Last Friday at the Mathematical Sciences Research Institute (MSRI) I met Gary Antonick, who writes a numbers blog for the New York Times, a fun blog that you must check out. I gave him a puzzle and he loved it so much he thought it was going to be one of the best he has posted on his blog.
We had gathered together at the Berkeley site of MSRI for an annual celebration to honor Martin Gardner, the famous gamer, puzzler and magician who wrote the column Mathematical Games for The Scientific American for 25 years. Many of us grew up loving his column and spent endless hours solving his posers and reveling in his tricks.
Elwyn Berlekamp, the well known author of several books on Nim-like games was there, playing Dots and Boxes with kids and other visitors. He runs a nifty outfit called Gathering for Gardner, whose website is here. Their logo reproduced below is really cool as it reads exactly the same when you turn it upside down – this is the sort of thing Martin Gardner loved.
In the spirit of Martin Gardner many math puzzles were presented to the attendees.
The 13-LInk Chain
I discussed the following 13-link chain puzzle with Gary – a puzzle that I have known since my grandfather posed it to me many, many years ago and one I had immensely enjoyed. Gary loved the puzzle and has posted it on the New York Times blog today. So check it out, see if you can solve it and send me the solution, or your comments or questions. I’m reproducing the puzzle below as well:
The 13-Link Chain Puzzle
You have a balance scale and a single chain with thirteen links. Each link of the chain weighs one ounce. How many links of the chain do you need to break in order to be able to weigh items from 1 to 13 ounces in 1-ounce increments?
As with all puzzles that have Martin Gardner’s elegance and spirit, this puzzle is very pleasing at many levels and encompasses some more general math patterns that are endlessly extended by people who then pose ever-more evolutionary puzzles based on this.
For those of you who want more of these fun puzzles try out the ten off-the-shelf classical (and simple) Martin Gardner puzzles here. (From Elwyn’s Gathering for Gardner site)
Happy Solving!
Need to break in 3 places. Between 1st and 2nd, between 3rd and 4th and between 7th and 8th
Hey Jay. Good attempt but you can get away with only ONE break. See solution below.
three pieces, 1,3 and 9 linka
OK, Here’s the solution:
You need to break the chain so that you end up with three subchains of length 1, 3 and 9 chains. With these three you can weigh out all the increments from 1 oz to 13 oz. As follows:
1-ounce == 1
2-ounce == 3-1
3-ounce == 3
4-ounce == 3+1
5-ounce == 9 – (3+1)
6-ounce == (9 – 3)
7-ounce == (9+1) – 3
8-ounce == (9 – 1)
9-ounce == 9
10-ounce == (9+1)
11-ounce == (9+3) – 1
12-ounce == (9+3)
13-ounce == (9+3+1)
Now here’s the fun part. How many links do you have to break to get 1,3,9? Only one. Break the fourth link and slip it out of the chain.
You’re right. But there’s another fine point. See solution above.
Break the chain in two places so you’re left with the following:
a 9-link remnant chain
a 3-link remnant chain
a single link
Use of these 1, 2 or 3 at a time (additive or subtractive) will allow weighing in 1-ounce increments from 1 to 13 oz.
Ajit,
You are right about needing 1,3 and 9 length links. See how to achieve this with only one link being broken.
BTW did you know that in general using weights of 1,3,9,27,81…. you can weigh out any increment up to the sum of the subchains. So using 1,3,9,27 you can weigh out 1 to 40 oz. Using 1,3,9,27,81 you can weigh out 1 to 121 and so on.
Ashok
Dear Ashok,
Very nice puzzle. My answer is – only 1 break is needed. You need to cut the 10th link. Now you have 3 weights to work with – the first of weight 9 links or 9oz, the second of weight 3 links or 3oz and the 3rd (broken link, but still usable) of weight 1 link or 1oz. Assuming you can put the pieces on either side of the weighing scale as needed (let us assume a plus sign on the left side scale and minus on the right side scale), the following combinations will yield all 13 link/oz weights on the left side.
1 oz : 1
2oz : 3 – 1
3oz: 3
4oz: 3+1
5oz: 9 -(3+1)
6oz: 9 – 3
7oz: 9 +1 -3
8oz: 9 – 1
9oz: 9
10oz: 9 + 1
11oz: 9 + 3 -1
12oz: 9 + 3
13oz: 9+3+1
I know of a couple very interesting puzzles from my days at IIT-B. I will email them to you sometime.
thanks for the enjoyment of this one,
Harsh Sharangpani
Harsh,
Great to hear from you! You are one of the few who got the puzzle completely. Would love to see your other puzzles from the IIT days.
Ashok
Dear Ashok,
I am first enclosing one of the puzzles (from my IIT-B days) that I was reminded of:
FIND THE ODD BALL OUT OF A DOZEN IN 3 WEIGHINGS:
You are given 12 balls and a balance scale. All 12 balls look identical. 11 of them weigh the same. But there is 1 ball that has a defect that makes it either heavier or lighter (don’t know which apriori) than the rest. Find the odd ball in 3 weighings. You can use a marker and label the balls as you wish. Will you also be able to tell whether it is lighter or heavier?
Enjoy,
Harsh
Thanks, Harsh I posted a solution and another one for you at my site. See you soon. Ashok
Sent from my iPad
Hi Harsh,
Thanks for posting the 12-ball puzzle. I too remember doing it 45 years ago! The solution:
Weigh any four balls against another 4. If they balance problem is easy. If not label the four on the heavy side H, those on the light side L, and the others O. Now weigh HHHL vs HOOO. If they balance one of the remaining 3 Ls is light and one weighing will identify it. If HHHL is heavy then one of three Hs is heavy – no prob. If HOOO is heavier then either that H is it or the L on the other side. In either case we are home.
Here’s one for you. You ave three boxes and they contain respectively either black balls, purely white balls or half-black and half white balls. You can’t see which is which. However the boxes are labeled: WHITE, BLACK AND BLACK-WHITE. None of the labels is correct, i. E. they are all three wrongly labeled. How many balls must you sample to identify the three boxes correctly?
Have fun and keep sending the posers.
Dear Ashok,
Hey, well solved on the 12 ball problem.
On the labelled boxes, you only need to sample 1 ball from the one labelled BLACK-WHITE. Since it is incorrectly labelled, it actually contains either only white balls or only black balls.
If you get a black ball, then this box is actually the BLACK Box, so the one incorrectly labelled WHITE is actually BLACK-WHITE (since it now cannot be BLACK) and the one incorrectly labelled BLACK is actually the WHITE Box. Meanwhile, if you get a white ball from the box labelled BLACK-WHITE, then this box is actually the WHITE Box, so the one incorrectly labelled BLACK is actually BLACK-WHITE (since it now cannot be WHITE) and the one incorrectly labelled WHITE is actually the BLACK Box.
I am currently in India, will be here for a couple more months. So there is a time zone difference. Will post another interesting one in a day or so – this was more complex, so some of the guys at school had simplified it to make it puzzle-worthy, but I first need to correctly formulate it before I post it.
Dear Ashok, Thanks for posting the 13-link chain puzzle and also the link for the Martin Gardner puzzles. Good brain teasers. Bala
A variation on this puzzle that we were familiar with as kids goes like this:
A merchant had a 40 lb stone which he used to weigh goods. One day the
stone fell and broke into 4 pieces. To his amazement the merchant found
that he could now use the pieces to weigh everything from 1 lb to 40 lbs.
What were the weights of the four pieces ?
(The solution has of course been discussed here in some detail already.)
Cheers,
Ashok based on how the puzzle is worded, i believed that the puzzle implies that only positive values can be used. In that case, there need to be three breaks between 2 and 3, 7 and 8, 10 and 11. this leaves us with 4 links of 2, 1, 4, 3, 3 ( the first break will give us a link of 2, and 1)
1
2
3
4
4,1
4,2
4,3
4,3,1
4,3,2
4,3,3
4,3,3,1
4,3,3,2
4,3,3,2,1
Great puzzle and links!
Pradip
Pradip: It states that it is a BALANCE scale. You can therefore place links on the same side as the item you are weighing. Hence the subtracted values.