The Magic of Logic (2)
Teaching young children the joy of thinking
There is a certain fascination that logical puzzles exert, even at a very young age. Like at six, when a bright child will display great joy while solving them, and even become hooked in the process. I reported previously on how it started with my grandson Enders. After a few months there was a pause, but now he is back at it again. I want to share the unique experience and tell you how you can do it with your own children.
The pause from pure logic and thinking was caused, to some extent, by the weather. When the sun is shining, the garden is in full bloom, running, jumping and bicycle tours are more appropriate than sitting indoors, pondering. In addition I had the impression that Enders was tiring of the strain the problems were causing, between his temples. Thinking is every bit as exhausting as running and chasing your brother around.
But now that temperatures have dropped Enders suddenly no longer reacted to my offers with the quick promise “Later.” In fact he came to me and said: “Okay, give me a puzzle.” I did, and promised him two episodes of Tom and Jerry if he solved the problem—that was additional motivation.
“You have three coins, Enders,” I said. “One is slightly heavier than the other two. You have scales like this one, and must find the defective coin. How many time do you have to weigh them?”
“Three times,” Enders blurted out, but then, after a few seconds, he corrected himself: “No, I can do it in two weighings!” And beamed with pleasure at having solved the problem so well. “How?” I wanted to know.
“Well, I put one coin in each pan,” Enders replied. “If the left pan goes down, that is the heavier coin. If the other side goes down, that is the coin. And if it balances, then both are good. Then I take the third coin and put it in one pan and a good coin…” There was a thirty second pause, after which he suddenly shouted out: “No, just one weighing!! If the scale balances then it is the third coin, and it is heavier. I don’t need to weigh it at all!” This time the joy on his face was much greater. He had solved something difficult. Now Tom and Jerry?
Not yet, Enders. Let’s say you have four coins, and one is heavier than the other three. How many times do you have to weigh them to find it? He thought about this for ten minutes before he told me: “You can’t do it in one weighing!” That was an interesting achievement — working out why it was definitely impossible in one.
So how do you find it in two? That was not a big problem and Enders quickly found the solution: put one coin in each pan for the first weighing. If one side goes down, you have found the heavier coin in one try. If it balances, then weigh the remaining two against each other. That will show you which is the heavier coin. He also came up with a second solution: weigh two against two coins, take the two that go down and weigh one against the other. The one that goes down is the heavier one.
Excellent! How about five coins? Enders sighed incredulously, but with a smile on his face. He got it in a minute or two: two against two, if they balance it is the remaining coin; if one side goes down it has the heavier coin, and you weigh the two coins against each other to find it.
Okay, six coins, Enders. “We can find it in two weighings??” he asked in disbelief. “You tell me,” I replied. This time he spent over five minutes thinking, a little embarrassed that it took him so long. But got it in the end: “Of course, it’s the same as five. Weigh two against two, if one side goes down, you can find the heavier coin in one more weighing; if they balance then the heaver coin is one of the two we have not weighed, and we can find it in the second weighing. We even found a second way: three against three, then one against one from the heavier side. Clever boy!
Seven! — “Seven?? In two weighings? That is impossible!” Enders said, but started thinking, and soon had the solution: three against three. If they balance, it is the remaining coin; if one side goes down we can find the heavier coin in one more try, “just like the first problem with three coins!” No more pestering about Tom and Jerry — Enders was waiting to see if more coins were possible.
Eight. Easy peasy: three against three, if they balance it is one of the remaining two, otherwise we find it with the three-coin method!
“Nine,” I said, and again the clever young boy, now equipped with experience, found the solution very quickly. Three against three, and one side goes down, or they balance. In each case we know which group has the heavier coin, and we can do the three-coin thing to find it.
I was very pleased with how he had solved the problems and offered to fire up Tom and Jerry on my Android tablet. But Enders wanted to know: can you do ten coins in two weighings? He spent some time working on that, and became a bit frustrated. So I told him it wasn’t possible — you need three weighings. He found a method, using admirable logic: weigh five against five, take the five that are heavier and use our method for five coins from earlier on to find the heavier one in two more weighings! Very good, Enders. T&J now? Nope, one more please!
Okay, the final problem: twenty-seven coins, with one heavier than the other twenty-six. How many times must we weigh? He threw back his head in disbelief. Twenty-seven? Really?? He started working on it without any chance of success, guessing it was probably six weighings. So I helped him: what is interesting about the number 27, Enders? After a number of guesses he at last said: “It is three time nine.”
He saw the big grin on my face, and in a few minutes he had found the solution: “Three weighings!!” How do you do that? Well, make three groups, nine in each. Weigh nine against nine. If one side goes down, or they balance, we know in which group of nine the heavier coin is. And then we can find the heavier coin in two more weighing, just like we did for nine coins previously. Perfect.
When his father got home Enders gave him the 27 coin problem and explained the whole solution to him — at great speed. Of course Martin could not follow everything immediately, since he had not gone through the entire series. But Enders insisted on explaining it again and again, until he did. Joy and pleasure — and at last he got his 20 minute Tom & Jerry reward.
The next time we picked him up from school his teacher told us that he had said, quite enthusiastically, that he was going home to solve more “weighing problems”— she had no idea what they could be. And back home he asked me for more. So I sat him down and gave him the following: “You have three coins, Enders, and one of them…” (“We already did that” he interjected) “Wait: … one of them is heavier or lighter than the other two. How many weighings?”
Enders spent a number of minutes working on that, mainly trying desperately to do it in one — until he convinced himself it was not possible. I confirmed it: you need two weighings.
That was easy: weigh one against one. If one side goes down either it has a heavier coin, or the other side has a lighter one! And then? Well, we take the third coin, which we know is good, and balance it against the one from the heavier side. If that still goes down it is the heavier coin; if it balances the defective coin is the one we took out of the other pan, and it is lighter.
And if it balances the first time? “Then it is the third coin, and we have got it in one. But we can also weigh it against one of the good ones to see if it is heavier or lighter.” Excellent.
Four coins, one heavier or lighter than the other three — how many weighings? This took quite a while to solve: two weighings! The simplest way: one against one, a against b. If they balance, replace the coin from one side with another one, a against c. If they also balance, then the fourth coin, d, is the defective one. We don’t know if it is heavier or lighter. But we were only asked to find the odd one. “What if one side goes down in the first weighing?” No problem: that means the other two are good, and we simply replace one from the first weighing with a good one. And if one side goes down in the second weighing? “Come on!” said Enders, “then that is the bad coin, and it is heavier.” Got it?
We have gone through five and six coins in this series, and Enders has worked them all out: three weighings, in each case. Out of pure mercy I did not proceed, and gave him a break. But he wanted to know: you can find the heavier or lighter one from how many coins in three weighings? Twelve, I told him, but it is very hard. I will keep that for when he is half a year older, at least. It is a problem that stumped tens of thousands of readers, years ago, on my chess newspage. I have written about it (see IQ link below), and no, I will not reveal the answer.
Related stories from my blog:
- The Magic of Logic (1)
- Remembering Martin Gardner
- Piazzi and the missing planet
- Lie to your kids — it’s good for them
- The Café Wall Illusion
- Prime number cicadas — really!
- Money problems without decimals
- IQ test: weighing the coins
- Tricking the brain
- Scamming programmers with Guess
- The art of skepticism
- He had a dream — a logical puzzle
