## Bar of Lead

For those of you that felt that past couple puzzles were too easy, here’s a much harder one:

A boy had a bar of lead that weighed 40 pounds, and he divided it into 4 pieces in such a way as to allow him to weigh any number of pounds from 1 to 40.

How did he manage the matter?

*[can submit answers below, will unmask this weekend] *

Daniel| Guest August 3rd, 2007 - 6:24 pmAnswer:

Divide the 40-pound bar into sections of 1, 3, 9, and 27 pounds. Using a two-sided balance scale, place the object to be weighed on the left side. Then place the “added” weights on the right side, and the “subtracted” weights on the left side; if the scale is balanced in the following combinations, then the left-side value is the weight of the object being weighed.

40 = 27 + 9 + 3 + 1 (i. e., if all four weights are on the right side while the object is on the left side of the scale and it is balanced, then the object weighs 40 pounds)

39 = 27 + 9 + 3

38 = 27 + 9 + 3 – 1

37 = 27 + 9 + 1

36 = 27 + 9

35 = 27 + 9 – 1

34 = 27 + 9 – 3 + 1

33 = 27 + 9 – 3

32 = 27 + 9 – 3 – 1

31 = 27 + 3 + 1

30 = 27 + 3

29 = 27 + 3 – 1

28 = 27 + 1

27 = 27

26 = 27 – 1

25 = 27 – 3 + 1

24 = 27 – 3

23 = 27 – 3 – 1

22 = 27 – 9 + 3 + 1

21 = 27 – 9 + 3

20 = 27 – 9 + 3 – 1

19 = 27 – 9 + 1

18 = 27 – 9

17 = 27 – 9 – 1

16 = 27 – 9 – 3 + 1

15 = 27 – 9 – 3

14 = 27 – 9 – 3 – 1

13 = 9 + 3 + 1

12 = 9 + 3

11 = 9 + 3 – 1

10 = 9 + 1

9 = 9

8 = 9 – 1

7 = 9 – 3 + 1

6 = 9 – 3

5 = 9 – 3 – 1

4 = 3 + 1

3 = 3

2 = 3 – 1

1 = 1

How to get the numbers {1 3 9 27}:

The maximum distance between weights must be observed to fill in all the ‘gaps,’ with the smaller weights closer together and the larger weights increasingly far apart. In other words, figure out the smallest units first, then the greatest combination they can reach with all the ‘gaps’ filled in; the difference between the next unknown number minus the sum of the known numbers will be one greater than the sum of the known numbers.

a must = 1, or else the weight 39 could not be achieved.

b – a = 2; therefore b = 3.

since a + b = 4, c – (a + b) = 5, therefore c = 9.

All combinations smaller than a + b + c can now be reached; a + b + c = 13. Therefore d – (a + b + c) must = 14, so d = 27.

Mark W| Guest August 4th, 2007 - 3:16 am1, 3, 9, 27 pounds.

I’ll explain my thought process:

a + b + c + d = 40

The only way he can measure 39 pounds in this situation is if he can remove one bar and get 39, so one of the bars must equal 1 pound (i’ll say d=1 pound).

a + b + c = 39

Putting the d=1 bar on the opposite side of the scale (the side with the object to be weighed), he can measure 38 pounds. Putting a weight on the opposing side is effectively subtraction:

a + b + c – 1 = 38

Now how can he possibly measure 37 pounds? It must be 3 bars, including the d=1 bar, so we’ll say

a + b + 1 = 37

Since a + b + c + 1 = 40, lead bar c=3 pounds. Now we can keep trying to calculate down the next numbers:

a + b = 36

a + b – 1 = 35

a + b – 3 + 1 = 34

a + b – 3 = 33

a + b – 3 – 1 = 32

Now how can he possibly measure 31 pounds? It must be three bars, including d=1 and c=3, so:

a + 3 + 1 = 31

This unfolds the final answer, because this means a = 27, so by elimination, b = 9. You can see that this is correct by continuing to calculate the rest of the numbers down the list using a combination of addition and subtraction.

Daniel| Guest August 4th, 2007 - 1:06 pmFascinatingly, the numbers of the answer correspond to successive powers of 3: 1 = 3^0, 3 = 3^1, 9 = 3^2, 27 = 3^3. If the pattern was to continue, the next number would be 81 = 2 times (27 + 9 + 3 + 1) plus 1 = 3 ^ 4, which would permit weights of up to 121. The next number would be 243, or 3 ^ 5…

Why does it work this way?

I wish I could draw a picture on here…I am thinking of a triangle. The triangle has three vertices (points). Imagine there is another ‘invisible’ vertex in the middle, which represents the basic unit (1 vertex). The numbers 1 through 4 can be reached through these points: 1, 3-1, 3, 3 + 1 (at 4, the whole thing is ‘lit up’. Now imagine this whole triangle is shrunk down into one vertex, at the center of another such triangle, in which each vertex is a ‘shrunken’ triangle. The interior vertex is still ‘invisible’ or ‘imaginary,’ representing some external basic unit (i.e. the numbers which aren’t part of this next number). The sum of the exterior vertices of the exterior triangles is now 9. This number plus the sum of the vertices of the interior vertex-triangle is 13, which corresponds to 9 + 3 + 1. Since you can imagine each of the vertexes ‘expanding’ into a triangle with its own vertices, you can concede that any number 1-13 could now be reached. Now by imagining THIS whole triangle as a vertex in another triangle, you can find that the next number is 27 (sum of exterior vertices of exterior triangles) and the sum reachable is 40. Sorry for the rambling, I’m just trying to reconcile to myself why the numbers are powers of 3. I guess another way to look at it is simply that the numbers must be in some constant ratio, where a factor of 2 is too close together (multiple sums equal the same number, since subtraction is possible) and a factor of 4 is too far apart (no sum equals some of the numbers).

PJ White| Guest August 5th, 2007 - 12:59 amBars of lead: 8, 9, 10, 13 lbs.

ken| Guest August 5th, 2007 - 11:10 am1,3,9,27 works.

It’s kinda weird that the answer is 3 raised to the 0,1,2, and 3 power…there’s probably some relation there. If you had a 121 pound weight, and were asked to divide it up 5 times, the answer would be (1,3,9,27,81).

John| Guest August 5th, 2007 - 10:06 pmCan you subtract one bar from another? If so then it’s: 1, 3, 9, 27. Otherwise, it is impossible.

abouseem| Guest August 6th, 2007 - 2:44 am1, 3, 9, 27

1=1

2=3-1

3=3

4=1+3

5=9-1-3

6=9-3

….

and so on

Alexey| Guest August 6th, 2007 - 7:12 am1, 3, 9, 27

RK| Profile August 6th, 2007 - 7:44 amAnswer: 1,3,9,27.

Great explanations, Mark W. & Daniel.

As Ken and Daniel point out, it is weird that the answer turns out to be successive powers of 3.

I just may try to get someone to draw out that triangle Daniel, I’m kind of wanting to see it now myself!

ken| Guest August 6th, 2007 - 9:31 amAnother way to look at it….

You take 1, double it, and add 1. Now you have 3.

You take (3+1), double it, and add 1. Now you have 9.

You take (9+3+1), double it, and add 1. Now you have 27.

Etc.

Thus, you can get 4 by adding 3 and 1, but to get 5, you subtract 3+1 from 9. You cover all your bases with a minimal number of different pieces of lead.

Daniel| Guest August 10th, 2007 - 6:34 pmI found a picture of the fractal I mentioned. It’s small, but it will give you the idea: http://www.miqel.com/images_1/.....actals.jpg

(on the left) The dark regions add up to the new number introduced; the white regions are the sum of the prior numbers. So in the first solid black triangle, 1 is introduced. In the second, the 1 moves to the middle, and 3 dark regions are added. In the next there are 9 dark regions and 4 white ones; 9 + 4 = 13 total. The next has 27 dark and 9 + 4 = 13 white, for a total of 27 + 13 = 40 triangles. Check it out!

RK| Profile August 10th, 2007 - 7:28 pmDaniel- Very Cool!

andi| Guest September 10th, 2007 - 9:12 pmI wanted to see if PJ White’s solution (8, 9, 10, 13) worked.

1 = 10 – 9

2 = 10 – 8

3 = 13 – 10

4 = 13 – 9

5 = 13 – 8

6 = 13 + 10 – 9 – 8

7 = 8 + 9 – 10

8 = 8

9 = 9

10 = 10

11 = 10 + 9 – 8

12 = 13 + 9 – 10

13 = 13

14 = 13 + 10 – 9

15 = 13 + 10 – 8

16 = impossible.

Damn.

Andrew| Guest October 15th, 2007 - 4:09 amI found Ken’s Definition to be most effective in describing the solution to this problem for the following reason.

I was curious about how to break up the weights if you wanted to measure by a smaller unit, say 0.5 lbs. Using his method of multiplying the base unit by two then adding the base unit again solves the problem very quickly (i.e. 0.5, 1.5, 3.5, 7.5, 15.5, 31.5). However, if you tried to work it as a successive power function it doesn’t work as easily.

Note:

After solving the weights using Ken’s method I was able to figure out a method using successive powers of two then subtracting 0.5 from each one. [(1,2,4,8,16,32…. ) – 0.5] which gives (0.5, 1.5, 3.5, 7.5, 15.5, 31.5 … lbs) calculated as follows,

2^0 – 0.5 = 0.5

2^1 – 0.5 = 1.5

2^2 – 0.5 = 3.5

2^3 – 0.5 = 7.5

2^4 – 0.5 = 15.5

2^5 – 0.5 = 31.5

and so on…

Bravo Ken!

Andrew| Guest October 15th, 2007 - 4:19 amworking the problem as fractions of one divided by powers of 2 (1/2, 1/4, 1/8, 1/16 etc.) yields very interesting patterns similar to the 0.5 I mentioned previously. However, this pattern changes with numbers like 0.1 lbs.

michaelc| Profile February 8th, 2008 - 5:29 pmInterestingly enough, if the scale is modified where you can only weigh one side, the answer has to be a power of 2. For instance, if you can’t use one side to take away from the other, to achieve every sum, the least amount of weights to hit every counting number is to use the powers of 2.

1, 2, 4, 8, 16, …

Makes me wonder if it’s possible to expand the scale and find some pattern for a power of 4… Nah, just coincidence I suppose.