All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Mixing Digits (Posted on 2004-12-29)
If this sum is correct, what digits do A and M represent? MA + A = AM

 See The Solution Submitted by Jim Rating: 2.1111 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 An attempt to generalization Comment 25 of 25 |
(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
 Posted by K Sengupta on 2009-01-22 06:27:30

 Search: Search body:
Forums (0)