### Thursday, February 09, 2006

## Divide 40 KG Stone

**Divide 40 Kg Stone into 4-Pieces such that using**

*Puzzle:*4-Pieces we can measure the all weights from 1 to 40 (only integer values).

I took half day to solve this problem.

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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?.

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