 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Get the bases! (Posted on 2018-10-17) 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.

 See The Solution Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) computers not needed | Comment 3 of 5 | Subtract the equations:

(b-1)(x(b+1)+y)=1962=2*(3^2)*109

The first solution is b-1=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 b-1=-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 2018-10-17 18:12:16 Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information