Brain Teasers are Back!

My workings are a bit more pedestrian !

First I have drawn a clock face to illustrate the solution and some of the key dimensions.

Then the arithmetic

I thought I’d better produce a second version of the arithmetic, just to illustrate that 45 minutes and 45 plus 5/7 seconds past one o’clock is the same as 2/7 hour before two.

Here’s a suggested brain teaser. I first came across it over 40 years ago so I hope I have recreated it well enough!

I Can’t See Myself Getting This Job

An interviewer sets a task for three bright candidates: Alice, Bill and Charlie. “I have five stickers, three are red and two are black. I will place one on each of your foreheads and keep two in my pocket. When I leave the room I will turn on the light so you can see each other’s stickers but not your own. When you work out what you have, come and tell me.”

The light goes on and all three stare at each other, poker-faced and not saying a word. After several minutes, Alice leaves the room and gives her answer.

What colour sticker did Alice have and why?

Alice has a red sticker as Bill and Charlie both have black. Neither Bill or Charlie can know what Alice has, as they see a red and black sticker. As would Alice if she had a black sticker.

Thanks for attempting this so quickly.

If Alice could see two blacks she would have got up straight away (she is a bright candidate and there are only two blacks so only red is left) but actually she took several minutes. So she can’t be seeing two blacks.

Ah, tricky. So then Bill and Charlie are in the same conundrum. Then Alice has a red sticker because Bill and Charlie both have red stickers and she realises that she must be red otherwise Bill and Charlie would have solved it straight away as they are bright too?

Yes, tricky! And yes, the answer relies on Alice using the fact that Bill and Charlie are bright.

Suppose Bill and Charlie both have red but Alice has black. How could either of Bill or Charlie solve it straight away as they would each see one red and one black.

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For clarification, maybe I should have said at the end of the teaser: “The interviewer is impressed with her answer and gives her the job.”

I have presumed that all three candidates are equally bright, tactically minded, and actually want the job.

Yes, that’s a fair assumption, broadly speaking. Maybe not equally bright but bright enough to be expected to solve a tricky brain teaser

My initial thoughts are as follows. They might help (or hinder) the thoughts of others?

There are three fundamental ways in which the stickers could be applied.

1 Red and 2 Black (first option)

2 Red and 1 Black (second and third options)

3 Red and 0 Black (fourth option)

The first option doesn’t need much thought.

We have already eliminated the first option. But we have the retained knowledge that nobody has seen 2 Blacks, because if they had, whoever saw 2 Blacks, would have got up immediately, (since, as Ravvie said, they are all bright candidates).

The second option needs a bit more thought.

Assume the single black sticker is on Alice.

She will see 2 Reds. Knowing instantly that there can’t be 2 Blacks (see option (1) above) she will have no certain knowledge as to whether she is Red or Black and will wait.

Both Bill and Charlie will see 1 Red and 1 Black. Knowing instantly that there can’t be 2 Blacks (see option (1) above) they will both know that they must have a Red sticker and get up almost instantly and simultaneously. This didn’t happen.

So now, Alice must know that she has a Red sticker and would get up almost instantly (she, like the other two, is bright !)

This didn’t happen. Alice has sat for several minutes !!

The third option also needs a bit more thought.

Assume the single black sticker is on Bill (this also works if the black sticker is on Charlie).

Alice and Charlie will both see 1 Red and 1 Black sticker. Knowing instantly that there can’t be 2 Blacks (see option (1) above) they will both know that they must have a Red sticker and get up almost instantly and simultaneously. This didn’t happen – Alice sat for several minutes

The fourth option also needs some thinking.

All three have red stickers.

All three see 2 Reds. Nobody gets up immediately (nobody sees 2 Blacks)

Nobody sees 1 Black (as per options two and three)

So they all must conclude there are 3 Reds including their own.

Alice just happens to be the first to have the courage of her conviction, and gets up.

(though why it took her several minutes, beats me !)

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Don

Bravo, an excellent answer.

I can’t really top it but I will set out my framework (because it’s a useful technique I use for problem solving) and maybe explain the “several minutes” bit.

The approach is one of cognitive empathy (that is, looking at a situation from someone else’s perspective) on increasing levels of complexity:

Level 0, e.g. anyone: “What can I see?”
Level 1, e.g. Alice: “What can Bill see?”
Level 2: e.g. Alice: “What does Bill think Charlie can see?
Level 3: e.g. Us: “How does Alice use level 2?”

Alice needs to leave it long enough for Bill or Charlie to use level 1, but before they advance to level 2. I suggested several minutes for this.

I think you have assumed that the step to level 1 is almost instant, maybe you are just brighter than them!

What I like about this sort of problem is that the subjects seemingly have insufficient information but we should still solve the problem with even less information (we can’t see what Alice sees).

Your solution is far better than when I first attempted it, though in my defence I was only 18 at the time!

Okay, maybe I will try an even harder one for my next teaser!

Mike,

Your first two cracks at this one made all the difference as to how I thought it through.

Without that heads-up start, I might still be struggling.

Good calls, thanks

Don

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That was an interesting teaser Ravvie.

They are sometimes more difficult to create than they are to solve !

The one about lifts, which you said you liked, is a case in point. Ian had to clarify a couple of aspects before we could be certain what the teaser was about, but it was worth it. A really interesting teaser, with a surprising outcome.

I’ve had to apologise for typos and grammatical errors that rendered some of mine grossly ambiguous.

Quite often the ones that look difficult on first read are the easiest to solve, whereas the simple-looking, casual ones are almost impossible - as IB has discovered once or twice :sunglasses:

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Well done to both Don and Mike for solving it between them.

Here’s an old favourite with a bit of 2020 window dressing.

The Impossible Problem

A professor sets his two best students a problem to keep them busy during lockdown. He thinks of two whole numbers greater than 1 (they may be the same). He gives the sum to Sam and the product to Pru. They need to determine the two numbers. (Just to reduce workings, assume the sum does not exceed 40.)

After many hours of finding the problem impossible, Sam sends a text to Pru:

“You must have found it frustrating not being able to solve this.”

A little while later Pru replies: “I have just solved it!”

Sometime later that day Sam replies: “Me too, what a relief!”

What are the two numbers that the professor was thinking of?

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I’ve seen this one before Ravvie, so I’m going to leave it for a while for others.

I might also need a bit of time to generate the two numbers because the conditions (> 1: could be the same: < 40: etc) vary from time to time !

Cheers
Don

Ravvie,
Please could you clarify:-
Can Sam tell Pru his sum, and can Pru tell Sam his product, or does each of them only have their own number?

Sam only has the sum and Pru only has the product. Their only contact is the three texts, nothing more.

Steve, it is solvable so worth persevering.

But it is a bit of a bug*er to work through the options and …


… You do need to be aware of what each person says …

I asked my son the follow up question of the diametrically opposite clock. He worked it out in his head in under half a minute. Impressive!

His solution was equivalent to mine, but a different logic. He said the hour hand would move from 1/7th of the way round to 3/14ths and the minute hand would be half an hour ahead, so 3/14ths+7/14ths gives 5/7ths past 1.

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