# Puzzler: The farmer and the 40 lb stone (10/22/11) (NO ANSWER GIVEN)

The Puzzle of Oct 22, 2011: A farmer has a 40 lb stone that he uses to measure out bales of hay on a 2 sided balance. He loans it to a friend who accidentally breaks it into 4 pieces. Instead of being angry, the farmer is quite happy. He says to the friend, “You managed to break it into just the right 4 pieces that will now let me weigh any weight between 1 and 40.” What are the weights of the 4 pieces?

I propose a second challenge: Are there any posters out there with black belts in Google who can find the solution on the internet? The solution is out there.

why does Google keep sending me back to cartalk’s page when I search?

Maybe because it’s a reused puzzler. Good title by the way. I hope anyone who wants to discuss possible solutions will start a new topic for that. I’m waiting for them to post the transcript so I can re"hear" the puzzler… streaming other stuff now.

why does Google keep sending me back to cartalk’s page when I search?

No…it’s probably because cartalk is the only thing google can find on the subject.

Here’s a hint…Hopefully it won’t give it away.

Suppose you have 3 stones (5lb, 2lb and a 1lb).

Someone brings in a bag of rice…that he says weighs 4lbs.

How would you weigh it? The challenge is you don’t have any combination of stones that weigh 4lbs.

Here’s how you’d do it.

Put the bag of rice on one side of the scale…and the 5lb stone on the other…Then add the 1lb stone to the side of scale with the rice…If they balance out…then the bag of rice weighs 4lbs.

MikeInNH — "No...it's probably because cartalk is the only thing google can find on the subject."
Not true. The Google quest must be properly phrased to return a solution.
MikeInNH — "Here's a hint....Hopefully it won't give it away."
That is not a hint at all, because one would still be left in choosing, presumably at random, four weights that could be placed on a balance scale to balance any integer weight between 1 and 40 lbs.
Warning! Hint given below.

Proceed more methodically. Find the smallest stone, of weight W(2) lbs, that can be broken into two parts, such that the two smaller broken pieces can balance any weight from 1 to W(2) lbs. Then proceed to find the next smallest weight, W(3) lbs, that can be broken in to three pieces, etc. … and proceed in that fashion to find the smallest weight, W(n) lbs, that can be broken into n pieces that can be used to balance any weight from 1 to W(n) lbs. (Obviously, W(4) = 40 lbs.) By solving initially the n=2 problem, a general solution will emerge.

Sorry mechaniker…That’s the best hint I can give you without actually telling you the answer…You should be able to figure out the rest.

If I said “I can’t believe i didn’t remember base three from elementary school”, would I be giving it away? I forgive the recycled puzzler since I forgot what I learned over Summer also

If I said “I can’t believe i didn’t remember base three from elementary school”, would I be giving it away? I forgive the recycled puzzler since I forgot what I learned over Summer also

I figured out the answer a while ago…And I do know how what base 3 is…And if you remember what Base 3 is…then this is a MAJOR clue.

I found the answer on the second page of google suggestions. Only reading search results, based on my google search, Less than 2 minutes, probably less than 1, hint up to 81 lbs you may need to place weights on both sides of the balance.

“Find the smallest stone, of weight W(2) lbs, that can be broken into two parts, such that the two smaller broken pieces can balance any weight from 1 to W(2) lbs.”

I think your attempt to be mathematically rigorous results in obfuscation. In the requirement above, W(2)=2 pounds.

“and proceed in that fashion to find the smallest weight, W(n) lbs, that can be broken into n pieces that can be used to balance any weight from 1 to W(n) lbs.”

For this one n =1 and W(n)=1.

While we’re giving out hints, nowhere in the puzzler does it say “in a single weighing” so once you’ve weighed 7 pounds, for example, you now have 7 pounds of stuff. Take all the stone weights away and put stuff into the empty basket until the two balance. Now you’ve got 14 pounds of stuff.

evil ape, are you still being sent back to the homepage? Are you using the Google box to search the forum archives, or are you using Google’s homepage?

For extra credit: how does the stone’s owner know the weight of each of the four pieces?

I seem to recall another puzzler that had someone trying to cut up a gold chain of x number of links into several shorter chains so that you could trade any multiple of one link’s worth of gold with someone. It was important to cut as few links as possible to create the shorter chains because cutting and closing up a link was an expensive or difficult process.

That puzzler is closely related to this one.

While we’re giving out hints, nowhere in the puzzler does it say “in a single weighing” so once you’ve weighed 7 pounds, for example, you now have 7 pounds of stuff. Take all the stone weights away and put stuff into the empty basket until the two balance. Now you’ve got 14 pounds of stuff.

While it doesn’t say that specifically…I think it was implied. And the 4 weights after the stone broke…you can do all the weighing from 1 to 40 lbs in one weighing.

I agree; The base-three based answer is elegant. The real puzzler is how the farmer knew the stone broke into the 4 perfect masses…hmm. Someone please post the answer to that one!

The base-three answer in a form that doesn’t give anything away: 1111 (base three) = 3³+3²+3¹+3º = 40 (base ten).

It might be fun to work it out the hard way. You can obviously weigh 40 pounds by using all four pieces, whatever individual sizes they are. To weigh 39 pounds, one of the weights has to be one pound so you can take that one off the pan. That also makes it possible to weigh one pound (obviously) by using only that weight. And if you put the other three weights (39 pounds total) on one pan and the bale of hay plus the 1-pound weight on the other, it balances a bale that’s 38 pounds. So there’s four of your forty possible weights accounted for (1, 38, 39 and 40).

Next step is to take another stone away. If you assume another one-pound stone (nobody said the four pieces had to be four different sizes), that lets you weigh 38 pounds again (by setting the two one-pound stones aside), 37 (the remaining two stones on one pan and the bale plus one one-pounder on the other), and 36 (the remaining two stones vs the bale and both one-pound stones). It also allows you to weigh two pounds (the two smaller stones against just the bale). But now you’ve only accomplished a total of seven out of forty weights, and you’re running out of pieces, so it’s time to backtrack.

Forget about the second one-pound stone and try again with a two-pounder instead. This allows you to weigh 1 pound (1-pound stone vs bale), 2 (2-pound stone vs bale), 3 (1- and 2-pound stones vs bale), 34 (last two vs bale plus 1- and 2-pounds), 35 (last two vs bale plus 2), 36 (last two vs bale plus 1), 37 (last two vs bale alone), 38, 39 and 40 (as described in the second paragraph). Ten out of forty. You can also get 36 by weighing the last two stones plus the 1-pounder against the bale and the 2-pounder; there’s something about having more than one way to measure the same weight that seems to waste combinations. A two-pounder isn’t going to hack it.

So make the second stone three pounds. Fiddling for a while, this allows you to measure 1, 2, 3, 4, 32, 33, 34, 35, 36, 37, 38, 39 and 40. That’s thirteen out of forty, so record that much information and move on to try weights for the last two stones. You know their combined weight is 36 pounds, so if you pick a weight for one you know what the weight is for the other. And you know that it won’t help you to make either of them three pounds or less; in fact you can pretty much bet that they’ll both be over four pounds because you’ve got that combination covered with the first two. So try four pounds for the third, five pounds, etc, knowing that you’ll know one way or another by the time you get to eighteen (at which point the “fourth” stone becomes smaller and you just end up running through the same combinations with the third and fourth stones reversed).