I'm confuse with thousand numerals. Especially for 4 distinct digits and 3 distinct digits. For example:  How to pronounce 1987?  a)  one thousand nine hundred eighty seven  b) nineteen "and" eighty seven  c) nineteen eighty seven  d) All correct  Is it always allowed to make a partition on a number so we can speak it easily? So, if i have 6 digits number, i can divide it 2 pieces or 3 pieces?. Example: 111222. I will pronounce it as   a) Eleven twelve twenty two, or  b) one hundred eleven two hundred twenty two  Those sound weird to be honest. Or maybe that rule only holds for thousand numerals?  Second question if i have 3 distinct digits on thousand numerals. Example:  2003  Which one is true between:  a) two thousand three  b) two thousand "and" three  c) twenty "o" (Read: oh) three, i heard this from facebook video.  d) All true?  Please explain to me with easy language, because my English isn't good enough. Thanks in advance.
58013
In German, we often use "Elfhundert" (literally, "eleven hundred") for 1100 or "neunzehnhundert" ("nineteen hundred") for 1900; but is this correct in English?
123512
A man counts numbers and skips the numbers that have digits divisible by 3. The first ten numbers are 1, 2, 4, 5, 7, 8, 11, 12, 14, 15. What is the 100th number?   Quick googling for solution gives an interesting alternative method using change in base:     Its actually base 6 system, So $100$ in base 6 is $244$ in decimal.   But as $3$ is not included in the system, 4 will actually be 3, 5 will be 4, so on, Hence $100th$ number is 255   Can someone explain this please?  I tried same method on similar question, to find $100th$ number for the sequence that skips digits divisible by 2 or 5.  100 in base 4 is 1210 and reconverting it back yields 1310. Which is wrong.  Many thanks in advance,  Chris
121901
501 can be pronounced 'five hundred one' or 'five zero one'. How do I tell someone how it was pronounced without saying the number. The best I've come up with is 'normal style' vs 'zip code style'. How about some proper terms? What about the distinction between 'five zero one' style and 'five oh one' style? I know this bit is a long shot but what the heck.
50411
I still remember the feeling, when I learned that a number is divisible by $3$, if the digit sum is divisible by $3$. The general way to get these rules for the regular decimal system is asked/answered here: . Now I wonder, what divisibility rules an alien with $12$ (or $42$) fingers would come up with?  So let $n=\sum_k c_k b^k$ be the representation with base $b$. Looking at some examples shows indicate that $n$ is divisible by $b-1$, if $\sum_k c_k$ is. This seems to be a poor man's extension to the decimal divisibility by $9$ rule.  The answer to the above mentioned question, says that "One needn't memorize motley exotic divisibility tests. ". Motley exotic divisibility tests are very welcome here!
115474
Is it OK to use the amount of with countable nouns? I have come across the usage of it with people, but I am confused whether to use it with countables.   What about the proportion of? How different is it from the amount of and the number of ?  Thank you in advance for the answer.
63266
\num{100000} prints 100 000  I want to have 1 00 000.  Question: How to set siunitx package to do the above?  Some more  \num{1000} => 1 000  \num{10000} => 10 000  \num{100000} => 1 00 000  \num{1000000} => 10 00 000  \documentclass[12pt]{article} \usepackage{siunitx} \sisetup{group-separator = {,}}  \begin{document}  \num{1000}  \num{10000}  \num{100000}  \num{1000000}  \num{10000000}  \num{100000000}  \num{1000000000}  \SI{1000000000}{m}  \end{document}   The exact rules are described on the :     The Indian Numbering System uses separators differently from the Arabic system; besides the three least significant digits of the integer part, a comma divides every two rather than every three digits.
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I have troubles writing a fraction whose items are words not numbers.  In a nutshell I should write:  minaccia/vulnerabilità   --&gt; this is the numerator ----------------------  contromisure             --&gt; this is the denominator   I simply do this, but it doesn't work.  \[ \frac{minaccia/vulnerabilit\'a}{contromisure}  \]   Can anyone help me?
119442
A lot of people will interchange an "o" for a zero when speaking them aloud, like with phone numbers and postal codes, so I'm thinking this is the same here, but with nicknames like "threepio", I'm not so sure.  I'm looking for the canon answer here.
37379
As a non-native speaker, I feel that it is ridiculous that can and can't could sound so similar in American accent.  Just now, I was just listening to a video in which the speaker with an American accent says     Fat cells can't reproduce themselves.   The speaker dragged the word "can't" longer for emphasis, but perhaps as a result, what I heard at first was     Fat cells can reproduce themselves.   Of course from the context I later figured out what he was really saying, but I felt extremely surprised by how these two words which literally mean the opposite could sound so similar in American accent. How is that not affecting American people's daily conversations? As a non-native speaker, is there a way for me to further clarify them when my listeners are confused? (Like when you say "one-eight" to clarify that you were trying to say "eighteen" as opposed to "eighty".)
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My friend thought of a system in which each number $n$ (I will first restrict my question to positive integers $n$) is represented by a digit string $(d_l,...,d_1)$ as follows $\forall n \in \mathbb{N}, \exists l \in \mathbb{N}: n = (d_l, d_{l-1}, ..., d_2, d_1) = \sum_{i=1}^l (d_i*(i!))$ where $\forall i \in \mathbb{N}, d_i \in \mathbb{N}: d_i \leq i$ and specifically $\forall i&gt;l, d_i = 0$ ; note that each integer $d_i$ would be represented in decimal with at least 1 digit (in general, these factorial digits can be quite long in decimal representation)- but this is not an issue at the moment [because his system uses a finite alphabet via balanced quinary for each $d_i$].  [In this post, any number shown is in decimal.]  For example, $$1729=2*6! + 2*5! + 2*4! + 1=(2,2,2,0,0,1)$$ and $$1729^3 = (10*12!+9*11!+5*10!+3*9!+6*8!) + 1729 = (10, 9, 5, 3, 6, 0,2,2,2,0,0,1)$$.  My question is: how would this digit string $(d_l,...,d_1)$ be computed most efficiently for general positive integer n? I have mostly just been trying to guess at the largest $l \in \mathbb{N}: l*(l!) \leq n, (l+1)! &gt; n$ and then (once I have found it), I just go through digit-by-digit, working my way down in values of candidate $d_i$ (mostly by squeeze theorem with the goal of finding out that $d_i + 1$ is too large first and then next testing $d_i$ and finding that it works). This process can be quite arduous. What would you recommend doing, algorithmically and efficiently? Bragging rights for Mathematica code that automates the process.  (Using Stirling's formula might be helpful, but it still requires a lot of playing around.)  What modifications would need to be made for general real numbers (including nonintegers)? After the radix point, it works similarly, but with $(i!)^{-1}$ (where, I think, $d_i&lt;i \forall i \in \mathbb{N}$ and where the first such base is $\frac{1}{2!}$).  More generally and philosophically, are there any spiritual problems with this representation? It is meant to be rather compact and alien. What are some cool properties that it exhibits?
120624
Base $10$ uses these digits: $\{0,1,2,3,4,5,6,7,8,9\};\;$ base $2$ uses: $\{0,1\};\;$ but what would base $1$ be?    Let's say we define Base $1$ to use: $\{0\}$. Because $10_2$ is equal to $010_2$, would all numbers be equal?  The way I have thought Base 1 might be represented is tally marks, $0_{10}$ would be represented by nothing. So, $5$ in Base 1 would be represented by $00000$? Or we could define Base 1 to use: $\{$|$\}$ and $5$ would be |||||?
23450
How many 6-digit numbers contain exactly 4 different digits?  My approach is: For any 3 digis same and the remaining 3 different(aaabcd)  4*9*8*7*6  For any 2 duplicate digits(aabb) and the remaining 2 different (aabbcd)  9*8*7*6*3  But I don't know how to add cases where there can be 0's can be as any digit but not the initial ones.  Can someone please help me if there is any simpler approach?
55075
I saw a post on  which links to a video where a person . Is he saying a large portion of the numbers wrong? Back in high school my algebra teacher was extremely picky.  If we were going to say the number 135, we would say one hundred thirty-five.  If a person said one hundred and thirty-five my teacher would interpret that to mean 100.35.  Was my algebra teacher wrong, or did Jon just waste over 80 hours counting to 100,000 incorrectly. I tried searching the web, but searching for ‘numbers and’  doesn’t work very well.  Can you point me to a reference or style guide that describes how to properly say large numbers?
12867
How many $N$ digits binary numbers can be formed where $0$ is not repeated. Note - first digit can be $0$.  I am more interested on the thought process to solve such problems, and not just the answer. If anyone can cite some resources for learning how to solve such problems would be great.
122418
How many words of $13$ letters of the English Alphabet can we constructed which contain $4$ vowels and $9$ different consonants?    This is what I did: Choose the consonants (diff.) in $\binom{21}9$ ways. Arrange these consonants in $9!$ ways. Now we have $4$ empty spots remaining for the vowels, which can be chosen in $5^4$ ways.    My question is: Is this the correct way to think about it? I'm unsure whether the $5^4$ arranges the vowels or not,
63034
If you have a fraction such as $$\frac{7}{26}=0.269230\overline{769230}$$ where there are a number of digits prior to the repeating section, how can you tell how many digits there will be given just the fraction?  I believe I could run through the standard long division algorithm until I come across the same remainder for the second time and then use the location of the first instance of this remainder to calculate the number of digits before the repeating section, but this feels very cumbersome.  After lots of reading online, I came across what looks like a formula for it from :     When a rational number $\frac{m}{n}$ with $(m,n)=1$ is expanded, the period begins after ${s}$ terms and has length ${t}$, where ${s}$ and ${t}$ are the smallest numbers satisfying $10^s\equiv10^{s+t}\pmod{n}$.   I know how to calculate the length of the period of a fraction, and so in the case of my original fraction we have $10^s\equiv10^{s+6}\pmod{26}$, but I don't know how to solve for ${s}$ in this equation!  So there are really two questions here - the one in the title, and a sneaky one about how to take logs in a modulo arithmetic equation.
78773
The decimal (base 10) number "2" can also be represented as the binary (base 2) number "10".  Let's use binary "10" (equivalent to decimal "2") as an example. I could see a few different ways to go here. Assume that the base doesn't need to be specified, and is understood from the context of the conversation (e.g. two programmers talking about memory addresses would understand that they were using hexadecimal).  It could be "ten", since that is what it looks like. One might even argue that ten, as a concept, refers to a one followed by a zero irrespective of the radix. In other words, ten means "a quantity exactly equal to the base it's represented in".  On the other hand, you could argue that "ten" refers specifically to the quantity; in other words, "1010" in binary, "10" in decimal, and "12" in octal would all be pronounced "ten," and "10" in binary should be pronounced "two".  So how would you pronounce the following numbers?  "10" binary ("2" decimal)  "10" octal ("8" decimal)  "10" hexadecimal ("16" decimal)  "1F" hexadecimal ("31" decimal)