Braingle: 'Mathematicians' Brain Teaser

Mathematicians

Math brain teasers require computations to solve.

Fun:	(2.09)
Difficulty:	(3.15)
Category:	Math
Submitted By:	rajeshwar

There are 4 mathematicians - Brahma, Sachin, Prashant and Nakul - having lunch in a hotel. Suddenly, Brahma thinks of 2 integer numbers greater than 1 and says, "The sum of the numbers is..." and he whispers the sum to Sachin. Then he says, "The product of the numbers is..." and he whispers the product to Prashant. After that, the following conversation takes place :

Sachin : Prashant, I don't think that we know the numbers.
Prashant : Aha! Now I know the numbers.
Sachin : Oh, now I also know the numbers.
Nakul : Now I also know the numbers.

How did they know the numbers?

Answer

The numbers are 4 and 13.

As Sachin is initially confident that they (i.e. he and Prashant) don't know the numbers, we can conclude that -
1) The sum must not be expressible as sum of two primes, otherwise Sachin could not have been sure in advance that Prashant did not know the numbers.
2) The product cannot be less than 12, otherwise there would only be one choice and Prashant would have figured that out also.

Such possible sum are - 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97, 101, 107, 113, 117, 119, 121, 123, 125, 127, 131, 135, 137, 143, 145, 147, 149, 155, 157, 161, 163, 167, 171, 173, 177, 179, 185, 187, 189, 191, 197, ....

Let's examine them one by one.

If the sum of two numbers is 11, Sachin will think that the numbers would be (2,9), (3,8 ), (4,7) or (5,6).

Sachin : "As 11 is not expressible as sum of two primes, Prashant can't know the numbers."

Here, the product would be 18(2*9), 24(3*8 ), 28(4*7) or 30(5*6). In all the cases except for product 30, Prashant would know the numbers.

- if product of two numbers is 18:
Prashant : "Since the product is 18, the sum could be either 11(2,9) or 9(3,6). But if the sum was 9, Sachin would have deduced that I might know the numbers as (2,7) is the possible prime numbers pair. Hence, the numbers must be 2 and 9." (OR in other words, 9 is not in the Possible Sum List).

- if product of two numbers is 24:
Prashant : "Since the product is 24, the sum could be either 14(2,12), 11(3,8 ) or 10(4,6). But 14 and 10 are not in the Possible Sum List. Hence, the numbers must be 3 and 8."

- if product of two numbers is 28:
Prashant : "Since the product is 28, the sum could be either 16(2,14) or 11(4,7). But 16 is not in the Possible Sum List. Hence, the numbers must be 4 and 7."

- if product of two numbers is 30:
Prashant : "Since the product is 30, the sum could be either 17(2,15), 13(3,10) or 11(5,6). But 13 is not in the Possible Sum List. Hence, the numbers must be either (2,15) or (5,6)." Here, Prashant won't be sure of the numbers.

Hence, Prashant will be sure of the numbers if the product is either 18, 24 or 28.

Sachin : "Since Prashant knows the numbers, they must be either (3,8 ), (4,7) or (5,6)." But he won't be sure. Hence, the sum is not 11.

Summerising data for sum 11:
Possible Sum PRODUCT Possible Sum
2+9 18 2+9=11 (possible)
3+6=9
3+8 24 2+12=14
3+8=11 (possible)
4+6=10
4+7 28 2+12=14
3+8=11 (possible)
4+6=10
5+6 30 2+15=17 (possible)
3+10=13
5+6=11 (possible)

Following the same procedure for 17:
Possible Sum PRODUCT Possible Sum
2+15 30 2+15=17 (possible)
3+10= 13
5+6=11 (possible)
3+14 42 2+21=23 (possible)
3+14=17 (possible)
6+7=13
4+13 52 2+26=28
4+13=17 (possible)
5+12 60 2+30=32
3+20=23 (possible)
4+15=19
5+12=17 (possible)
6+10=16
6+11 66 2+33=35 (possible)
3+22=25
6+11=17 (possible)

7+10 70 2+35=37 (possible)
5+14=19
7+10=17 (possible)
8+9 72 2+36=38
3+24=27 (possible)
4+18=22
6+12=18
8+9=17 (possible)

Here, Prashant will be sure of the numbers if the product is 52.

Sachin : "Since Prashant knows the numbers, they must be (4,13)."

For all other numbers in the Possible Sum List, Prashant might be sure of the numbers, but Sachin won't.

Here is the step by step explanation:

Sachin : "As the sum is 17, two numbers can be either (2,15), (3,14), (4,13), (5,12), (6,11), (7,10) or (8,9). Also, as none of them is a prime numbers pair, Prashant won't be knowing numbers either."

Prashant : "Since Sachin is sure that both of us don't know the numbers, the sum must be one of the Possible Sum List. Further, as the product is 52, two numbers can be either (2,26) or (4,13). But if they were (2,26), Sachin would not have been sure in advance that I don't know the numbers as 28 (2+26) is not in the Possible Sum List. Hence, the two numbers are 4 and 13."

Sachin : "As Prashant now knows both the numbers, out of all possible products - 30(2,15), 42(3,14), 52(4,13), 60(5,12), 66(6,11), 70(7,10), 72(8,9) - there is one product for which list of all possible sum contains ONLY ONE sum from the Possible Sum List. And also, no such two lists exist. [see table above for 17] Hence, two numbers are 4 and 13."

Nakul figured out both the numbers just as we did by observing the conversation between Sachin and Prashant.

It is interesting to note that there are no other such two numbers. We checked all the possible sums till 500!!!
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Comments

Posted by electronjohn on Feb 11, 2003

Very interesting, although extremely hard to solve. That has got to be one of the longest answers posted so far. My hats off to you for all the work you did on this one.

Posted by Codammanus on Feb 12, 2003

I'm not a fan of teasers involving primes. However, the level of thought involved in constructing this one is astonishing.

Posted by Gizzer on Feb 18, 2003

He didn't work hard on this one at all. Check out the Mathematicians post under Teasers without answers. Mad-Ade pointed him to another site with this teaser!

Posted by smileysteve on Apr 02, 2003

Huh?

Posted by gerdmain on Nov 05, 2003

Is this really the only solution? What about a sum of 65? It can be presented as (2,63)...(32,33). All possible products except (4,61) = 244 have multiple answers.

Also a sum of 89 and the product 1168(16,73) is another solution, and there are many more.

Posted by karanw on Nov 20, 2003

Oh...u cant solve this...!

Posted by knbrain on Dec 24, 2004

I got bored reading the answer

Posted by Sane on Mar 23, 2005

Posted by Gizzer Feb 18, 2003

He didn't work hard on this one at all. Check out the Mathematicians post under Teasers without answers. Mad-Ade pointed him to another site with this teaser!

Owned

Posted by yevgen on Sep 02, 2006

In fact, sum 29 = 13+16 is also a solution (the smallest solution above 13+4 = 17). The way to make solution unique is to say that the product is less than 200.

Also, the solutoion is very tediously presented. A good way to solve this is the following:

1) the fact that Sachin knew Prashant cannot know the sumbers means that the sum is ODD and also sum munis 2 is composite. Indeed, every even number = sum of two primes (famous Euler's conjecture, easily verified for small numbers).

2) the fact that Prashant knew the answer after that means that his product is uniquely decomposable into two numbers whose sum is ODD. It is easy to verify that this means his product = prime times a power of two (except for 2 itself, since then Prashant would know right away).

3) The fact that Sachin knows as well means that his sum S is uniqulely decomposable as prime+power of two (except for 2 itslef), and also S-2 is not prime.

Now, we go through all numbers of the form (composite+2), and check if they are uniquely decomposable as prime+power of 2.

This way we see 15 (composite) + 2 = 17 = 13 (prime) + 4 (power of 2), but 17-8 = 9 is composite, and 17-16 is 1 which is disallowed. Going thorough other numbers, the next sum like that is 27 (compsite)+2 = 29 = 13 (prime) + 16 (power of 2), but 29-4=25 (composite) and 29-8 = 21 (composite). However, 13*16=2-8, which is greater than 200, so it can be rules out. Going a bit further we see that next number is indeed 63 (composite) + 2 = 65 = 61 (prime) + 4, since 65-8 = 57 (composite), 65-16=49 (composite), 65-32 = 33 (compsite). Once again, though, 61*4 = 244 > 200, so it's out too.

Posted by rachayl on Sep 13, 2006

I'm adding this to my favorites. This is exactly the kind of teaser I was hoping to find when I joined this site. I like thinking about numbers, though I don't have any background in number theory, and solving a prob like this gets me thinking about numbers in all sorts of new ways. Like, I didn't know about Euler's conjecture, but I just spent 20 pleasant minutes conjecturing it on my own!
Hope to find more teasers like this!! Thanks!

Posted by teasermaster on Dec 19, 2006

I thought this one was unsolvable. Way too hard and there should've been a hint!

Posted by on Feb 02, 2007

To earlier comment - hint probably wouldn't have helped much...

Yah this is crazy hard, but interesting none the less

Posted by lolliepup on Feb 16, 2007

way...way...too hard.

Posted by pating on Jun 25, 2007

What just happened there?

I got lost!!!

Posted by javlad27 on Jul 28, 2007

That's the longest answer I've ever seen!

Posted by sftball_rocks13 on Aug 27, 2007

wow

um...the answer was longer than the teaser

but...wow
i *never* would have gotten that

Mathematicians

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Mathematicians

Answer

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Comments

To post a comment, please visit the Full Site

Copyright © 1999-2007 | Green | Privacy | Conditions You are using the TV formatted version of Braingle. For more functionality, please visit the Full Website.

Copyright © 1999-2007 | Green | Privacy | Conditions
You are using the TV formatted version of Braingle. For more functionality, please visit the Full Website.