Puzzle Corner

[superlion]

How does this sound.... take prime numbers of rivets from the drums... 1, 3, 5, 7, 9, 11, 13, 17, 19, 23 (think I got those right)... Then you can't add them up and have the number from another drum. I think. (keep in mind I'm sick so that may have come from the fever and not my brain)

[Jay]

I am sorry to hear you are sick and have a fever. Take care of yourself. Use whatever helps you to keep the fever under control and see a doctor, if necessary.

No, it is not prime numbers. (9 is not a prime number, by the way. It can be evenly divided by 3.) As one example of why prime numbers will not work: If the weighing was off by 19, you do not know if it was one drum that had lighter rivets (the drum where you took 19) or if three drums had lighter rivets (drums where you took 3, 5, and 11, since 3+5+11 = 19).

But that is the type of thinking that is needed to determine a set of numbers that work! (So maybe the fever is helping you. )

[Capt.Rutlinger]

is it like this? take 1from the first, then 2 from the second, then 4 from the 3rd then 4*2=8 from the 4th and so one

[Jay]

Congratulations, that is correct. And it is the most efiicient answer as well.

Here is a very long (and boring) explanation. But I like to be thorough.

There are many possible solutions. The important thing is that the rivets taken from the next drum has to be greater than the sum of all of the rivets taken from the previous drums. Here is one possible solution: Take 1 from drum one, 10 from drum two, 100 from drum three, etc. Then, which digits were 1 in the difference between the expected and actual weights (counting from the right) would indicate which drums had lower weight rivets. For ten drums with the heavier rivets, the expected weight would be 6666666666. If all drums had lighter rivets, the weight would be 5555555555. If the third drum had lower weight rivets, the weight would be 6666666566. (The 5 is the third digit from the right). 6666666655 would mean drums one and two had lower weight rivets. 6666666565 would mean the first and third drums had lower weight rivets.

The problem with this solution is that a billion rivets are needed from the tenth drum, which is very unlikely. The diameter of each rivet would have to be less than 1/609 of an inch. [That shows you how much of a nerd I am. The calculation is: square root of (55 gallons per drum * 231 cubic inches per gallon * 4 / pi / 6 inch length / 1 billion rivets) assuming perfect rivet packing.]

The best solution is the one given by Capt.Rutlinger: take 1 rivet from drum one, 2 rivets from drum two, 4 rivets from drum three, 8 rivets from drum four, ..., doubling the amount of rivets from each additional drum. This approach only needs 1023 total rivets (Here is the nerd in me again: 1023 is 2 to the # of drums power - 1. 512 will come from drum ten. The rivets could therefore be anything up to nearly 2.3 inches in diameter.)

This approach is actually the same as the previous approach, as any real computer professional (nerd?) should know, since it is how computer memory works. (1 in base 2 is 1, 2 in base 2 is 10, 4 in base 2 is 100, etc.) In fact, the original question could have been asked differently: "How does computer memory work?" The only difference is that computers (instead of scales) measure electrical current (instead of weight) to determine an "on" position (low weight rivet) from an "off" position (high weight rivet).

[Jay]

And the new puzzle is... the original puzzle again, well with a couple slight twists. When that puzzle was discussed, balance scales came up. So, for this puzzle, we will use a balance scale instead of a scale that gives a weight reading. There are 9 drums this time, with 1 of those 9 drums containing lighter rivets. Using the most efficient solution, how many times will the balance scale need to be used to find the drum with lighter rivets? And, be detailed of how the scale would be used.

[Capt.Rutlinger]

first all can you weigh the drums entirely or do you have to do the 1.2.3.4.5.6.7.8 thingy again?

[Jay]

It actually does not matter. But balance scales are not usually big enough to hold drums. So take 1 rivet from each of the 9 drums. Of these 9 rivets, use the balance scale to find which rivet is the lightest one. The goal is still the same. Using the most efficient solution, how many times will the balance scale need to be used to find the lightest rivet? And, be detailed of how the scale would be used.

[Capt.Rutlinger]

you need to use it 3 times:
1. divide the nine drums in 3 groups of 3
2. place one group on 1 side of the balance and 1 on the other.
3. then if the scale is in balance you know that the drum with the lighter rivets is in the other group, replace on of the groups on the scale with the one containing the lighter one but before you start weighing go to step 4 , if the scale is not in balance then the drum with the lighter rivets is already on the scale.
4. take one drum from each side; then if the scale is in balance then you know you took of the one with the lighter rivets, if the scale is not in balance then take another one of on both sides. If the scale is now in balance you again know you took of the lighter one, if the scale is not in balance then the lighter drum is the one remaining.

[Jay]

If the scale is in balance on the third weighing, you know you took off the lighter one. However, you took off two. So you do not know which one of the two is lighter. Therefore, using that approach, you need more than 3 weighings. That is not the most efficient method.

However, your first two steps are correct. We need something more efficient after that.

[Capt.Rutlinger]

1. divide the nine drums in 3 groups of 3
2. place one group on 1 side of the balance and 1 on the other.
3.1if the scale is in balance take one group out and take two rivets from the group you're leaving on of the scale (you know they are not the from the wrong drum)
then place one of the rivets from group 3 on the scale
1. if it is in balance replace the rivet with another one from group 3, (go to 3 when the scale is in balance again, otherwise go to 2)
2 if it is not in balance you know you've placed the lighter rivet on the scale
3. when the scal is still not unbalanced when you've placed the second rivet from drum 3 on the scale, you now know the lighter one is the one you did not place on the scale.

2if the scale is not in balance take one rivet away from each side of the balance
1. if the scale is now in balance you know you took of the lighter one
2. if the scale is not in balance take one from each side again, if the scale is in balance you took of the lighter on, or if the scale is not in balance the lighter rivet is now visible because of the scale

[Jay]

That solution is better. Sometimes you know the lighter rivet in 2 weighings; other times in 3 weighings. However, that is still not the most efficient solution. With the most efficient solution, you will always know the lighter rivet in 2 weighings.

[Jay]

Do not let this puzzle beat you. It is not as complicated as it might seem. Think about what your first two steps did. Hopefully that will tell you what the second weighing should do.

[Capt.Rutlinger]

1. divide the nine drums in 3 groups of 3
2. place one group on 1 side of the balance and 1 on the other.
3. if the scale was in balance, take the two groups of and place one rivet from the third on each side of the scale, so there is one left. Then if the scale in balance, the lighter rivet is the one you did not put on the scale, if the scale is not in balance you can see wich one is lighter
4. if the scale was not in balance take the heavier group off and take two rivets from the lighter group off, place one of them on the other side of the scale.

[Jay]

That is correct. Congratulations on getting the answer and not giving up!

Here is the same answer in a chart. The 9 rivets are A, B, C, D, E, F, G, H, and I. For the first weighing: A, B, and C are on one side; D, E, and F are on the other. The result of the first weighing indicates which 2 rivets to weigh next.

Code: Select all

ABC<DEF      or      ABC>DEF      or      ABC=DEF
|                    |                    |
V                    V                    V
A<B or A>B or A=B    D<E or D>E or D=E    G<H or G>H or G=H
|      |      |      |      |      |      |      |      |
V      V      V      V      V      V      V      V      V
A is   B is   C is   D is   E is   F is   G is   H is   I is
light  light  light  light  light  light  light  light  light
one    one    one    one    one    one    one    one    one

Or, for those that do not have a wide browser:

weighing 1 -> weighing 2 -> answer
ABC<DEF -> A<B -> A is light one
ABC<DEF -> A>B -> B is light one
ABC<DEF -> A=B -> C is light one
ABC>DEF -> D<E -> D is light one
ABC>DEF -> D>E -> E is light one
ABC>DEF -> D=E -> F is light one
ABC=DEF -> G<H -> G is light one
ABC=DEF -> G>H -> H is light one
ABC=DEF -> G=H -> I is light one

[superlion]

Here's a riddle I just solved that a friend left in an away message:

Pronounced as one letter, but written with three, two different letters there are and two only in me. I'm double, I'm single, I'm black, blue, and gray. I'm read from both ends and the same either way. What am I?