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        riddles >> medium >> An All Possible Value Puzzle
        
(Message started by: K Sengupta on Nov 3^rd, 2006, 7:05am)

Title: An All Possible Value Puzzle
Post by K Sengupta on Nov 3^rd, 2006, 7:05am

A, B and C are positive integers.
Determine all possible values of C, for which the equation:
2^A – 5^B = C , possesses exactly two distinct solutions.

Title: Re: An All Possible Value Puzzle
Post by THUDandBLUNDER on Nov 3^rd, 2006, 12:07pm

Nice problem!

So (A₁, B₁) and (A₂, B₁) are not distinct?

Title: Re: An All Possible Value Puzzle
Post by Icarus on Nov 3^rd, 2006, 3:59pm

The question doesn't come up, for if the values of B and C are the same, the values of A must be as well.

Title: Re: An All Possible Value Puzzle
Post by THUDandBLUNDER on Nov 3^rd, 2006, 4:27pm

on 11/03/06 at 15:59:03, Icarus wrote:

The question doesn't come up, for if the values of B and C are the same, the values of A must be as well.

I understand now. C is constant (a positive integer) and we need to find what values of this constant allow exactly 2 distinct solutions.

Title: Re: An All Possible Value Puzzle
Post by Barukh on Nov 3^rd, 2006, 5:59pm

One solution (to start with): [hide]C = 3[/hide].

Are there any others? :-/

Title: Re: An All Possible Value Puzzle
Post by Eigenray on Nov 4^th, 2006, 12:27am

Hmmm...

2^x(2^4r-1) = 5^y(5^2s-1).

We'd like to conclude r=s=1.

Suppose, for example, that 2|r. Then 17|2^4r-1, so 17|5^2s-1, so 8|s. Then 11489|5^2s-1, so 11489|2^4r-1, so 1436|r. Etc.?

The number 3 might be important. The number of times 3 divides r is the same as the number of times 3 divides s, for example. We also know the number of times 5 divides r is y-1, and the number of times 2 divides s is x-3, but I don't know if that helps.

Title: Re: An All Possible Value Puzzle
Post by Sameer on Dec 18^th, 2006, 9:14am

Actually thanks for T&B's link in another post. I think this riddle just got buried. So this way it will be bumped!! :D

Title: Re: An All Possible Value Puzzle
Post by Eigenray on Dec 18^th, 2006, 11:14pm

Well, 2^x(2^4r-1) = 5^y(5^2s-1) has only one solution with 0<r,s<10⁶ (much past that I run out of memory cause I'm too lazy to write an efficient program). But I haven't made any real progress.

Title: Re: An All Possible Value Puzzle
Post by Sameer on Dec 19^th, 2006, 10:41am

anyway to do this analytically? :o

Title: Re: An All Possible Value Puzzle
Post by Eigenray on Jun 23^rd, 2007, 8:10pm

K Sengupta, did you have a solution?

Title: Re: An All Possible Value Puzzle
Post by K Sengupta on Jun 26^th, 2007, 1:49am

on 06/23/07 at 20:10:23, Eigenray wrote:

K Sengupta, did you have a solution?

I truly do not have the solution to this one.

Prima facie, this problem occurred to my mind having it's unmistakable similarity with the famous 2^x - 3^y = 7 problem which possesses an elementary solution.

Having regard to its simplicity, I naively assumed that the 2^x - 5^y case would also lend itself to a similar treatment, a premise which has since proved erroneous.

In conclusion, I would like to thank you for your brilliant treatment of the various facets corresponding to the problem under reference, with a hope that someday, an analytic solution to the foregoing problem could be found.