Easy math puzzles
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51 members have voted
6210001000.
Quote: WizardCan you find the one 10-digit "autobiographic number." Such a number describes itself with the number in each position, starting with 0, equaling the count of that digit in the whole number.
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Quote: charliepatrickThe logic I used was N210...010...0 works if you know how many zeroes there are, so there is one N. two 1s, and one 2.
6210001000.
link to original post
I agree!
Quote: WizardThis one is an old classic that may have been asked before.
Can you find the one 10-digit "autobiographic number." Such a number describes itself with the number in each position, starting with 0, equaling the count of that digit in the whole number.
For example, 1210 is one because:
There is 1 in in the 0 position and there is 1 numeral 0 in the entire number.
There is 2 in in the 1 position and there are 2 numeral 1's (that's an improper use of an apostrophe, Wiz) in the entire number.
There is 1 in in the 2 position and there is 1 numeral 2 in the entire number.
There is 0 in in the 3 position and there are 0 numeral 3's in the entire number.
Yes, the number is in base-10.
link to original post
I think this works:
6210001000
6x0, 2x1 1x2, 1x6
Is the answer really unique?
ETA: I guess I'm late with this answer. But I'm curious to see a uniqueness proof.
Quote: SkinnyTonyETA: I guess I'm late with this answer. But I'm curious to see a uniqueness proof.
link to original post
The source I got it from indicated it was the only 10-digit answer. I think I could prove no other first digit works.
D: 0 1 2 3 4 5 6 7 8 9
N: 6 2 1 0 0 0 1 0 0 0
P: 0 2 2 0 0 0 6 0 0 0
Note that the sum of Row N is 10 because it says how often each number appears, and so there are ten numbers in total.
Working along the N row , there will be six 0's, then two 1's, then one 2 etc. This also has to total to the ten digits. Hence the sum of the products, as in Row P, also has to be 10.
I suspect there's some kind of contradictory proof, for instance there cannot be 5 zeroes as that would mean there's a 1 under 5 and a total of 4 under numbers 1-4, so that the product is 5: 5xyz010000 etc. don't work.
You are trapped in a circular tunnel of unknown length. Equally spaced apart are at least two switches, which can be either up or down. You may flip the switches as much as you like. You may not leave any markings other than adjusting the switches. How can you tell with certainty how many switches are in the tunnel?
Quote: WizardI'll call this one the Tunnel of Unknown Length Puzzle.
You are trapped in a circular tunnel of unknown length. Equally spaced apart are at least two switches, which can be either up or down. You may flip the switches as much as you like. You may not leave any markings other than adjusting the switches. How can you tell with certainty how many switches are in the tunnel?
link to original post
For i = 2 to infinity:
flip current switch up
walk cw i switches
flip current switch down
walk ccw i switches
if current switch is down return i
Quote: SkinnyTony
For i = 2 to infinity:
flip current switch up
walk cw i switches
flip current switch down
walk ccw i switches
if current switch is down return i
link to original post
I agree. Welcome to the forum and to this thread in particular! We could use some new blood.
BTW, I think I should have said there are at least two switches in the up position.
You are trapped in a circular tunnel. At regular intervals are switches. The switches are far apart enough to be unable to see two at once. The switches can be in either position, except you're told at least two are in the "on" position. You may switch the switches as much as you wish. Otherwise, you may not leave markings of any kind. How can you determine with certainty how many switches are in the tunnel?
The following is not the most efficient solution, but probably the easiest to explain.
1. Go either direction until you arrive at the first switch. If it is in the OFF position, turn it ON.
2. Walk clockwise until you get to the first switch in the ON position. Then turn it OFF. Keep track of how many switches you travel (not counting the starting switch).
3. Turn back the opposite direction the same number of switches from step 2.
4. If you encounter the switch you return to in the ON position, then go back to step 2.
5. If you encounter the switch you return to in the OFF position, then the number of switches equals the number of switches you counted from step 2.
I'm open to any challenges of my solution or comments if it wasn't clear.
Does the starting state (at least two switches up) matter? Is there a nicer (faster, less brute force) solution?
You're mailing a package to a recipient in another country. Because the mail system is corrupt, the contents of any unlocked package will be stolen. You can definitely prevent contents theft by putting a lock on the box. However, the recipient can't open your lock without the key. How can you get the contents to the recipient?
Here are clarifications about thingsI wondered about:
• There is nothing special about the lock or the key. It's not a combination lock. It's not a puzzle lock. The recipient has no ability to manufacture a key.
• You cannot carry the package to the recipient yourself nor use a courier.
My whole family failed this one. I'm not a good problem-solver. I imagine forum regulars will figure it out quickly, though.
[edited for clarity]
Quote: MichaelBluejayThis comes from a Japanese puzzle book. I searched and didn't find it posted here before.
You're mailing a package to a recipient in a corrupt country. If the package is unlocked, it's likely the contents will be stolen. You can definitely prevent contents theft by putting a lock on the box. However, the recipient can't open your lock without the key either. How can you get the contents to the recipient?
Here were the clarifications I wondered about:
• If you mail the key separately in an unlocked package, you can presume the key will be stolen.
• There is nothing special about the lock or the key. It's not a combination lock. It's not a puzzle lock. The recipient has no ability to manufacture a key.
• You cannot carry the package to the recipient yourself nor use a courier.
My whole family failed this one. I'm not a good problem-solver. I imagine forum regulars will figure it out quickly, though.
link to original post
Is there communication between the two parties?
The two parties can communicate as much as they like. I hear they're partial to WhatsApp.
Since AM put an idea in a spoiler (which doesn't work), I'll clarify that one in a spoiler.
God. Damn.
About 20 years ago when IBM's early AI "Watson" played as a contestant on Jeopardy and won, I told the Wizard that that marked the beginning of the end. He replied, "When the robot overlords come for me, my dying words will be, 'Bluejay was right.' "
Quote: MichaelBluejayYou're mailing a package to a recipient in a corrupt country. If the package is unlocked, it's likely the contents will be stolen. You can definitely prevent contents theft by putting a lock on the box. However, the recipient can't open your lock without the key either. How can you get the contents to the recipient?
link to original post
1. Put your own lock on the box and mail it to the recipient. Hold onto the key.
2. Recipient adds his own lock to the box, keeping the key, and mails the box back to you.
3. You take off your own lock and mail it back to recipient.
4. Recipient removes his own lock and reveals the contents.
ChatGpt: 1
Wizard: 1
AutomaticMonkey: 0.5
Bluejay: 0
Quote: MichaelBluejayI asked ChatGPT and it got the right answer.
God. Damn.
About 20 years ago when IBM's early AI "Watson" played as a contestant on Jeopardy and won, I told the Wizard that that marked the beginning of the end. He replied, "When the robot overlords come for me, my dying words will be, 'Bluejay was right.' "
link to original post
That's not so surprising. It's just an advanced search engine plus a very good language engine.
ChatGPT didn't figure out the answer. It found the answer online somewhere. The impressive part (which ChatGPT is very good at) is that it was able to interpret your question and figure out which problem it was referring to.
But it didn't solve it in the way that you would expect an intelligent entity to. It has no intelligence. It just looked the answer up. It doesn't understand the question or the answer.
Quote: MichaelBluejayThis comes from a Japanese puzzle book. I searched and didn't find it posted here before.
You're mailing a package to a recipient in a corrupt country. If the package is unlocked, it's likely the contents will be stolen. You can definitely prevent contents theft by putting a lock on the box. However, the recipient can't open your lock without the key either. How can you get the contents to the recipient?
Here were the clarifications I wondered about:
• If you mail the key separately in an unlocked package, you can presume the key will be stolen.
• There is nothing special about the lock or the key. It's not a combination lock. It's not a puzzle lock. The recipient has no ability to manufacture a key.
• You cannot carry the package to the recipient yourself nor use a courier.
My whole family failed this one. I'm not a good problem-solver. I imagine forum regulars will figure it out quickly, though.
link to original post
I haven't read the Wizard's answer. I'm a little unsure about what will and won't be stolen though. A key will be stolen (despite it being worthless without the lock that it opens). A lock without a key (also worthless) will also be stolen. But a locked package (worthless without the key) will not be.
So my question is, how do I determine what will and won't be stolen? It seems kind of random. Is anything that's not specifically a locked box stolen?
Quote: SkinnyTony
But it didn't solve it in the way that you would expect an intelligent entity to. It has no intelligence. It just looked the answer up. It doesn't understand the question or the answer.
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Most people just want an adequately correct answer, and don't care how it is achieved. If they could look up all the answers to life's problems in the back of the teacher's book, they probably would.
That approach to problems certainly leaves more time to watch reality television.
Quote: DieterQuote: SkinnyTony
But it didn't solve it in the way that you would expect an intelligent entity to. It has no intelligence. It just looked the answer up. It doesn't understand the question or the answer.
link to original post
Most people just want an adequately correct answer, and don't care how it is achieved. If they could look up all the answers to life's problems in the back of the teacher's book, they probably would.
That approach to problems certainly leaves more time to watch reality television.
link to original post
To be clear, I'm not questioning ChatGPT's usefulness. I'm questioning its intelligence (artificial or otherwise).
It's severely limited in that it can only answer questions that have already been answered (as opposed to an actually intelligent entity, who is capable of reasoning and answering new problems). So, maybe this is also about how useful it is.
Quote: SkinnyTonyTo be clear, I'm not questioning ChatGPT's usefulness. I'm questioning its intelligence (artificial or otherwise).
It's severely limited in that it can only answer questions that have already been answered (as opposed to an actually intelligent entity, who is capable of reasoning and answering new problems). So, maybe this is also about how useful it is.
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ChatGPT is less that 3 years old. It's still a baby.
Problem:
You have 3 burger patties and a frying pan that can hold only 2 patties at a time.
Each side of a patty takes 1 minute to fry.
You need to fry both sides of each patty.
Question:
What is the minimum time required to fully fry all the patties?
We are going to roll a single 6-sided die repeatedly until a 6 is rolled, and keep track of how many rolls it takes.
So for example, if you rolled a 2, 3, 1, 6, that would be 4 rolls. If you rolled a 6 on the first roll, that would be 1 roll.
What is the average number of rolls needed, conditioned on the fact that all rolls are an even number? In other words, if we throw out all sequences that contain an odd number, what's the average length of the remaining sequences?
So for example, 2, 2, 6 would be 3 rolls. 2, 2, 3, 6 would not count since it includes an odd number (3). Similarly 2, 1, 3, 6 would not count.
Quote: AnotherBillI've seen this puzzle great many years ago in some printed book. (I asked ChatGPT to translate and adapt it into English (from memory).)
Problem:
You have 3 burger patties and a frying pan that can hold only 2 patties at a time.
Each side of a patty takes 1 minute to fry.
You need to fry both sides of each patty.
Question:
What is the minimum time required to fully fry all the patties?
link to original post
Number the patties 1, 2, 3, and label the sides A and B. Then fry:
1st minute: 1A & 2A
2nd minute: 1B & 3A
3rd minute: 2B & 3B
Quote: SkinnyTony3 minutes
Number the patties 1, 2, 3, and label the sides A and B. Then fry:
1st minute: 1A & 2A
2nd minute: 1B & 3A
3rd minute: 2B & 3B
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Well done.
I'm not so sure about that. From what I've read, AI can indeed solve original problems without having seen a solution.Quote: SkinnyTonyThat's not so surprising. [AI] is just an advanced search engine plus a very good language engine. ChatGPT didn't figure out the answer. It found the answer online somewhere. link to original post
(1) The contents of an unlocked package will always be stolen.Quote: SkinnyTonyI'm a little unsure about what will and won't be stolen though....It seems kind of random.
(2) The contents of a locked package will never be stolen.
(3) A locked package itself will never be stolen.
Quote: MichaelBluejay(1) The contents of an unlocked package will always be stolen.
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Or, the contents of an unlocked package may be stolen but you trust no one and don't want to take any risk :)
Quote: SkinnyTonyWhat is the average number of rolls needed, conditioned on the fact that all rolls are an even number? In other words, if we throw out all sequences that contain an odd number, what's the average length of the remaining sequences?
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I'm not a native speaker, but isn't there any contradiction?
Quote: SkinnyTonyHere's one that I got from a YouTube video.
We are going to roll a single 6-sided die repeatedly until a 6 is rolled, and keep track of how many rolls it takes.
So for example, if you rolled a 2, 3, 1, 6, that would be 4 rolls. If you rolled a 6 on the first roll, that would be 1 roll.
What is the average number of rolls needed, conditioned on the fact that all rolls are an even number? In other words, if we throw out all sequences that contain an odd number, what's the average length of the remaining sequences?
So for example, 2, 2, 6 would be 3 rolls. 2, 2, 3, 6 would not count since it includes an odd number (3). Similarly 2, 1, 3, 6 would not count.
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I'm having some trouble understanding what is being asked. Wouldn't 2-3-1-6 not count, since there is an odd number in there?
Could this be reworded to say that if an odd number precedes the six, then that six doesn't end rolling, but resets the roll history?
For example 2-3-1-6-1-2-3-4-5-6-2-4-6 would be an example of a concluded trial. Note how there are all even number between the last and second to last 6.
Quote: SkinnyTonyThanks!
Does the starting state (at least two switches up) matter? Is there a nicer (faster, less brute force) solution?
link to original post
I don't think I needed the condition about two up switches, but instead could have just said any two switches. Let me think about it.
Also, instead of each time going to the next up switch, you could double the number of switches you passed the last time.
Quote: WizardQuote: SkinnyTonySimilarly 2, 1, 3, 6 would not count.
I'm having some trouble understanding what is being asked. Wouldn't 2-3-1-6 not count, since there is an odd number in there?
That's what I said. 2,3,1,6 would not count.
Quote:Could this be reworded to say that if an odd number precedes the six, then that six doesn't end rolling, but resets the roll history?
For example 2-3-1-6-1-2-3-4-5-6-2-4-6 would be an example of a concluded trial. Note how there are all even number between the last and second to last 6.
But that would be a very long sequence, and we are concerned with the average length of the sequences.
What we do is roll until we get a 6, and record that as a sequence. We can do this repeatedly. In this case the expected length of a sequence is 6. (This is well known I think)
But now we throw out all the sequences that contain at least one odd number. What's the expected length of a sequence that remains?
Quote: SkinnyTony
That's what I said. 2,3,1,6 would not count.Quote:Could this be reworded to say that if an odd number precedes the six, then that six doesn't end rolling, but resets the roll history?
For example 2-3-1-6-1-2-3-4-5-6-2-4-6 would be an example of a concluded trial. Note how there are all even number between the last and second to last 6.
But that would be a very long sequence, and we are concerned with the average length of the sequences.
What we do is roll until we get a 6, and record that as a sequence. We can do this repeatedly. In this case the expected length of a sequence is 6. (This is well known I think)
But now we throw out all the sequences that contain at least one odd number. What's the expected length of a sequence that remains?
link to original post
Thanks, but I still don't understand what's being asked. Maybe somebody smarter than me can rephrase it.
Quote: WizardQuote: SkinnyTony
That's what I said. 2,3,1,6 would not count.Quote:Could this be reworded to say that if an odd number precedes the six, then that six doesn't end rolling, but resets the roll history?
For example 2-3-1-6-1-2-3-4-5-6-2-4-6 would be an example of a concluded trial. Note how there are all even number between the last and second to last 6.
But that would be a very long sequence, and we are concerned with the average length of the sequences.
What we do is roll until we get a 6, and record that as a sequence. We can do this repeatedly. In this case the expected length of a sequence is 6. (This is well known I think)
But now we throw out all the sequences that contain at least one odd number. What's the expected length of a sequence that remains?
link to original post
Thanks, but I still don't understand what's being asked. Maybe somebody smarter than me can rephrase it.
link to original post
I’ll give it a try by starting the list…
SkinnyTony, let us know whether I’ve got the gist…
(rhyme unintentional!)
(partial) List of POSSIBLE sequences:
6
1,6
2,6
3,6
4,6
5,6
1,1,6
1,2,6
1,3,6
1,4,6
1,5,6
2,1,6
2,2,6
2,3,6
2,4,6
2,5,6
3,1,6
3,2,6
3,3,6
3,4,6
3,5,6
4,1,6
4,2,6
4,3,6
4,4,6
4,5,6
5,1,6
5,2,6
5,3,6
5,4,6
5,5,6
…
Now, we eliminate those with an odd number to make the…
(partial) List of INCLUDED sequences:
6
2,6
4,6
2,2,6
2,4,6
4,2,6
4,4,6
…
Now, I’ll attempt a new definition. Allowed sequences will include any number of 2’s and 4’s (including 0 of either or both) and exactly one 6, which is strictly at the end of the sequence.
Quote: SkinnyTonyHere's one that I got from a YouTube video.
We are going to roll a single 6-sided die repeatedly until a 6 is rolled, and keep track of how many rolls it takes.
So for example, if you rolled a 2, 3, 1, 6, that would be 4 rolls. If you rolled a 6 on the first roll, that would be 1 roll.
What is the average number of rolls needed, conditioned on the fact that all rolls are an even number? In other words, if we throw out all sequences that contain an odd number, what's the average length of the remaining sequences?
So for example, 2, 2, 6 would be 3 rolls. 2, 2, 3, 6 would not count since it includes an odd number (3). Similarly 2, 1, 3, 6 would not count.
link to original post
Hasn't this one been asked already? I remember providing an incorrect solution, but I can't remember if it was here or somewhere else.
The probability for each number of rolls needed:
1 roll: 1/6 x 1
2 rolls: 1/3 x 1/6 x 2
3 rolls: (1/3)^2 x 1/6 x 3
...
The sum is 1/6 x (1 + 1/3 x 2 + (1/3)^2 x 3 + ...)
= 1/6 (1 + 1/3 + (1/3)^2 + ...)^2
= 1/6 x (3/2)^2
= 1/6 x 9/4
= 3/8
The probability of a valid sequences is 1/6 + 1/3 x 1/6 + (1/3)^2 x 1/6 + ...
= 1/6 (1 + 1/3 + (1/3)^2 + ...)
= 1/6 / (2/3)
= 1/4
The expected number of rolls over all valid sequences = 3/8 / 1/4 = 3/2.
This has been confirmed by simulation.
Update: this has been asked here before, in February - by me, in fact
You are trapped in a circular tunnel by an evil wizard. Inside the circular tunnel are evenly-spaced switches. You're told there are at least two switches. Each switch can be in either the On or Off position. You can at most one switch at a time only. You're allowed to change the position of the switches as much as you like, but may not leave any other markings in the tunnel.
The wizard will set you free if you correctly tell him the number of switches. However, if you're wrong, you'll be trapped forever in the tunnel. What do you do?
Simple Answer
Here is the simplest answer to explain, but not the most efficient.
1. Go to the nearest switch and turn it on.
2. Go clockwise to the nearest on switch and turn if off. Keep a count of the number of switches you travel to (not counting the starting switch).
3. Turn around and go counterclockwise the number of switches from step 2.
4. If the switch you come to is off, then the number of switches from step 2 is the answer.
5. Otherwise, if the switch you return to in step 4 is on, go back to step 2.
Here is a more efficient answer. Note the first four steps are the same as above.
1. Go to the nearest switch and turn it on.
2. Go clockwise to the nearest on switch and turn if off. Keep a count of the number of switches you travel to (not counting the starting switch).
3. Turn around and go counterclockwise the number of switches from step 2.
4. If the switch you come to is off, then the number of switches from step 2 is the answer.
5. Otherwise, if the switch you return to in step 4 is on, then decide on a multiplier factor greater than 1. For example, 5.
6. Multiply the number of switches you traveled in one direction the last time by the multiplier in step 5.
7. Travel clockwise the number of switches calculated in step 6.
8. Turn off all up switches as you go.
9. As you go, remember the number of switches traveled between the starting switch and the last up switch seen.
10. When you get to the number of switches from step 6, turn around.
11. Go back the same number of switches from step 6.
12. If the switch you come to is off, the number of switches is the total from step 9.
13. If the switch you come to is on, then go back to step 6.
I think it makes for good discussion for the best multiplier in step 5. I think we need a mean distance to properly answer the question. Without that, but assuming the tunnel could be really long, like a google number of switches, the factor should be about 45.
As it's past wine o'clock here I haven't seen an obvious logical reason!
Overall Result: W: 3748231 2497716 Avg: 1.500663406087802 L: 11256251 7502284 Avg: 1.5003765519940329
Parms: sh:10000000 Time:21:39:42:422
Quote: camaplQuote: WizardQuote: SkinnyTony
That's what I said. 2,3,1,6 would not count.Quote:Could this be reworded to say that if an odd number precedes the six, then that six doesn't end rolling, but resets the roll history?
For example 2-3-1-6-1-2-3-4-5-6-2-4-6 would be an example of a concluded trial. Note how there are all even number between the last and second to last 6.
But that would be a very long sequence, and we are concerned with the average length of the sequences.
What we do is roll until we get a 6, and record that as a sequence. We can do this repeatedly. In this case the expected length of a sequence is 6. (This is well known I think)
But now we throw out all the sequences that contain at least one odd number. What's the expected length of a sequence that remains?
link to original post
Thanks, but I still don't understand what's being asked. Maybe somebody smarter than me can rephrase it.
link to original post
I’ll give it a try by starting the list…
SkinnyTony, let us know whether I’ve got the gist…
(rhyme unintentional!)
(partial) List of POSSIBLE sequences:
6
1,6
2,6
3,6
4,6
5,6
1,1,6
1,2,6
1,3,6
1,4,6
1,5,6
2,1,6
2,2,6
2,3,6
2,4,6
2,5,6
3,1,6
3,2,6
3,3,6
3,4,6
3,5,6
4,1,6
4,2,6
4,3,6
4,4,6
4,5,6
5,1,6
5,2,6
5,3,6
5,4,6
5,5,6
…
Now, we eliminate those with an odd number to make the…
(partial) List of INCLUDED sequences:
6
2,6
4,6
2,2,6
2,4,6
4,2,6
4,4,6
…
Now, I’ll attempt a new definition. Allowed sequences will include any number of 2’s and 4’s (including 0 of either or both) and exactly one 6, which is strictly at the end of the sequence.
link to original post
That's not wrong, but the method to get the sequences is important. We are repeatedly rolling a fair die, stopping on a 6, and eliminating those sequences that contain an odd number. There are other ways to obtain the same sequences that might change the answer.
(1) It takes 4 workers 4 days to produce 4 units. How many workers are required to produce 100 units in 100 days?
(2) One person enters an empty concert venue 1 minute after the doors open, and every successive minute, the number of people in the venue doubles. After 12 minutes, the venue is full. After how many minutes is the venue half-full?
Thankfully, I got both of those right, so I kind of redeemed myself. Like a coupon.
Dog Hand
Quote: MichaelBluejayHere are two more from the Japanese puzzles book. These are much easier.
(1) It takes 4 workers 4 days to produce 4 units. How many workers are required to produce 100 units in 100 days?
(2) One person enters an empty concert venue 1 minute after the doors open, and every successive minute, the number of people in the venue doubles. After 12 minutes, the venue is full. After how many minutes is the venue half-full?
Thankfully, I got both of those right, so I kind of redeemed myself. Like a coupon.
link to original post
1. 4
2. 11
Answer: The average rolls to get a six or odd is 6/(1+3) = 1.5. Call this a sequenceQuote: SkinnyTonyHere's one that I got from a YouTube video.
We are going to roll a single 6-sided die repeatedly until a 6 is rolled, and keep track of how many rolls it takes.
So for example, if you rolled a 2, 3, 1, 6, that would be 4 rolls. If you rolled a 6 on the first roll, that would be 1 roll.
What is the average number of rolls needed, conditioned on the fact that all rolls are an even number? In other words, if we throw out all sequences that contain an odd number, what's the average length of the remaining sequences?
So for example, 2, 2, 6 would be 3 rolls. 2, 2, 3, 6 would not count since it includes an odd number (3). Similarly 2, 1, 3, 6 would not count.
link to original post
1/(1+3) chance the six came first and the string ends there. If odd came first (3/4 chance) then an average of six more rolls will be needed to get a six. So 1.5 average rolls to get a six when the six comes first and 7.5 average rolls to get a six when an odd comes first
Check: 1/4 * 1.5 + 3/4 * 7.5 = 6 average rolls to get a six irrespective of even/odd
One would initially assume that you can just consider 2,4,6 which would imply an answer of 3 average rolls to get a six. Assuming this, there is a 1/3 chance of ending a sequence with each roll. In actuality, there is a 4/6 chance of ending a sequence with each roll. The reciprocal (6/4) being the answer
Well, if you can solve the package/key problem, then obviously these two were trivial for you. I mean, if *I* can get them right...Quote: WizardQuote: MichaelBluejayHere are two more from the Japanese puzzles book. These are much easier.
(1) It takes 4 workers 4 days to produce 4 units. How many workers are required to produce 100 units in 100 days?
(2) One person enters an empty concert venue 1 minute after the doors open, and every successive minute, the number of people in the venue doubles. After 12 minutes, the venue is full. After how many minutes is the venue half-full?
Thankfully, I got both of those right, so I kind of redeemed myself. Like a coupon.
link to original post
1. 4
2. 11
link to original post
Four workers working four days is (4x4=) 16 person-days. They produce 4 units, so it takes 16 person-days ÷ 4 units = 4 person days to make 1 unit.
So 100 units should take 400 person-days.
We know we have 100 days, so 400 person-days ÷ 100 days = 4 persons.
To me this is a units-of-measurement problem. "Man-hours" is a related unit, how many hours are worked by all workers in a given situation. Here I just changed the work time from "-hours" to "-days", because that's what the problem called for.
The most useful thing I learned in college was in freshman Chemistry, where we covered unit conversion. I like units of measurement so much I made one.
(2) The venue question is basically a concert Martingale.
An evil wizard puts you in the middle of a desert. He informs you he hid a treasure directly north of you. Yes, you have a compass. To escape, you must find the treasure and bring it back to the starting point. You have absolutely no idea how far north the treasure is. There is a condition that you must declare how far north you will go each trip before leaving. If you don't find the treasure, you must turn back at that point. If you do find it, you must still travel that distance before returning. You may make as many trips as you wish.
What is the most efficient strategy to find the treasure to find it with the least total distance traveled?
Let me know if you have any questions or find the rules unclear.
I say, "I will walk north distance X, and then return," then I walk north distance X and return.
The treasure is hidden in such a way that I am guaranteed to find it if I walk to, or past it. For example, if the treasure is 16.8 km north, and I declare that I am walking any distance of 16.8 km or longer, then I will find it.
Note that if I do find the treasure before reaching distance X, then I must continue north until I am distance X from my starting point, then return.
If I do not find the treasure (because it is farther away than X), then I choose a new X and repeat the process.
Quote: ThatDonGuyLet me see if I understand the problem correctly.
I say, "I will walk north distance X, and then return," then I walk north distance X and return.
Yes!
Quote:The treasure is hidden in such a way that I am guaranteed to find it if I walk to, or past it. For example, if the treasure is 16.8 km north, and I declare that I am walking any distance of 16.8 km or longer, then I will find it.
Yes!
Quote:
Note that if I do find the treasure before reaching distance X, then I must continue north until I am distance X from my starting point, then return.
If I do not find the treasure (because it is farther away than X), then I choose a new X and repeat the process.
link to original post
Yes, exactly. Every time you return you may set a new distance to anything you wish.
Since the treasure is "due north", it is between the starting point and the North Pole.
Let D be the distance to the North Pole.
Choose a positive integer N, and declare distances of D/N, 2D/N, 3D/N, and so on.
Assuming the treasure's location is uniformly possible anywhere between the starting point and the north pole, the expected distance traveled is:
1/N x 2 (D / N) + 1/N x 2 (2D / N) + 1/N x 2 (3D/N) + ... + 1/N x 2(ND/N)
= 2D / N^2 (1 + 2 + 3 + ... + N)
= 2D N (N+1) / (2 N^2)
= D (N+1) / N
This decreases as N approaches positive infinity, so choose as large an N as you feel you can handle.
In other words, choose as small of a distance as you can as your first distance, then after each unsuccessful trip, increase the distance by the initial distance.
Fair warning that as I double check my work, I find there are an infinite number of solutions that are all equally good, which I find surprising.
Never mind crossed-out statement above. I was wrong, I think.
