Let a be the base
(231)a * (ABA)a=(BA4AA)a
Final equation Iam getting is a^3(2A-B)+a^2(A+B)+2a(A+B+2)+ 2A+B=0
Not able to find value of a ?
(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
(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
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