The number 1987 can be written as a three digit number xyz in some base
b.
If x + y + z = 1 + 9 + 8 + 7=25, determine all possible values of x, y, z, b.
Source: 1987 Canadian Mathematical Olympiad.
Subtract the equations:
(b1)(x(b+1)+y)=1962=2*(3^2)*109
The first solution is b1=18 which leads to 20x+y=109 and (x,y,z,b)=(5,9,B,19). 5*19*19+9*19+11=1987 and 5+9+11=25.
The second solution is sneakier. If b1=18, 16x+y=109 and (x,y,z,b)=(7,3,F,17). 7*17*17+3*17+15=1987 and 7+3+15=25.

Posted by xdog
on 20181017 18:12:16 