Solve Enjoyable Enigmas with Mr.E's Challenge

[zoobyshoe]

Spoiler

fruit

All you want, yes!

 

[Travis_King]

It was time to send the kids to camp, and Sally and Jim were shopping for supplies. They spent half of the money they had plus $4 on socks for the kids; half of what was then left plus $3 on name tapes; and half of what was then left plus $2 on a small wallet for each child. They found themselves with $3 left over. How much did they start with?

FWIW: The book says only 70% of Mensa members who tackled this one got it right. I don't know why. It didn't seem to be that tricky to me.

Spoiler

It's much easier backwards, like 60 times easier

 

[zoobyshoe]

Spoiler

It's much easier backwards, like 60 times easier

Spoiler

That is correct. I'm not sure how you'd do it the other way round.

 

[Travis_King]

Someone sent me this a while back, I'm sure it's searchable, but don't cheat! It's a tough-ish one but not so bad.

You are given eight cards with numbers written on them: 4,4,3,3,2,2,1,1. Your task is to arrange the cards in such a way that the ones are separated by one digit, the twos are separated by two digits, the threes are separated by three digits, and the fours are separated by four digits. You must use all the cards. What is the number you create?

 

[drizzle]

Spoiler

23421314...

 

[collinsmark]

Someone sent me this a while back, I'm sure it's searchable, but don't cheat! It's a tough-ish one but not so bad.

You are given eight cards with numbers written on them: 4,4,3,3,2,2,1,1. Your task is to arrange the cards in such a way that the ones are separated by one digit, the twos are separated by two digits, the threes are separated by three digits, and the fours are separated by four digits. You must use all the cards. What is the number you create?

Spoiler

23421314...

Spoiler

I think Gad's way is valid (as is the symmetrical counterpart 41312432).

But I think the problem, as stated, is ambiguous in what is meant by being separated by a certain amount of digits.

One way to interpret being "separated by one digit" is the way Gad did, in that the as in xxxxaxax are separated by by one 'x' (where x is some other number besides a.

But another way to interpret "separated by one digit" is being adjacent to each other. In other words they are one digit away from being right on top of each other. For example in xaxaxxxx the as are separated by 2 (as in the second a is positioned two units away from the first). If that's the case, a valid solution could be 42324311, or its symmetrical counterpart 11342324.

 

[lisab]

Spoiler

I think Gad's way is valid (as is the symmetrical counterpart 41312432).

But I think the problem, as stated, is ambiguous in what is meant by being separated by a certain amount of digits.

One way to interpret being "separated by one digit" is the way Gad did, in that the as in xxxxaxax are separated by by one 'x' (where x is some other number besides a.

But another way to interpret "separated by one digit" is being adjacent to each other. In other words they are one digit away from being right on top of each other. For example in xaxaxxxx the as are separated by 2 (as in the second a is positioned two units away from the first). If that's the case, a valid solution could be 42324311, or its symmetrical counterpart 11342324.

I interpreted it the way Gad did, I think she got it . Travis, is that right?

 

[consciousness]

I had to write out and solve 3 equations. What's "hit and trail"?

Spoiler

The trick is to do everything backwards. Start with 3, add 2 and double to get 10. Then add three and double...oh and I meant "trial and error"

 

[Enigman]

Another FRUITY puzzle...
And no, you can't eat them Gad, not yet.
You have been given two oranges one unpeeled the other peeled and an aquarium which is full of water- nothing else.
Your job- make them both float. You can't use anything other than the things given.
---------------------------------------------------------------------------------
mmm...the thread got quite a few enigmas while I was away...delicious...

 

[Travis_King]

I took it the way Gad did as well, so we'll call that the correct answer. Though I admit when I first read the problem I too wondered if they meant that "separated by one digit" meant that the two 1's were adjacent. However, after reading it a couple times, I think it was clearly intended to mean that the two 1's must be separated by another digit between them. Congrats

 

[drizzle]

*cuts both pealed and un-pealed oranges into two half, and eats half of each leaving the other halfs for PFers to solve*

 

[Office_Shredder]

OK, let's roll with Gad's solution. I eat both oranges entirely, then do a backfloat in the aquarium (which I assume was designed to hold dolphins).

A more real attempt at a solution which is probably wrong

Spoiler

OK, I didn't even know which one was supposed to float normally so I looked it up on youtube. I found a couple videos showing the peeled orange is the one that sinks. They both claimed that the reason why is that the rind is less dense than the orange - for the purposes of this solution I disbelieve their claims because neither showed the rind itself floating, so really they shouldn't be calling themselves science videos if they don't even test their hypothesis.

Instead I'm going to assume the rind is less dense than water, but the orange itself has air pockets in it.. When peeled, the orange's air holes are filled by water and it sinks. So if you peel half of the unpeeled orange, and place that under the peeled orange, that orange should be able to float as long as it only sinks half of its volume into the water. If the peel is heavier than water, instead place the peel on top of the orange to prevent air from escaping and then place two oranges with rind-hats in the water.

If I had an orange I would test to see if either of these work, but alas I do not.

 

[Enigman]

OK, let's roll with Gad's solution. I eat both oranges entirely, then do a backfloat in the aquarium (which I assume was designed to hold dolphins).

You violated the first rule of fruit puzzles:
Gad always gets the fruit!

Spoiler

the rind is less dense than water, but the orange itself has air pockets in it.. When peeled, the orange's air holes are filled by water and it sinks. So if you peel half of the unpeeled orange, and place that under the peeled orange, that orange should be able to float as long as it only sinks half of its volume into the water. If the peel is heavier than water, instead place the peel on top of the orange to prevent air from escaping and then place two oranges with rind-hats in the water.

Correct.

Spoiler

The first assumption is correct. And if they are the same size (a real snug fit) it doesn't matter where you place the rind it will always return to rind on top position.

Spoiler

I probably should have mentioned the size constraints- unpeeled orange cannot be too much smaller than peeled one.

 

[collinsmark]

So we don't have a complete answer yet to the oranges; is that right?

Spoiler

The question that remains in my mind is how does one attach the previously peeled orange to the half rind, without the peeled orange falling out, and down to the bottom of the aquarium. (The orange/rind combo will naturally tend to "capsize.")

 

[Enigman]

So we don't have a complete answer yet to the oranges; is that right?

Spoiler

The question that remains in my mind is how does one attach the previously peeled orange to the half rind, without the peeled orange falling out, and down to the bottom of the aquarium. (The orange/rind combo will naturally tend to "capsize.")

Spoiler

As I said size constraints- the peeled orange has to be slightly larger than unpeeled orange...(should have mentioned it) then you can just stretch it to fit the orange and conversely if its larger you can manage it with some clever balancing- you will need to use slightly more than half a rind...isn't very stable though.

 

[lendav_rott]

Spoiler

Throw the peeled and the unpeeled orange into the water tank, they will float.

 

[inotyce]

Does adding salt into water help float the fruits ?

 

[Enigman]

Spoiler

Throw the peeled and the unpeeled orange into the water tank, they will float.

peeled one sinks- already have done this at home. Office shredder gave the correct anser.

Does adding salt into water help float the fruits ?

Yes, it would but you can't use salt, sorry.

 

[Enigman]

Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough light left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and it’s only strong enough to support two people at any given time. Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slow poke takes 10 minutes to cross. How do the campers make it across in 17 minutes?

 

[Ibix]

Deleted because spoiler tags didn't work right - I'll post again later.

 

[Ibix]

Try again.

Edit: fluff! Because the first few characters of the post shown in the post list does not respect the spoiler tag - in the app at least.

Spoiler

2 & 1 cross, 2 returns (4 minutes)
5 & 10 cross, 1 returns (11 minutes)
2 & 1 cross (2 minutes)
Total - 17 minutes

That's about the third time I've seen the problem, but the first time I've solved it. :D

 

[zoobyshoe]

Spoiler

2 & 1 cross, 2 returns (4 minutes)
5 & 10 cross, 1 returns (11 minutes)
2 & 1 cross (2 minutes)
Total - 17 minutes

Spoiler

I think this must be right! Excellent job! It had me going in circles all morning.

 

[Ibix]

I puzzled over it for hours - fifteen years ago. For some reason it clicked in about two minutes today - maybe because I'd had time to think about it.

Here's one that took me ages (then I asked a mathematician friend who stared at me for about five seconds before reeling off the answer, which made me feel a bit dumb).

I have eleven dalmations. Prove that it is always possible to select some or all of them such that the sum of the spots of the selected dogs is an integer multiple of eleven.

You may consider eleven bunches of grapes, if that makes the problem more familiar.

 

[zoobyshoe]

What's an "integer multiple," as opposed to just a multiple?

 

[Enigman]

Try again.

Edit: fluff! Because the first few characters of the post shown in the post list does not respect the spoiler tag - in the app at least.

Spoiler

2 & 1 cross, 2 returns (4 minutes)
5 & 10 cross, 1 returns (11 minutes)
2 & 1 cross (2 minutes)
Total - 17 minutes

Correct.
And the app's buggy- it doesn't show spoilers properly...

 

[Ibix]

What's an "integer multiple," as opposed to just a multiple?

The total number of spots is 11n, where n is an integer.

"Integer" is a redundant modifier in this context because the problem is trivial if you interpret "multiple of eleven" as 11x where x is real. Sorry.

 

[Ibix]

Correct.

Woohoo!

And the app's buggy- it doesn't show spoilers properly...

So I realized. You need to reload the page after posting to not see them (as it were), but then you can't always see the full text when you want to.

 

[collinsmark]

I have eleven dalmations. Prove that it is always possible to select some or all of them such that the sum of the spots of the selected dogs is an integer multiple of eleven.

You may consider eleven bunches of grapes, if that makes the problem more familiar.

Stayed up till 1:00 AM last night trying to figure this one out. Much of that time was brushing up on Galois fields, since I suspect the subject might be related, together with the "characteristic" of a finite field. I haven't been able to prove anything yet though.

[Edit: I also started coding up a program to prove that the assertion is true (I don't doubt that it is true) or show that it is false. But then I realized that the inner loop would need to go through about 2¹¹11¹¹ iterations (584,318,301,411,328 iterations). Even with my fast computer, it would take a very long time. I could knock that number down significantly by taking advantage of symmetry, but even then it would still take a long time (just not quite as long). So I stopped that effort.]

 

[consciousness]

I also stayed up for this problem, managed to simplify it somewhat but the "basic" logic behind the question eludes me. I have abandoned it for now, because the solution might require some number theory theorem which I don't know and have little chance of stumbling upon.

 

[Enigman]

...I have postponed solving it for a week...got tests coming up...Last idea was it holds even if the 11s are exchanged by 1, 2 or 3...perhaps its a general trend?...Need to focus on physics now.
Sayonara