Minimize the bases P and Q such that each of the following alphanumeric equations has at least one solution:

(A) (GO)_{Base P} + (GO)_{Base P} = (GIG)_{Base P}

(B) (GO)_{Base Q}*(GO)_{Base Q} = (GIG)_{Base Q}

Note: Solve each of the alphanumeric equations separately and remember G, O and I must be distinct and G can't be zero.

I think K also wants an analytical solution. With 3 digits, the minimum base this would work is base 3.

For the first part, it is clear G must be 1 as a three digit number is twice a two digit number. Thus, the base must be odd, and trying 3 implies O=2, and then trying I=0 gives a solution: 12+12=101 (5+5=10) So P=3

For the second part, G must be small compared to Q, since the result is a 3 digit number with the same carry digit. For example, in base 4: 23*23=1321, even 13*13=301. So this means G=1.

Using G=1 and converting to base 10 gives (Q+O)(Q+O)=Q^2+1+IQ or O^2+2OQ=IQ+1. Solving for Q gives Q=(O^2-1)/(I-2O)

I must be less than Q so I-2O must be at least Q-5 to avoid reusing the digit 1, thus Q>=6. If Q=6, I=5, O=2. However this implies 12*12=151 (8*8=87) which is not true. So Q>=7

If Q=7, and O<Q, then for O^2-1 to be divisible by 7, O=6, but this value of O produces a negative I-2O. So Q>=8.

If Q=8, and O>3, a similar problem develops, and if O<3, O^2-1 is less than Q. So try O=3, and thus I=7. It ends up working -- 13*13=171. (11*11=121)


So the final solutions are 12 + 12 = 101 (base 3), 13 * 13 = 171 (base 8)

Posted by Gamer on 2007-05-25 20:17:23