
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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July 11th, 2011, 12:53 PM  #1 
Senior Member Joined: Jun 2011 Posts: 164 Thanks: 0  a very long tricky problem...
I am not sure if it should be included in this section, if not, it is ok for moderators to remove it because this problem might be too hard for middle school students but the technique school students should know, it is just hard to think it that way. The question is: "A (a person) had 2 Integers in his head between 2 and 99 inclusive, he told B (a person) the sum of two numbers, and told C (of course, another human being) the product of those two numbers, and then B and C had this following conversation: C: I don't know the numbers. B: I know you don't know the numbers, I don't know it either. C: Then I know the numbers. B: Then I do too! What were the two numbers?" This question is given from a friend of mine; I thought about it for a long long time, did 10 pages of work, 3 months of thinking 10 minutes everyday later (because I couldn't solve it), and finally, I figured out the answers, which should've used 3 pages of work at most. It is fun question to do if you think about it. Note: both B and C are perfect logicians, which means they can calculate infinite amount of times in their head without making a mistake in 1 second. Now try it if you feel like it, if you can't solve it, I am gonna do some hints. Now hint #1: [hide]this problem consists 4 big steps, which first step has to do with prime numbers and prime factorization and a little pattern finding[/hide] hint #2 (still for first step): [hide]if you prime factorize a number that is a product of 2 numbers, you will get 2 and only 2 prime numbers as its prime factors[/hide] hint #3 (still for first step, but after hint #2) : [hide]if you divide a prime number by a number that isn't one or itself, you will have decimal solutions, this hint and last hint eliminate lots of possibilities.[/hide] hint #4(finishing step 1, conclusion after hint #2 and 3) : [hide]you will eliminate "that ANY of the two numbers cannot be a prime number above 50" because if it multiplies by any other number from 299, the only prime factorization of the product that is possible to be the solution to this problem is that prime number and the other number. And it cannot be 2 prime numbers, which means 5 and 7 is not possible, 11 and 17 doesn't work, etc. (little work could work the second "prime elimination out"[/hide] hint #5(starting step 2)[hide]you have to think from the B side, be a logician, he said "i know you don't know it", which means he knows that it is not possible to get to the conditions of "that ANY of the two numbers cannot be a prime number above 50" and "And it cannot be 2 prime numbers"[/hide] hint #6(deep into step 2)[hide]so make a list of sums which is not possible to get any of the two conditions in hint 5, you will find a pattern, and keep the possible sums, which I will tell you next.[/hide] hint #7(conclusion of step 2) [hide]the only possible sums are 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53. For example: 15 can be eliminated because 2 and 13 are possible to be the sum of the two numbers and the product 26, which C would know (back into hint 4);it cannot be 75 because 53 and 22 gives 75 but if you give the product of 53 and 22 to C, then he would've known the numbers, think closely if you don't get it, look from both angles of view.[/hide] hint #8(start of step 3)[hide]so C knew the numbers after that, but you can not solve it right away, I can tell you C must have thought what B thought (they are perfect logicians!).[/hide] hint #9(conclusion of step 3) [hide]eliminate all the answers that comes from 2 different pairs of numbers, for example: eliminate "5 and 6" and "3 and 10" because otherwise C would not know the numbers, it has to be unique in those products[/hide] hint#10 (start of step 4, final step)[hide]after you eliminate all the "double products" (products that appeared twice in the sum possibilities and product possibilities), you will find something interesting...[/hide] hint #11(final conclusion draws near!)[hide]after you eliminate all the "double products" you find that one of those sum possibilities will have only 1 product possibility, and if you are lazy, and want to know the answer, see the final hint.[/hide] hint#12(final solution!)[hide]the answer is 4 and 13, the sumis17 row's products appeared twice in different places in this possible product list.[/hide] final work check: [hide]A told B the sum of numbers, which will be 17, and A told C the products of the numbers, which will be 52, C does not know because 2 and 26 is possible to be one of the solutions, B knew C doesn't know because no two numbers that have a sum of 17 that gives a unique product (to this question, ranging from 299). So he doesn't know the numbers and knows that C doesn't know the numbers because all the pairs of numbers that gives a sum of 17 could have a second answer that are products of two other numbers (possible sums again strikes) then C made the same calculation as what B did in his head in 1 second , and made the same list, to C's surprise, 52 was a part in productofsumof17 list, and he just solved xy=52 and x+y=17, and he found 4 and 13 are the numbers. Then B knew the product has to be unique from the possibleproduct list, so he spent an infinitely small time to calculate what numbers there are left, and found that the possibleproductof17 row has only 52 left, and the sum was 17, so he solved xy=52 and x+y like C did, and found the answers, and lived happily ever after. ()[/hide] If you solved this problem on your own, Great job, you might be a logician in the future. 
July 11th, 2011, 12:54 PM  #2 
Senior Member Joined: Jun 2011 Posts: 164 Thanks: 0  Re: a very long tricky problem...
looks like [hide][/hide] does not work here, is there anyway that can hide something you are going to say?

July 11th, 2011, 11:22 PM  #3 
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: a very long tricky problem...
As far as I'm aware, the only codes available are the ones listed at the top of the reply box, although even some of those will not work until you have 10 posts. If you were trying to hide the hints... you can always just leave a big space and put HINTS or SPOILER above it. People can just not scroll. (I did not read your hints  it wasn't that difficult to figure out whether I chose to read them or not. ) Another trick I know is putting the print in white, which blends it in with the background, and the reader has to highlight it to make it show. However, this has the light blue background and I don't know how well we could match that. [color=#BFFFFF] Testing hiding print.[/color] Well, there's nothing that matches, but I used the bottom blue to the right on the text above and it's very difficult to read, so that would work well enough. (But, you also can't edit until you have 10 posts.) 
July 11th, 2011, 11:38 PM  #4 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: a very long tricky problem...
I don't think it's really necessary, but you can use the tags color=#e1ebf2 or color=#ecf1f3 depending on whether your post number in the thread is odd or even: [color=#e1ebf2]Like this[/color] or [color=#ecf1f3]like this[/color]. 
July 16th, 2011, 07:49 PM  #5 
Newbie Joined: Jul 2011 Posts: 16 Thanks: 0  Re: a very long tricky problem...
[color=#e1ebf2]Surely you meant to say ECF3F7, right?[/color]

July 26th, 2011, 04:20 PM  #6  
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: a very long tricky problem... Quote:
 

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