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Ah yes. Revision: all prime numbers and numbers that are the product of primes will be tails. Primes are flipped once. Products of n prime numbers have have 2n factors, but as they are not flipped at 1 they are flipped an odd number of times (2n-1) and thus tails.Quote:
Originally Posted by Xei
A 9.6% chance of the results being false means 9.6% of the people with positive results have no cancer. But this is different from 9.6% of people with no cancer having positive results.Quote:
Originally Posted by Somii
I recommend you draw a rectangle representing a population of 10,000 typical people, then subdivide it into people with cancer with positive tests, people without cancer with positive tests, etcetera.
Your approach is the closest yet, though your answer is not. All numbers are the product of primes, so as far as I can tell your current argument has all coins tails up.Quote:
Originally Posted by Photolysis
I would have said that that for cut n, the maximum number of pieces that can be intersected with a straight cut is also n, thus making the maximum number of pieces the nth triangle number. Or the sum of the geometric series 1+2+...n, k=1.
Question: does the pizza have to retain its shape whilst being cut? If not, then it's 2^n pieces.
Yeup, true. To revise it some more.Quote:
Your approach is the closest yet, though your answer is not. All numbers are the product of primes, so as far as I can tell your current argument has all coins tails up.
All square numbers are heads, every other number is tails. All square numbers have an odd number of factors and are thus flipped an even number of times. If that's not right then I shouldn't do this stuff so late :P
Right. Stupid me. Thanks for seeing through my problem with word selection.Quote:
Originally Posted by Xei
The reason is inherent in the answer...?Quote:
And of course, the point of the coins question is to provide a reason. Anybody can assert an answer after checking a few cases.
A coin n will be flipped one time for each divisor of n that's greater than 1?
That's fairly obvious once you look at a few cases if not before?
I'll think about the friends problem but might get bored.
1. all of the squares (I did it with 10 coins)
2. 24 (?)
3. 85.2% (?)
4. 8/103
5. 2 & red
6. the age of the beer drinker and the drink of the 16 yearold.
7. true (?)
(?) = guess
edit: #5 fuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuck. (I knew better... I've read this problem in books before, lol)
I wasn't trying to be sarcastic if that's what you mean... friendship could plausibly be reflexive.Quote:
Originally Posted by PhilosopherStoned
Of course, if your answer is 'coins with an odd number of divisors', then it follows immediately. But that's not far off from trivially answering 'coins which are flipped to heads by the given process'. The meat of the question is in explaining why odd number of divisors is equivalent to the more concise condition you stated of being a square.Quote:
The reason is inherent in the answer...?
A coin n will be flipped one time for each divisor of n that's greater than 1?
That's fairly obvious once you look at a few cases if not before?
It's quite a quick and linear question (it took me thirty seconds or so, but I'm used to these kinds of things).Quote:
I'll think about the friends problem but might get bored.
1. Yep, why though? (Nobody's nailed this one yet).Quote:
Originally Posted by IndieAnthias
2. Close, try to think of how you'd answer it if it were 100 cuts.
3. Nope, but try calculating the probabilities for various things in question 4. :P
4. Ya.
7. Maybe. :P
7. True.
Here's my answer to why.
Let n be group size
Let x be the number of friends a person has
Let z be the number of people in the group with 0 friends
Let d be the number of discreet values x may take
For this to be false, group size n must have a d value of n.
A person can only have a maximum of n-1 friends, x cannot be negative, and a person with 0 friends cannot increase the value of x for someone else therefore:
0<= x <= (n-(1+z))
If z=0, d=n-1, therefore for the statement to be false, z must =1.
Since the number of people with zero friends must be positive
0<= z <= n
However, a person with no friends cannot increment the friend count of someone else so for z=1, d = n-z
for z=0 and z=1 d>n, therefore statement is true.
The 2 card and the blue card.Quote:
5. I have four cards with numbers on one side and colours on the other. I claim that cards with even numbers are red on the other side. I put the cards in front of you. They are
1, 2, red, blue.
Which cards do you need to turn over to check my claim?
Since a person without cancer can only have a 9.6% chance of testing positive, no person without cancer can have a >9.6%> chance of testing positive. If a person is tested positive, which assumes a person is 100% positive, then it follows that with exclusion to this assumption, I may have a < 90.4% chance of having cancer. (This was actually the answer before the "original" one I tried to force. Albeit a bit broad, it's still possible. Who needs such silly complex math when you can make fenced guesses!) I win :party:Quote:
Originally Posted by Xei
I'm an engineer undergrad. How sad :(
Thanks Xei, this was fun.
I'm answering these in Notepad before reading any of the other answers. I'm using a pencil, paper and calculator, not looking anything up online or anywhere.
1. Any y such that y = x^2, where x is an integer and y is between 0 and 1000. So 1, 2, 4, 16, 25, 36, etc. These are the only ones that will be flipped an even number of times. But I don't know why. I arrived at the conclusion through induction and recognized the pattern.
2. 22. I made a table:
Spoiler for Table:
The total resulting slices gives 22. That's assuming it's always possible to cut through every previous edge. I drew it out to make sure of this. And all of this is hypothetical, In real life, some of the slices would be so ridiculously small.
Spoiler for Circle:
3. This is unfair because I've taken STAT courses and know how to do this. I doubt
I'd be able to figure it out if I hadn't.
Spoiler for Cancer diagram:
4. T = tree-huggers
S = pretend to be tree-huggers
genuine tree huggers = T/(T+S) = 80/(80+950) = 80/1030 = 8/103
5. 2 and red. That was too easy. Is there some trick I'm missing?
6. If I take your question literally, you didn't specifically say what the minimum drinking age is. If you can drink at 18, maybe the minimum age is still 14 or something so neither would be breaking the law. So I guess you'd have to check the minimum drinking age, and then make sure the one drinking beer is at least that age. (Assuming the one drinking coke isn't drinking rum and coke or something). Also, I suppose you'd have to check everything about both of their lives, since they could be breaking the law in some way other than drinking alcohol. There are other possible word-play problems too, like maybe you just mean drinking anything, not just alcohol.
7. Do the friends have to be within that group? If they don't, then false. Otherwise, true. If none of them are friends, they they all have 0 friends. If there's at least 1 pair of friends, yes. I'm not sure why, I just don't see any possible arrangement which might make it false. It might be fun to try to prove it but I don't have much time to think about it lol.
Impressive stuff Dianeva. The pizza thing is right, the key insight of course that a straight line can only intersect another straight line once. If you fiddle about for a while you can find a nice way of doing it:
Spoiler for :
And 3 and 4 are right too of course.
The funny thing about 5 is that it is technically easy, yes, but your answer isn't correct. :P
The vast majority of people get this wrong.
Just take 6 at face value.
I won't comment on 1 or 7 because along with the pizza one they are the best. You've done the first half of what a mathematician does, which is 'experiment with the question and try to find a pattern'. The second part, and really the part that requires almost all the thinking, is in finding an iron cast reason for your suspected answer.
The main focus of my degree is to 'learn how to think well'. It's quite a subtle process by which they teach you so it's interesting to me to see how well others fare at pure, hard thought. I suggest mulling them over when you have time because they're quite satisfying... the friends one is probably the one to do first. The only thing I'll say is not to get lost in drawing complex diagrams, the answer can be distilled into a few lines of plain English and can be arrived at in a pretty linear, general fashion (I didn't need to draw anything).
I think I'm okay at reasoning, but you like to point out how horrible I am at it. Actually you like to point out how horrible everyone is at it.
Well, I tried reading the questions, but I shut down when I figured out they were math problems.
Okay my final answers for all except #6 which I still haven't looked at, along with the explanations.
1. All square numbers are heads, rest are tails.
For a natural number n, where n>1, all non-square numbers have an even number of factors. Square numbers have an odd number of factors because at least one factor is multiplied by itself.
A number has f factors, and because they are not flipped at 1, they are flipped (f-1) times.
Where f-1 is odd, the coin lands on tails. Where f-1 is even, the coin lands on heads. Therefore only square numbers are heads.
(Much easier if you look at the problem instead of trying to be clever as I did and jumping straight in. Though it was late. Yes that's my excuse!)
2. A straight line can only be intersected once, so cut n can only intersect n pieces.
The number of pieces you can obtain is the nth triangle number+1 (since you start with one piece already) , aka the result of the geometric series (1+2...+n, k=1) +1
for n=6, the number of pieces is 1+1+2+3+4+5+6 = 22
(Note I must have miscounted with my diagram... should have caught this one earlier.)
(Note 2: Also do not go to parties with mathematicians!)
3. You have to look at the number of people getting a positive result in total (including false positives)
0.01*0.8% get true positive results
0.01*0.2 get false negatives
0.99*0.096 get false positives
0.99*0.904 get true negatives
The chance of actually having cancer is therefore 0.01*0.8 / ( [0.01*0.8] + [0.99*0.096]. Works out at 800/10304 or ~7.76%
200/89696 or ~0.22% get a false negative.
Because the number of false positives is much higher than those who have cancer and get a genuine positive result, there is a very good chance that even a positive result means you don't have cancer, which is why doing multiple tests is so important in the real world.
4. It's similar maths here except the numbers are worked out.
80 people claim to hug trees who actually do
950 claim to who don't
80/(950+80) = 80/1030
5. You flip the 2 and blue cards.
You do not say that all red cards have an even number on the other side, only even numbers must have red. Therefore you don't need to flip 1 because the colour doesn't matter, and you don't flip the red card because the number doesn't matter (it could be odd).
If the two is not red, or if the blue has an even number on the other side this would disprove your claim.
7. For group size n, where n>=2, the maximum number of friends a person can have (x) is n-1 because a person cannot be friends with themselves. Let f be the number of friends a person has and z the number of people in the group with zero friends.
For the statement to be proven false, a group size n must have n different values for f. Since the maximum value for f is n-1 the only way this would be possible is if someone had zero friends.
However a person with zero friends cannot increment the friend count of someone else because friendships are two way, and so every person with zero friends further reduces x by 1.
Therefore the number of values f may take is always less than n, therefore the statement cannot be proven false therefore it is true.
No, I just disagree with you frequently because you are often wrong. What a whiny thing to say.Quote:
Originally Posted by DeeryTheDeer
Only three of seven. Cool excuse though.Quote:
Well, I tried reading the questions, but I shut down when I figured out they were math problems.
I'm often wrong on what? Opinions?
Anyway, I'll take an actual stab at some of the questions, don't know if I'll get any of them right though.
3. 70.4%
4. 80 out of 1050
5. 2 and red (my first guess was all of them)
6. The one drinking beer
7. True....
I just realized that (5) is 2 and blue. What the hell was I thinking. I didn't even make a mistake before with the wording. I translated it in my mind to [even number] --> [red other side] and realized it didn't have to be the other way around, but still made that mistake somehow. The 2 has to be flipped to make sure it's red on the other side, and the blue has to be flipped to make sure its other side is an odd number. No need to comment on that unless for some reason it's wrong.
EDIT: I was trying to work out the last one and have so far failed, but I'll keep trying. I was thinking of using mathematical induction. Showing that in any random configuration in which at least two are shared, any added line must also produce a graph in which at least two are shared. (I've been thinking of it on paper like a graph consisting of dots and edges in between them to represent friendship). I have no idea if I'm on the right track or not. I like these problems that sound so simple but answering them is so difficult, but then in the end the answer is simple again.
Math can solve a lot of puzzles. But yeah, It doesn't make sense to judge anyone's intelligence on their abilities to answer the math ones. They require specific knowledge learned in University-level courses (or in one's free time if they're into that) to answer. The pizza one I came up with on my own, but for the many of the others, there's no way I'd be able to answer them if I hadn't taken a couple STAT courses, math courses and a discrete logic & data structures course. And I still can't seem to get the reasons for 1 and 7.Quote:
Well, I tried reading the questions, but I shut down when I figured out they were math problems.
Ah, I should have just tried to do this, this is how I imagined doing it.Quote:
Originally Posted by Xei
I skipped the statistics since I haven't learned how to do this yet but I'm really anticipating the answer to number 1.
For number 5 you just flip the 2, since you didn't say that an odd number can't be red. To completely check the claim I guess you would have to flip both colored cards too.
For 6 I suspect an underlying trick, but my best guess is you would have to check what the 16 year old is drinking and if the person drinking beer is 18+.
Number 7 kinda boggled my mind lol At first I was almost sure the answer didn't have to be true, but after reasoning it out with a few examples I couldn't make it to be false.
Ok, I realize for 5 flipping the red card may additionally confirm the claim but isn't necessary.
I thought the meat was in figuring out the odd factor thing. The squares thing is (imo)just a cute way to use a well known fact from number theory to cover it up and make it look cool. I hate it when people do that.Quote:
Originally Posted by Xei
At any rate, an integer has an odd number of factors if and only if it's a square.
proof:
(if)
Suppose it's a square, N=n2. We need to show that |S| is odd where S is the set of factors and |S| is the number of elements in S. Then for every divisor, d, N/d is also a factor. Hence we get a map f from S to itself given by f(d) = N/d. Note that f(f(d)) = f(N/d) = N/(N/d) = d. Hence we can partition S into distinct sets, S_d = {d, f(d)}. if d =/= n then d =/= f(d) and so each set formed from such d has two distinct factors. But n lays in S and so S_n = {n}. Hence S is partitioned into some numer k of sets with 2 elements and 1 set of 1 elements, and we can count it as |S| = 2k + 1. This is trivially an odd number as it is one greater than the even number 2k.
(only if)
Now suppose it has an odd number of factors. We need to show that it must be square. (This is the part that everybody seems to have forgotten so far.) Using the same labels as above, partition S into sets Sd = {d, f(d)} using the same f. Again, f(f(d))=d and so there is a maximum of two elements in a given partition. Let k be the count of partitions with 2 elements and p be the count of partions with 1 element. Then we have |S| = 2k + p. For |S| to be odd, p must be odd and hence not zero. Hence there is at least one d with |Sd| = 1, and f(d) = N/d = d and so N = d2.
Now that's quite excessively nailed :)
Yeah, you're right. Took me longer than thirty seconds and Photo made it to the comp first but I'll post my argument anyways as it's cleaner and actually correct rather than just being very close. ("it cannot be proven false, therefore it's correct"? :P Flirting with Goedel there...)Quote:
It's quite a quick and linear question (it took me thirty seconds or so, but I'm used to these kinds of things).
For a group of N people, there will be two with the same number of friends.
Proof:
For a person P, let f(P) be the number of friends.
First suppose that there are n people with zero friends. If there are two people with F(Pi) = F(Pj) among the N-n people with friends, then F(Pi) = F(Pj) among all N people as somebody with zero friends won't change the value of F. Hence we can assume that n=0, that is everyobdy as at least one friend.
Hence we have 1 <= F(P) < N for all P.(we're not counting friendship with oneself).
But then there are N values of F(P) to be chosen from among N-1 values. Hence by the pigeon hole principle, at least one value will occur at least twice.
I'm gonna say only the first one is still heads up, coz it never gets flipped and the rest flip on to tails.Quote:
Originally Posted by Xei
I didn't do this mathematically per se, I just drew it out, cutting through the maximum number of lines with each new line, I got 22.Quote:
Originally Posted by Xei
1%Quote:
Originally Posted by Xei
80/950Quote:
Originally Posted by Xei
The 2 card. EDIT: Just revising this, I thought I was smart by doing the smallest amount possible. You would also need to flip the red card to make sure it was even. And also the Blue card lol And also the 1 hahaQuote:
Originally Posted by Xei
Of course IRL it would depend on whether the 2 ended up having blue on the back or not, and so on. But yes, you need to check them all to make sure all reds are even and all evens are red.
Check if the bar keeper checked for ID for the 16 year old and the one drinking beer. And check if there's alcohol in the coke :lol:Quote:
Originally Posted by Xei
Not sure what N.B means, but....I'd say true. Since we only have to check with a group of 2 people, they could both either have 1 friend in that group, or none.Quote:
Originally Posted by Xei
Thanks Xei, that was fun. Now I'm off to check the other responses to see how far off I was :lol:
Now I feel like I'm in school again, waiting for my marks on a test :lol:
Don't be mean to Deery. His avatar his cool.Quote:
Originally Posted by Xei
Version #1) You just flip the number 2 card, for that is the only even number card in front of me that is respective to your claim. The "other" side, presumably, is strictly the side opposite of what's facing me.Quote:
5. I have four cards with numbers on one side and colours on the other. I claim that cards with even numbers are red on the other side. I put the cards in front of you. They are
1, 2, red, blue.
Which cards do you need to turn over to check my claim?
Version #2) If each card has strictly numbers on one side, colors on the opposite side, and some numbers are colored in themselves, then it follows in respect to your claim that we need only to check red and blue.
Version #3) 2 and red. 1 and blue are irrelevant, since we know at least one side for each card excludes a condition to the claim.
No inconsistencies of the law :)Quote:
6. In the UK you can drink at 18. There are four people drinking at the bar. One is 21, one is 16, one is drinking coke, one is drinking beer. What do I have to check to make sure nobody is breaking the law?
These are the only non-math ones, I think, that no-one else has answered correctly?
Apparently I suck... the only one I got without looking was the cards one...
It's late and I'm tired so I only attempted 2, think it's 22. The sequence formed is just adding one more each time as each time the wire can go through one more previous cut. So you get the sequence 1 2 4 7 11 16 22. Think that works out as, (n(n+1))/2 +1.
1.To be flipped at the end it must have an odd number of divisors not including 1, and since in non squares the divisors can all be paired off and must be even, all numbers that are not square or prime must end up being flipped an odd number of times and so tails up (squares cant be paired off as one number would be paired with itself and in this coin game you are only going through each number once.). All primes must be tails up as they are only flipped once. So all that is left is squares. That argument doesn't quite feel right, think I've done something stupid but I'm going to go with all squares heads up.
2. answered last night.
3. Well that's just the chance of having cancer and testing positive over the chance of having cancer and testing positive added to the chance of not having cancer and testing positive. Numerically that's
(0.01*0.8)/(0.01*0.8+0.99*0.096)
4. 8/103 not really sure that one needs any explanation.
5. 2 and blue, 2 so you can see if it's red and blue so you can see if it's even.
6. Same thing again, 16 to see if its alcoholic, beer to see if their 18.
7. May have to mull this one over a bit. Can't immediately think of any counter examples but not got any reason that it would be true yet.
EDIT: with 7, if you think of it as a network you are essential saying that with friends as nodes you must always have 2 or more nodes with the same order. Now for a network with n nodes, since each person can only be connected to each other person once. The maximum order of a node is n-1. Now they can only take integer values and with n nodes and only n-1 different possibilities at least 2 must have the same value. Now you could say that n-1 possibilities doesn't include 0 but then if a person has no friends in the group then you can just treat the group as having one less member and so the argument still holds.
Again probably made something wrong there, normally seem to overlook something but seems sound to me.
Actually, I got both of those correct. :PQuote:
Originally Posted by Somii
Does anybody have any more cool puzzles; especially like questions 1, 2 and 7?
I'm busy tonight but I'll try to post optimal answers to those three tomorrow for people to check against.