In the movie when Cumberbatch said the following phrase      There are 159 million, million, million possible Enigma settings. All   we had to do was try each one. But if we had 10 men checking one   setting a minute for 24 hours every day and seven days every week   the sentence      10 men checking one setting a minute   does it mean 10 men checking on single setting in one single minute, or that each of the 10 men is checking a setting in a minute, which will make 10 settings for one minute?
122005
In explaining how hard it is to crack the code of the Enigma machine, Cumberbatch' character says in The Imitation Game:     There are 159 million, million, million possible Enigma settings. All we had to do was try each one. But if we had 10 men checking one setting a minute for 24 hours every day and seven days every week, how many days do you think it would take to check each of the settings? Well, it’s not days; it’s years. It’s 20 million years. To stop an incoming attack, we would have to check 20 million years’ worth of settings in 20 minutes.   For no reason other than curiosity, I tried to calculate whether or not this "20 million years" is correct, and found that it is completely incorrect.  Let's just use simple math to do this: It's stated that the Enigma machine has 159 million million million different settings, that's 1.59×10^20. Let's assume that a man can check a code each second (which is much faster than what was stated in the movie), and while a year has 31,536,000 seconds, then in 20 million years period that would be: 10(men)× 31,536,000(seconds in a year) × 20 ×1,000,000 (20 million years) = 6.3×10^15 settings. Obviously that is so much less than the Enigma settings so there is a huge error in saying "20 million years" as it would take much much longer than that. 25,000 times more to be precise.  So is this a calculation mistake they made?
48441
32 people were invited at a party and started exchanging handshakes. Because of the confusion, each of them shook hands with each other multiple times: at least twice and up to X times. However, every two people exchanged different number of handshakes from every other two. What is the minimum possible number X, so that the above condition is met?
90983
The many-worlds interpretation of quantum mechanics has always been explained to me at a high level using examples of binary events (e.g. atom either did or did not decay at any given moment in time), which leads to a conceptually clean idea of "branching" into two distinct universes. But how does branching work if you have, say, a 70% probability of something happening (e.g. measuring the spin of an electron having gone through an SG apparatus oriented at an arbitrary angle)?  Do you say that 7 universes got to spin up and 3 that got spin down (to account for the 70% probability)? Does that mean those 7 universes are in-every-way identical copies of each other? But then what if you had something with a 71.87154819...% chance of happening? You would need an uncountably infinite number of branches to be able to represent arbitrarily precise probability ratios, and then subsets of those branches would contain uncountably infinite universes that are 100% degenerate and identical to each other. Is this what the standard many worlds interpretation assumes?  If not this, then what? You can't say that there's a 70% chance of universe A happening and a 30% chance of universe B happening if you're saying both happen. How does the many-worlds interpretation put a "weighting function" on different branches or outcomes?
123227
This is for self-study. This question is from Rosen's "Discrete Mathematics And Its Applications", 6th edition.  An arm wrestler is the champion for a period of 75 hours. (Here, by an hour, we mean a period starting from an exact hour, such as 1 P.M., until the next hour.) The arm wrestler had at least one match an hour, but no more than 125 total matches.  1 - Show that there is a period of consecutive hours during which the arm wrestler had exactly 24 matches.  2 - Is the statement in the previous exercise true if 24 is replaced by  a) 2? b) 23? c) 25? d) 30?  My solution to part 1 is the following, based on Rosen's solution to a similar problem given as example in the text:  1 - If I consider $a_i$ to be the number of competitions until the $i^{th}$ hour, then, $1\leq a_1&lt;a_2&lt;\cdots&lt;a_{75}\leq 125$, because there are no more than 75 hours and the total number of competitions is no more than 125. Now I will add 24 to all the terms of the above inequality: $25\leq a_1+24&lt;a_2+24&lt;\cdots&lt;a_{75}+24\leq 149$.  There are 150 numbers $a_1,\cdots,a_{75},a_1+24,\cdots,a_{75}+24$. By the inequalities above, these numbers range from 1 to 149. Then, by the pigeonhole principle, at least two of them are equal (in a list of 150 integers ranging from 1 to 149, at least two are equal).  Because all the numbers $a_1,\cdots,a_{75}$ are distinct, and all numbers $a_1+24,\cdots,a_{75}+24$ are also distinct, it follows that $a_i = a_j + 24$ for some $i &gt; j$. Therefore, from the $(j+1)^{th}$ hour to the $i^{th}$ hour, there were exactly 24 competitions.  Now, I will show the attempt at a solution for part 2:  2 - a) The same reasoning as above can be used:  $1\leq a_1&lt;a_2&lt;\cdots&lt;a_{75}\leq 125$. Adding 2 to all terms:  $3\leq a_1 + 2&lt;a_2 + 2&lt;\cdots&lt;a_{75} + 2\leq 127$  So, there are 150 numbers that range from 1 to 127. Therefore, by the pigeonhole principle, at least $\left \lceil \frac{150}{127} \right \rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 2$. This guarantees that there is a period of consecutive hours during which there were exactly 2 competitions.  b) The same reasoning as above can be used:  $1\leq a_1&lt;a_2&lt;\cdots&lt;a_{75}\leq 125$. Adding 23 to all terms:  $24\leq a_1 + 23&lt;a_2 + 23&lt;\cdots&lt;a_{75} + 23\leq 148$  So, there are 150 numbers that range from 1 to 148. Therefore, by the pigeonhole principle, at least $\left \lceil \frac{150}{148} \right \rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 23$. This guarantees that there is a period of consecutive hours during which there were exactly 23 competitions.  c) $1\leq a_1&lt;a_2&lt;\cdots&lt;a_{75}\leq 125$. Adding 25 to all terms:  $26\leq a_1 + 25&lt;a_2 + 25&lt;\cdots&lt;a_{75} + 25\leq 150$  In this case, there are 150 numbers that range from 1 to 150. So, there are not necessarily two equal numbers. Therefore, we can't conclude anything directly.  But it is possible to show that the statement is not true for 25, because an explicit counter-example (suggested below in the comments) can be found. Suppose the number of matches until each one of the 75 hours is, respectively: $\{1,2,\cdots,25,51,\cdots,75,101,\cdots,125\}$. Here, there are no pairs of numbers whose difference is 25.  d) $1\leq a_1&lt;a_2&lt;\cdots&lt;a_{75}\leq 125$. Adding 30 to all terms:  $31\leq a_1 + 30&lt;a_2 + 30&lt;\cdots&lt;a_{75} + 30\leq 155$  In this case, there are 150 numbers that range from 1 to 155. So, we can't apply the pigeonhole principle here, similarly to the above situation. It is not easy to find a counter-example in this case; I think that, for 30, the statement is always true (that is, there is always a period of consecutive hours during which there were exactly 30 matches). But I'm not sure how to prove it.  Edit: I think I found a way to prove it, based on the suggestion given by Lopsy; I've included it as an answer to this question.  Thank you in advance.
109675
I encountered this sentence when I was learning another language. I have never used such a sentence in English nor seen one, but it seems it exists.  What idea does this sentence trying to convey? What time should we relate to?
25547
Problem (1.8): There are ten girls and four boys in Mr. Fat's combinatorics class. In how many ways can these students sit around a circular table such that no boys are next to each other.   My Attempt: Total number of ways in which both boys are girls can be arranged on a circular table is $13!$ and the number of ways in which any two boys can be together is $2!*12!$. Therefore the answer is $13!-2*12!=11*12!$. Is this answer correct? I am asking this because I am confused about the exact meaning of the line "no boys are together." Does this mean that no two/three/four boys are together?   Note: Meanwhile I have made another attempt on the problem. We can place the $10$ girls on the circular table in $9!$ ways. Then we are left with $10$ spots which need to be filled with $4$ boys. This can be done in $P^{10}_4$ ways. Thus, the number of ways is $9!P^{10}_4$.
31335
I've just gotten started with Nielsen and Chuang's text, and I'm a little stuck. They mention that theoretically, it would be possible to store an infinite amount of information in the state of a single qubit. I'm not sure I completely comprehend this.    Here's how I rationalized it: You take all the information you want to store, put it in binary form, and make it the real component of $\alpha $ or $\beta$ (the coefficients of the computational basis states).   Now I'm not sure if I've understood it right, but since it's still fuzzy in my head it would be great to get some kind of ELI5 explanation or possibly a more detailed picture of how this would, even theoretically, be possible.    Apologies if the question doesn't meet standards. I'm new to the forum and would be open to feedback regarding asking questions or answering them.
55410
There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? Is the answer 1800 or 3600 ?  I have asked this question before () in this site itself to understand the right answer - 3600 or 1800? @Ross Millikan has given me a convincing answer and explained idea of overcounting and why 3600 is wrong. Posting it again here to assert the answer is 1800 and not 3600, mainly because almost all websites give 3600 as answer and only one website gives answer as 1800. Listing the websites for reference  (1) Ans is 3600 see() (2) Ans is 3600 see question 15 in () (3) Ans is 3600 - see 8th question in () (4) Ans is 3600 - see example 5 in () These sites as well as almost all reputed sites give the answer as 3600. But answer is 1800 according to   Sorry for asking the question again. Please tell me which is the right answer. 1800 or 3600. Since I am preparing for my exams, I don't want to be in the wrong path and hence clarifying.
68177
There are five boxes consisting of several identical coins. The first box contains 4 coins, the second box contains 9 coins, the third box contains 7 coins, the fourth box contains 10 coins, and the fifth box contains 3 coins. Some coins are taken from the boxes such that at least a coin taken from each box. In how many ways in taking the coins, such that the numbers of the coins taken are 12?  My attempt was solving it using repetition combination of which, there are only seven coins remaining considering that at least a coin taken from each box. By using the repetition combination, i.e. taking 7 from 9 (five boxes and four borders), resulting in 36 ways. However, there is no 36 ways in the choices.
11113
Context: I was looking at . I didn't know the result referred to in Brian M. Scott's answer is hard enough to require a separate proof   ― .  ― .  (1)  ― What's the intuition? Is Bezout Identity supposed to be intuitive?  (2) — Why doesn't this try work? Let $c$ be any common divisor of $a, b$. Therefore $c\mid a, c\mid b \iff cj_1 = a, cj_2 = b$ for some integers $j_1, j_2$.     Call $g = \gcd$. By definition $\gcd(a,b)\mid a$ and $\gcd(a,b)\mid b$ and $c \le \gcd(a,b)$. Therefore $g\mid a, g\mid b \iff gk_1 = a, gk_2 = b$ for some integers $k_1, k_2$.   The two paragraphs together precipitate $cj_1 = a = gk_1,  cj_2 = b = gk_2$ but this only proves $c\mid gk_1, c\mid gk_2$?    Origin - Elementary Number Theory, Jones,  p9, Exercise 1.8
116249
Here's the question:     In an organization there are $80$ people, $40$ men and $40$ women.   In how many ways can we choose, from those $80$ people, a $31$ member management so that there is a majority of women?    My answer was to first choose $16$ women (to assure the majority) out of the $40$ meaning $\binom{40}{16}$, and then choose the rest $15$ from the rest of all people meaning $\binom{64}{15}$. So the final answer is ${40 \choose 16} \cdot{64 \choose 15}$ but that isn't the right answer so I really don't know.. help? Thank you.    Note $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, the binomial coefficient.
97907
I am able to solve this sort of problem pretty easily.     An arm wrestler is the champion for a period of 75 hours. The arm   wrestler had at least one match an hour, but no more than 125 total   matches. Show that there is a period of consecutive hours during which   the arm wrestler had exactly 24 matches.   However this homework problem I can't seem to figure out.     Tyler has five weeks to prepare for his driver’s test. His mother   volunteers to ride with him for either 15 minutes or one half hour   every day until the test but not for more than 15 hours in all. Show   that during some period of consecutive days, Tyler and his mother will   drive for exactly 8.75 hours.   If I try doing it the same way as the first problem. I get $70$ pigeons and $95$ pigeon holes. I honestly don't know where to start with this problem. I thought that maybe I could assume that they only drove $.25$ hours on each day and prove that there is a period of consecutive days where they drove for exactly $8.75$ hours. Then I could say that we assume there is one day that they drove for $.5$ hours and work my way up, but I can't seem to get that to work either.
9936
Of course, we've all heard the colloquialism "If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet."  I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact, which got me thinking... Is this really true?  Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite.  Therefore, it isn't necessarily given that the monkeys would write Hamlet.  Could someone who's better at this kind of math than me tell me if this is correct?  Or is there more to it than I'm thinking?
101044
I found the following sentence in an American novel (this one:  you can read the whole paragraph). A woman broke her wrist and just had her cast removed. She thinks she can start playing piano again, but her wrist still hurts too much. I'm not sure I understand the meaning of the final sentence:  But a body can only heal so much, so fast.  Can someone help me, please?
74169
There is a common problem in which a long line of N mathematicians are each given a hat that is either red or blue.  They cannot see their own hat but can see all in front of time and can hear any response. They must state their own hat color from the back of the line forwards and no information may be passed besides their guess.  They may have a planning session in advance.  What is the maximum number of correct guesses can they guarantee?  The answer is (look away if you don't know) $N-1$.  If, instead of 2 colored hats, we use $c$ different colored hats, how many can we guarantee?  Assuming $N&gt;&gt;c$  I believe the answer to be $N-c+1$ and have confirmed this for $c=2, 3$, and $4$.  I believe this should be simple to prove by using the concept of "$x$ formulas allow you to solve for $x$ variables" but I am not well versed enough in pure mathematics to answer this.
75339
From  by Richard Brualdi  We have a chess master. He has 11 weeks to prepare for a competition so he decides that he will practice everyday by playing at least 1 game a day. To make sure that he's able to take a couple of breaks, he also decides that he wont play more than 12 games per week.  My question is, how do you prove/dis-prove that there will be a succession of (consecutive) days in which he will play k games, $k \geq  22$?  Specifically, I don't quite understand the use of the pigeonhole principle here.  Thanks in advance! :D  EDIT  I'll try to clarify what I'm asking:  The goal is to prove that for every possible practice schedule that the chess master makes, there will always be a consecutive sequence of days such that he plays exactly k games.  So for example, if k = 22, how do you prove that no matter what his schedule is like, he will always play exactly k games in r days, $1 \leq r\leq  77$.  My main question is: what is the general method for proving different values of k? Are there values of k (from 1 to 132 inclusive) that are impossible to obtain?  Note that he does not necessarily have to play 132 games within 77 days! Also, keep in mind the bounds of k.
63463
I was just wondering, what is the most normal way of interpretating in the last two hours.  If somebody asked you: What have you been doing in the last two hours?  What events from what time period would you mostly cover in your answer if it was already 19:40?     A. I would mention all important things since 17:00 until 19:40(now)   or       B. I would mention all important things since 18:00 until 19:40   (because 40 minutes can be understood as almost hour, but I am not sure if people would count it if it is not full 60 minute hour yet)     C. I would mention all important things since 17:00 until 19:00   Personally, I would go with choice A. But I am not sure if this is the most common way people use it. It might seem that it is almost 3 hours (2h 40m) but so far it is still just 2 hours + some minutes because i would say in the last 3 hours ONLY if 3x60m was completed and would not care if it was more than 3 hours but it still would have to be less than 4 hours.  Does somebody understand me? Which is the most common way of understanding?