at first sight, but a bit of logical thought soon leads you to the correct answer. This appeared ( worded differently ) as the Enigma puzzle in New Scientist a few years ago. Here goes.....

I have two numbers in mind, both positive integers.
I tell two people the first number, then reveal the following information about the second:

I tell both the two people above that the second number is less than or equal to the first number, and then whisper to the first person ( so that the second can not hear) the letter that the second number begins with. I then whisper to the second person ( so that the first can not hear ) the letter that the second number ends with.

The two people then have the following conversation:
1st : I don't know what the second number is.
2nd: Neither do I.
1st: I still don't know what it is.
2nd: I know what the second number is now.
1st: So do I.

What are the two numbers ? Note: words used are not important, merely information conveyed.

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No best answer has yet been selected by nescio. Once a best answer has been selected, it will be shown here.

Surely when the 1st says he doesn't know for the first time then this means the number has to be in one of the pairs : Four/Five
Two/Three
Six/Seven

When the 2nd says he doesn't know this tells the 1st that the number must end in E (the only common last letter) and that therefore the number must be Five or Three depending on the first letter that the 1st has already been told ?

dsg

The first person has no way to distinguish between 4,5,14,15,40 etc
or 6,7,16,17,63,etc
or 2,3,10 ,12 13, 20,21,22, etc
ther second person cannot distiguish between
3,5,9,12,23, etc
or 7,10,11,13,14,15, etc

they both know the first number, which sets an upper bound for the second number; you don't yet.
you have to determine both numbers -the solution is unique.
Oops !! I was assuming the integers had to be one digit !!

How about 12 & 10 for the numbers ?
To narrow it to one possibility we must assume the lowest number that qualifies.

We are given that #2 <= #1.

First and last letters of 1-10:

TN
NE
ET
SN
SX
FE
FR
TE
TO
OE

Here�s a hypothetical logical analysis of the conversation:
(within quotions logically inferred)

If #2 = 1 this would have been obvious to P1 from the start if <100 due to the exclusive first letter �o�.

P1: �#2 is > 1�

If #2 = 2 this would likewise have been obvious to P2 from the start if <22 due to the exclusive final letter �o�.

P2: �You would have known if #2 was 1, and I know it isn�t 2, therefore #2 > 2�

P1: �You would have known if #2 was 2, and I know it doesn�t begin with �t� therefore #2 > 3�

P2: �#2 is 5�

4 would also have been obvious to P2 from the start due to the exclusive �r�.

P1: �Since you now know what #2 is I now also know that #2 must be 5�

#1 must also be 5, otherwise there would be an alternative value for #1 and its value could not be resolved within the information provided?
mibn2cweus

why have you restricted your argument to the numbers 1-10?
The problem as stated said that the numbers were positive integers- the first in particular, since it sets an upper bound for the second, could be as high as you like.
If the solution is unique and my previous answer of 12 & 10 works then it must be right.

A will know the number begins with T & so is 12,10,3 or 2.

B knows the number ends in N & so is 11,10 or 7 and now knows from A (as A doesn't know the number) that it must begin with T,E,S or F (the only letters which start more than one number in the range.

A now knows by the same logic that the number must end in E or N & so can eliminate the number 2 but is still left with a choice of three numbers 12,10 or 3.

B now knows, as A is still uncertain, that A is left with a choice of numbers that start & end with the same letters - this only happens with numbers starting with T - and so B can now conclude that the number is 10 !!

A having heard that B now knows the answer can conclude that B has worked out the starting letter and it matches only one of his choices - this can only happen with numbers ending in N - A now knows the number is 10 !!

Question Author
Dsg.

Yes 12 and 10 are the numbers.

To other AB ers, the first step in the solution is finding an upper limit for the first number.
Because all final letters repeat E,O,E,R,E,X,N,T,E,Y after 20, it becomes impossible to distinguish them, therefore the first number must be less than 20.
Also, all the teens end in N, so 13 is an absolute maximum possible value.

The first person does not know the second number, so we can eliminate 1 and 9, since these begin with unique letters.
The second person still does not know the second number , so we can eliminate 2,4,6,8.
The first person still does not know the second number , so we get rid of 5,7.
We now know that the first person heard the letter T, and are left with 3,10,12 ( and possibly 13)
The second person now knows the second number, so must have heard N, which determines it as 10, and eliminates 13, since this would duplicate N
Note this is the step you must take, since the people in the problem already know the second number is <= 12
The first person now knows that the second person must have heard N, so now knows the number is 10 too.

Hope that clears it up.
Nightmare, If I only needed 1-5 for the solution I saw no point in posting more numbers, stopping at ten was arbitrary.
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In testing my hypothesis I found it to be fallacious:

Suppose they were both told the first number (and therefore the upper limit) was five.

P1 knows the second number begins with �f� and so could be four or five.
P2 knows the second number ends with �e� and so could be one, three or five.

Conversation results:

P1 doesn�t know if #2 is 4 or 5.

P2 knows #2 isn�t 1 because P1 would have known it was 1 from the unique first letter �o� but P2 still cannot determine if #2 is 3 or 5.

P1 now knows prematurely #2 is five because P2 would have known it was 4 by the unique last letter �r�.

P2 still cannot determine whether #2 is 3 or 5.

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JEANALICE

macca81