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

 Leave no stone unturned (Posted on 2017-03-19)
There are three virtual piles of stones. In one operation one may add to, or remove from, one of the piles the number of stones equivalent to the quantity in the other two piles combined, leaving the numbers in those two piles unchanged.
Thus, e.g., (12,3,5) can become (12,20,5) by adding 12+5=17 stones to the second pile, or (12,3,5) can become (4,3,5) by removing 3+5=8 stones from the first pile.

Assume a starting state (1111,111,11).
Is it possible, by a sequence of such operations, reach a state where one of the piles is empty?

 See The Solution Submitted by Ady TZIDON Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Other cases | Comment 3 of 5 |
(In reply to Other cases by Steve Herman)

I probably picked a bad example.  It is obvious that (1110,110, 10) cannot be reduced to zero.  Just divide by 10 to reach (111,11,1) and then apply Charlie's parity argument.

A better example is (1110, 108, 10), which can be reduced to (48, 50, 10) but no further.

I do not see an obvious test for whether any set can reduced so that one of its terms is zero, other than:
a) divide each pile by their greatest common divisor
b) if all piles are odd after the division, it is impossible
c) Otherwise, repeatedly subtracted the two smallest piles from whatever is the largest at each step, until zero is reached or until the two smaller piles have a sum greater than the largest pile.

 Posted by Steve Herman on 2017-03-19 14:29:54
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 (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information