Thursday, February 09, 2006
Divide 40 KG Stone
Puzzle: Divide 40 Kg Stone into 4-Pieces such that using
4-Pieces we can measure the all weights from 1 to 40 (only integer values).
I took half day to solve this problem.
4-Pieces we can measure the all weights from 1 to 40 (only integer values).
I took half day to solve this problem.
Comments:
<< Home
As you have stated the puzzle, a solution is not possible. Please provide more details on the method of weighing.
I assume you mean weighing using a balance, where you can put the unknown mass (which you want to determine) on one side, and put any combination of stones on either side of the balance ?
I assume you mean weighing using a balance, where you can put the unknown mass (which you want to determine) on one side, and put any combination of stones on either side of the balance ?
anonymous said:
> As you have stated the puzzle,
> a solution is not possible.
You're wrong. The weights 1, 3, 9 and 27 work perfectly fine.
> As you have stated the puzzle,
> a solution is not possible.
You're wrong. The weights 1, 3, 9 and 27 work perfectly fine.
Hello Pradeep,
Maybe you can guide us: do you want people to not post answers? Are you interested in programs that solve the puzzles? I solved the 40kg puzzle in Lisp. What languages are other people using?
Maybe you can guide us: do you want people to not post answers? Are you interested in programs that solve the puzzles? I solved the 40kg puzzle in Lisp. What languages are other people using?
Yes! I am interested in solving puzzles programmatically.
Please post your code in lisp. I am familiar with C++ so i will try to convert your code to C/C++.
Please post your code in lisp. I am familiar with C++ so i will try to convert your code to C/C++.
Lets call n the first weight.
The next consecutive even/odd integer would have to be (n+2)
So the 4 weights would look like :- n, n+2, n+4, n+6
Now set their sum equal to 40 and
solve for n.
n+(n+2)+(n+4)+(n+6)=40
4n+12=40
4n=28
n=7
Therefore 7,9,11,and 13 are the four numbers.
The next consecutive even/odd integer would have to be (n+2)
So the 4 weights would look like :- n, n+2, n+4, n+6
Now set their sum equal to 40 and
solve for n.
n+(n+2)+(n+4)+(n+6)=40
4n+12=40
4n=28
n=7
Therefore 7,9,11,and 13 are the four numbers.
"The next consecutive even/odd integer would have to be (n+2)". How did you arrive at this conclusion?.
Post a Comment
<< Home