Logic Puzzle: "Rock On"

[The Armchair]

Before I leave work for the day:

You have of those counter-balance things and a big, 40-pound rock. You use the balance and the rock to measure out 40 pounds of hay, which he sells at the market.

Your annoying neighbor comes by and asks to borrow the rock. Begrudgingly, you agree, figuring not even he could mess up your 40 pound rock.

Bzzt.

He comes back a day later, hands in his face. He tells you that he broke your rock.

As you go for your hoe (the garden tool, you fool!), hoping to drive it into his back, he yells, "But wait! It's an improvement." A reasonable person, you hear him out.

"You now have four rocks. And I tested them on my super-accurate bathoom scale [which you've seen, and you trust], and it's really cool. With the four rocks, you can now find any weight on your counter-balance thingy, 1 to 40!"

He offers to let you weigh each of the rocks on his bathroom scale, to prove it to you. If he's telling the truth, what should each rock piece weigh?

[aloiz]

Assuming that we are only want weights to an integer. <font color="white">
2,4,11,23
Didn't go through all 40 but I'm pretty sure that you can get all weights with these.
</font>

aloiz

[AndysDaddy]

The correct answer is:
<font color="white">1, 3, 9, &amp; 27</font>
But I "cheated". I wrote a c++ app to find it for me.

To aloiz: How can a get a measurement of 1 from your combo?

[jdl22]

Are you allowed to use a rock twice somehow? Otherwise I don't think it's possible. Clearly 1 must be there. If you can't somehow use it twice then you also need 2. Similarly you also need 4 and 8 and 16 and 32 which means you need more than 4 rocks and that their weight adds up to more than 40.

[AndysDaddy]

You can measure a weight of 1 with no rock that weighs 1 (in white):
<font color="white">Put a 3 rock on one side, and a 2 rock and the object to be weighed on the other. If the object weighs 1, the scale will balance.</font>

[aloiz]

I assumed you wouldn't be weighing anyting less than 1. So if anything is lighter than 2 pound rock it must be 1.

aloiz

[jdl22]

clever. I'll have to give it some thought then.

[Oski]

The answer should be 3, 4, 12, and 21.

Still no good: Can't get a 23.

[AndysDaddy]

Nice. I hadn't thought of that. But what about 3? More in White:
<font color="white">My guess is that you'll say that in two weighings if its more than 2, but less than 4 you know its 3. I believe an unstated assumption is that you can weigh any amount up to 40 in one weighing.</font>

[aloiz]

Yea I would use two weighs, and you're probably right only one was to be allowed, so yours would be the only correct answer.

aloiz

[The Armchair]

Edit: They're integers, and you want integer values, including 1. (For clarity's sake!)

[jwvdcw]

Think I got it:

<font color="white"> My thought process went something like this:

You need 3 smalls numbers which would enable you to get any smaller numbers. Then you'll need one big number. The 3 small numbers that seemed to work for everything(just used trial and error) are 9,3, and 1. With these numbers you can get anything from 1-13. Then I looked for a number halfway between 13 and 40 and got 27. Since I can get any number from 1-13, if I add in 27, I can get anything from 14(27-13) to 40(27+13). So the numbers are 9,3,1, and 27.</font>

[Oski]

looks good.

[GuyOnTilt]

The answer is <font color="white">1,3,9,27. Why these are the series of the first 4 non-fractional exponents of 3, I have no idea, but they work. Is there only one answer?

I started at the top. It seems easiest if all 4 rocks add up to exactly 40, and to get 39 one rock must be 1. 38 is just putting that 1 lb rock on the other side. 37 would have to be a 3 pounder. Then 36, 35, 34, 33, and 32 all take care of themselves with just moving those 2 rocks around. 31 isn't possible without a rock of a different value, so the 3rd rock is 9. And since they all have to add up to 40, the last is 27. I tried it and it worked. First try!!!</font>

GoT

[jwvdcw]

[ QUOTE ]
The answer is 1,3,9,27. Why these are the series of the first 4 non-fractional exponents of 3, I have no idea, but they work. Is there only one answer?

I started at the top. It seems easiest if all 4 rocks add up to exactly 40, and to get 39 one rock must be 1. 38 is just putting that 1 lb rock on the other side. 37 would have to be a 3 pounder. Then 36, 35, 34, 33, and 32 all take care of themselves with just moving those 2 rocks around. 31 isn't possible without a rock of a different value, so the 3rd rock is 9. And since they all have to add up to 40, the last is 27. I tried it and it worked. First try!!!

GoT

[/ QUOTE ]

seems so easy when you say it like this.