“And the 4 weights after the stone broke…you can do all the weighing from 1 to 40 lbs in one weighing.”
If you’re replying to this: “For extra credit: how does the stone’s owner know the weight of each of the four pieces?” I think you misunderstood my point. If not, nevermind.
I arrived at the same answer that Ray gave; it is a unique solution. But to imply, as he does, that the exponential series is the key is more than a little misleading. As our beloved puzzlemeister freely admits, he stumbled upon the correct answer.
Suppose instead that you had a 31 pound rock that broke into 5 pieces. Would you automatically conclude that the unique solution would be 1+2+4+8+16=31? It works - but so does 1+2+4+9+15=31. Anyone?
And yes, Evil Ape, I too wrestled with the question you pose. Until I decided that Ray was more interested in the mathematical than the logical aspects of the problem. . .
mookie911 — "I arrived at the same answer that Ray gave; it is a unique solution. But to imply, as he does, that the exponential series is the key is more than a little misleading. As our beloved puzzlemeister freely admits, he stumbled upon the correct answer."Stumbled indeed. An MIT diploma should come with an expiration date.
littlemouse: “For extra credit: how does the stone’s owner know the weight of each of the four pieces?”
You mean how can one prove the stones weigh 1,3,9,27 pounds using only the equipment available?
If you take the one pound stone and use it to weigh out one pound of feed, and then do it again, two pounds of feed weighed out in two weighings plus the one pound stone would balance the three pound stone and thus prove that it weighed three times the one pound stone.
A similar process can be used to prove the 9 pound stone weighs 3 times as much as the three pound stone etc.
Why not use a single weighing to weigh out two pounds of feed to prove the weight of the 3 pound stone? That would be a case of using circular logic to prove something.
B.L.E. - If I read littlemouse’s question correctly, he is suggesting that the farmer cannot know the weight of any of the stones.
You write “If you take the one pound stone. . .”. He (and I) would stop you right there and ask how you know the smallest stone weighs one pound. And supposing you did, there would be no need to proceed further, since you could then in effect create your own set of weights simply by measuring out first one, then two, then three. . . as many discrete pound weights of feed as necessary to determine the weights of all the stones.
Again, there is no disputing the (only) correct answer, only the assumption that you know anything beyond a+b+c+d=40. As I hypothesized in my earlier post, the base 3 ‘proof’ collapses in the face of analogous problems with base 2. How about base 4? Can you generate the entire whole number series with 1, 4, 16, 64, 256? Of course not.
Clearly, Ray presents a foregone conclusion, i.e., that the puzzler’s hero had ready-made referents (grain, guano, gold bricks, whatever) against which he could compare the stone fragments and thereby determine their weight. A boon in itself, inasmuch as he could now use the stones as his standardized system of weights instead of, say, trotting out the guano.
It would have been simpler had the puzzler asked: what four whole numbers, adding up to 40, can generate the whole number series 1. . . 40? But then, I suppose, we wouldn’t have had this delightful discussion, would we?
"B.L.E. 6:47AM
littlemouse: “For extra credit: how does the stone’s owner know the weight of each of the four pieces?”
“If you take the one pound stone…” WHAT one pound stone?
How does the stone’s owner know the weight of each of the four pieces?
How does the stone’s owner know the weight of each of the four pieces?
This puzzler is a theoretical logic exercise. It is ONLY used for problem solving skills. Knowing or NOT knowing HOW the stone’s owner knew the weight of the four pieces has NOTHING to do with solving the problem.
OK, Mike, riddle me this: have you in fact solved the problem by discovering a formula to find a unique solution, or do you consider it solved after numerous trials and errors? If the former, please enlighten littlemouse and me. If not, I’ll put my crack cadre of sliderule-wielding capuchin monkeys back in their cubicles.
OK, Mike, riddle me this: have you in fact solved the problem by discovering a formula to find a unique solution, or do you consider it solved after numerous trials and errors?
There are many ways to solve the problem…Trial and error is one…So is analytical thought…Trial and error will get you there…Again…so what. The point I made is that it doesn’t matter HOW the Stone owner knew the weight of the four pieces to solve the problem…What the Stone owner knew or didn’t know has NOTHING to do with solving the problem.
“Knowing or NOT knowing HOW the stone’s owner knew the weight of the four pieces has NOTHING to do with solving the problem.”
That’s why I said “for extra credit”. No need to shout.
littlemouse "B.L.E. 6:47AM
littlemouse: “For extra credit: how does the stone’s owner know the weight of each of the four pieces?”
“If you take the one pound stone…” WHAT one pound stone?
How does the stone’s owner know the weight of each of the four pieces?
OK, how about “the smallest stone” instead of the “one pound stone” since the farmer doesn’t know yet that it weighs one pound. Using the procedure I outlined, he can then prove that the next larger stone weighs three times as much as the smallest stone etc, etc, until he has proven that the stones have weights of one, three, nine, and 27 “units”. Now with prior knowledge that the original stone weighed 40 pounds, it becomes obvious that the units are pounds.
That’s why I said “for extra credit”. No need to shout.
Who’s shouting?? When a whole sentence is capitalized it’s shouting…when one word or a phrase is capitalized it’s considered EMPHASIZING.
As as to how can they tell…Does it matter?? And there is more then one correct answer…so not really worth the effort…The puzzler is a math question…let’s stick with math.
Higher math, the math of math majors, is mostly about proofs. In math, “proof” does not mean “beyond reasonable doubt” or “supported by a preponderence of evidence”, in math, proof is an absolute, like checkmate.
“How can the farmer prove using only the equipment at hand that the stones weigh 1,3,9,27 pounds?” is indeed a math question.
What equipment is at hand? does that include a 40 lb hay bale? Cut the bale in quarters so you have 10 lbs = a 1 and 9 lb weight? then 3/4 so 30 lbs = 27 + 3?
At first you are tempted to run up the powers of 2. Weights of 1, 2, 4, and 8 pounds, in various combinations, can weigh everything from 1 to 15 pounds, but that’s nowhere near 40. The trick is, some of the weights can sit on the left pan, with the box. Each weight can be on the left, the right, or not used at all. This suggests powers of 3.
Using 1 and 3 pound weights, you see immediately how to measure 1, 3, and 4 pounds. If the box weighs 2 pounds, put the 1 pound weight on the left pan with the box, and the 3 pound weight on the right, and everything is in balance. Next, bring in a 9 pound weight and put it on the right. We already showed we can add a net weight of 1 through 4 to the box, so if the box weighs anywhere from 5 to 8 pounds, we’ll be able to balance it against the 9 pound weight. Of course there is no trouble if the box weighs 9 pounds, and if it weighs anywhere from 10 to 13 pounds, we achieve balance by adding (net) 1 to 4 pounds to the right. Now we’ve covered everything from 1 to 13 pounds, and the last weight to bring in is 27 pounds. This lets us weigh everything from 27-13 pounds up to 27+13 pounds, or 40 pounds. An inductive proof shows successive weights should always be powers of 3.
To weigh a box from 1 to 81 pounds, double the weights. Thus the weights are 2, 6, 18, and 54 pounds. If the weight of the box is even, you can determine it using the procedure outlined above. If the weight is odd, the balance will always tip one way or the other. Determine that the box is heavier than n and lighter than n+2. For instance, the box might be more than 30 pounds and less than 32 pounds, whence it is 31 pounds.
Higher math, the math of math majors, is mostly about proofs. In math, “proof” does not mean “beyond reasonable doubt” or “supported by a preponderence of evidence”, in math, proof is an absolute, like checkmate.
“How can the farmer prove using only the equipment at hand that the stones weigh 1,3,9,27 pounds?” is indeed a math question.
NOT all higher math is about proof…I have a MS in Applied Mathematics…I think only 1 class had to deal with proofs…
Linear Math, Discrete Math, Numerical Analysis…No proofs in those classes.
And proof is a mathematical way to PROVE something is correct…The question is not to prove the answer…
The base-3 solution is elegant. The puzzle can be recast into the following: Find the base, “a”, of the string “1111”, given that the value of the string is 40 (in decimals). Trial and error, starting with a=2, takes only one more step to get the right solution, a=3.
If you like to play with algebra instead, using the formula for the sum of geometrical series gives (a^4-1)/(a-1)=40. The solution is unique, with a=3.
Questions raised by mookie911 are interesting. In the case of 1,2,4,9,15, the issue is that one cannot CONSISTENTLY represent this as a string of 1’s and 0’s (as in a digital computer). The computer can only handle 9 as 8+1 and 15 as 16-1. In fact, this is not efficient to do (see last PP.)
As for 1,4,16,64,256, since there are only 5 places (or digits), there is not enough room to hold all the information that is necessary to represent all non-negative integers up to 256 or higher. The best one can do is 121 (40+81) under base-3.
At this juncture, the natural question to ask is “Why don’t we use base-3 in computers, as base-3 can hold more information than base-2 in each digit?” The answer is that some extra information is also needed for base-3 to represent all non-negative integers. Consider 2 (base-10) is simply “10” in base-2. In base-3, it is “10”-“01”. We need to take care of both “10” and “01”, as well as the minus sign; so there is no saving at all. Sorry, there is no free lunch.
(Not a mathematician or a computer scientist; but do hold an advanced MIT degree from decades ago.)
"Consider 2 (base-10) is simply “10” in base-2. In base-3, it is “10”-“01”. We need to take care of both “10” and “01”, as well as the minus sign; so there is no saving at all. Sorry, there is no free lunch.
(Not a mathematician or a computer scientist; but do hold an advanced MIT degree from decades ago.)"
Maybe MIT should have a confidentiality agreement with all their graduates, not just Tom and Ray. 
2 in base 3 is “2”. In the 1960s we looked at this in grade school, albeit briefly. No MIT needed.
Decimal Base 3
0 0
1 1
2 2
3 10
4 11
5 12
6 20
…
10 101
etc.
The reason we don’t use base 3 in computers is a lot simpler than you’ve painted it.
The system that works with the two-pan balance is what’s called “balanced ternary”. That’s base three but with the digits {-,0,+} (shorthand for -1, 0, +1) instead of {0,1,2}. + corresponds to putting a weight on the side of the scale opposite the thing you’re trying to weigh; 0 to not using that weight, and - to putting it on the same side as the mystery weight. Works like this with four places (representing 27, 9, 3 and 1 pounds in that order):
Decimal/Balanced Ternary
0 0000
1 000+
2 00±
3 00+0
4 00++
5 0±-
6 0±0
7 0±+
8 0+0-
9 0+00
10 0+0+
11 0+±
12 0++0
13 0+++
14 ±–
15 ±-0
16 ±-+
17 ±0-
18 ±00
19 ±0+
20 ±±
21 ±+0
22 ±++
23 +0–
24 +0-0
25 +0-+
26 +00-
27 +000
28 +00+
29 +0±
30 +0+0
31 +0++
32 +±-
33 +±0
34 +±+
35 ++0-
36 ++00
37 ++0+
38 ++±
39 +++0
40 ++++
Littlemouse is correct that in the normal base-3 system, 2 (base-10) is written as 2, except that the solution to the puzzle does not allow us to have a 2, only (3-1). So the string “1111” is a binary notation but the base is 3: 13^0+13^1+13^2+13^3. This is certainly not a standard way of doing things but is one way of dealing with this puzzle.
It’s not a binary notation.