New Batches at TathaGat Delhi & Noida!               Directions to CP centre
Re: Base system
by Nikhil Sinha - Friday, 12 November 2010, 01:33 AM
  Hey Amit,

This one is a tough problem, often discussed in various forums.. (Hope such probs dont come my way :D)

P.S.: U got the start for this one though.. smile

Here is the soln to the problem:

(231)n * (ABA)n=(BA4AA)n

(2n+1)(n+1)(An^2 + Bn + A) = BA4AA
BA4AA is divisible by n+1, hence (A+4+B) - (A+A) = 0 or (n+1)*k,where k is an integer.

OR (another way of saying the same thing)

231 = 11*21 (in any base system)
so we know that BA4AA has 11 as a factor
applying divisibility rule of 11 we get B+4-A= 0 or n+1(here n is the base of the system)

so B+4-A= n+1 ---------(1)

now take the product
we get 231*ABA= 2An^4+(2B+3A)n^3+(3A+3B)n^2+(3A+B)n+A
Replace all Bs with n+A-3 (from (1)) then compare it with BA4AA

so we get 2A+2= B comparing power of n^4 and
4A-3=A
so A=1 and B =4
therefore, n =B-A+3 =6


Nikhil cool