(In reply to
Problem Solution : Method III by K Sengupta)
We will try to extend this problem to positive integer bases (including base 10). Let the said base of expression be denoted by p.
At the outset, we will assume non-leading zeroes.
Further in consonance with the spirit of a typical cryptarithmetic addition problem, we will assume that M and A correspond to different digits. In my opinion, the italicized stipulation should have constituted part of the problem text.
By the problem,
MA + A = AM
-> pM + 2A = pA + M
-> (p-1)M = (p-2)A
-> M/A = (p-2)/(p-1) .....(i)
Since we cannot have leading zeroes, it follows that M >=1, so that p>=3 .......(ii)
Since, p-1 and p-2 are consecutive, it follows that they must be relatively coprime, so that from (i), we must have:
(M, A) = ((p-2)r, (p-1)r), for some positive integer r.(since r=0
would force A=M, contradicting the italicized stipulation
Since A is a base p digit, it follows that:
A <= (p-1)
-> (p-1)r <=1
-> r<=1. Since r is positive, this is possible only when r=1, so that: (M, A) = (p-2, p-1)
Consequently, we can now assert that, in general, for any positive integer base p>=3:
(M, A) = (p-2, p-1) is the solution to the given problem.
Edited on January 22, 2009, 1:27 pm
An attempt to generalization
