1/7 is 1x 6digits repeating, 1/13 is 2x 6 digits repeating. Why? I can see this pattern of repeating digits, I can calculate it for most primes but I don't have an intuitive understanding of why it happnes. I know there are certain rules like 10 ^ k = 1 mod(p) and so on but I am after a plain English understanding if there exists one :) I would really appreciate it if you could help me with this or point me to resources that explain this. Fun fact: 2/23 = 22 digits repeating 0.222222222222222/23 = first 15 digits of 2/207 followed by 22 repeating digits of 6/23. True and is similar with different numbers but no clue why?
123609
The number $142,857$ is widely known as a cyclic number, meaning consecutive multiples are cyclic permutations, i.e.  $1 &#215; 142,857 = 142,857$  $2 &#215; 142,857 = 285,714$  $3 &#215; 142,857 = 428,571$  and so on.   142857 is the repeating unit of $\frac{1}{7} = 0.\overline{142857}$ and in fact, every prime for which $10$ is a primitive root will generate a cyclic number (if we allow $0$ as a first digit, for example $0588235294117647$ which is the repeating unit of $\frac{1}{17}$). These primes are called full reptend primes.  From what I've read, it seems that there is a bijection between full reptend primes and cyclic numbers: a number is cyclic if and only if it is the repeating unit for the reciprocal of a full reptend prime.  I was able to find a proof for the if part. I was wondering if anyone can provide a proof for the only if part.  Edit: It's been a while since I posed this question and I still haven't found a proof. I've added a bounty in hopes of prompting some interest.
61995
What's the $n$-th digit in the sequence $S$ of numbers formed by the natural numbers, i.e.,  $n$-th digit in the sequence 1 2 3 4 5 6 7 8 9 10 11 12... ? For example 11th digit in the sequence is 0.
72410
Now I have proved the following two problems:      (1) Prove that a number and the sum of its digits have the same remainder upon division by 9.    (2) Given an integer, consider the difference between the sum of the digits in odd positions (counting from the right) and those in even positions. Show that the difference has the same remainder as the number itself when divided by 11.    Then, I need help on the following problem:      Partition the digits of a number into groups of two and three, respectively, starting from the right. Treat these as two- and three-digit numbers, respectively, and proceed similarly as in the two problems above. What divisibility rules arise?    In fact, I did not quite get what this problem asks for and not clear about this process. Could anyone give me some hints on this, please?  Thanks in advance.
69457
So a couple of days ago the $17$ million digit number $2^{57885161}-1$ was beaten by the $22$ million digit number $2^{74207281}-1$ at being the largest known prime number.  Are there any specific (purely mathematical) implications of the fact that this particular number is prime? In particular, does this resolve something other than the question if $2^{74207281}-1$ is prime, in the style of some newly discovered theorem resolving previous conjectures? Is it perhaps a counterexample to something so far believed to be true?  Edit: I'm aware of the many non-(strictly mathematical) practical implications and reasons to continue the search for large primes (cryptography etc), but I'm not aware of any strictly mathematical uses of particular numbers being prime, which is why (and what) I'm asking.
11305
How many integers from 1-999 do not have any repeated digits?  The answer is explained in this , but why is the last set 9*9*8? Why not 9*9*9?
90336
$13/92=0.14\overline{1304347826086956521739}$  In this example, the length of nonrepeating part is $2$. The length of repeating part (repeating period) is $22$.  I collected some properties related to repeating decimals.   The repeating period must be less than divisor.-- It doesn't give me the exact period.   If divisor d is a multiple of $2^m 5^n$, the length of non-repeating part of $1\div d$ is $\max(m,n)$. -- It only applies to $1\div d$. If $1 \leqslant  b &lt; a$, and $a$ is not a multiple of $2$ or $5$, and $a$ and $b$ are  relatively prime, then the repeating period of $b\div a$ equals $\operatorname{min}\left \{ e\in \mathbb{N}:10^e\equiv 1 \pmod{a}  \right \}$. -- It has many conditions.   Without calculating the decimal, can we determine the length of the two parts of any dividend and divisor?
112498
(I am aware there are similar questions on the forum)  What is the Question?  A positive integer (in decimal notation) is divisible by $11$ if and only if the difference of the sum of the digits in even-numbered positions and the sum of digits in odd-numbered positions is divisible by $11$.     For example consider the integer 7096276.       The sum of the even positioned digits is $0+7+6=13.$ The sum of the odd positioned   digits is $7+9+2+6=24.$ The difference is $24-13=11$, which is divisible by 11.      Hence 7096276 is divisible by 11.   (a)  Check that the numbers 77, 121, 10857 are divisible using this fact, and that 24 and 256 are not divisible by 11.  (b)  Show that divisibility statement is true for three-digit integers $abc$. Hint: $100 = 99+1$.  What I've Done?  I've done some research and have found some good explanations of divisibility proofs. Whether that be 3,9 or even 11. But...the question lets me take it as fact so I don't need to prove the odd/even divisible by 11 thing.   I need some help on the modular arithmetic on these.   For example... Is 77 divisible by 11? $$7+7 = 14 \equiv ...$$  I don't know what to do next.  Thanks very much, and I need help on both (a) and (b).
82160
If we reverse the digits of $12$ we will get $21$. $12^{2}=144$. If we reverse its digits we will get $441$ which is $21^{2}$.  Here is the puzzle. How many such two digit numbers are there? Digits must be different.  We got an algebraic method to solve it if the square is a three digit number.It given here :  Let $a&gt;b$  $(10a+b)^{2}=100x+10y+z$ and $(10b+a)^{2}=100z+10y+x$  Difference between these two equations will lead to $a^{2}-b^{2}=x-z$  $x-z \leq 8$.Then  $a^{2}-b^{2} \leq 8$.  Since squares are of three digits $a &lt;4$.But since $a&gt;b ,a \neq 1$.  $b=1 \Longrightarrow a^{2} \leq 9$.Then $a=1,2,3$ but only possibilities are $2 $ and $3$.  $b=2 \Longrightarrow a^{2} \leq 12 \Longrightarrow a= 1,2,3$. But possibility is $a=3$. So next such pair will be $13,31$.  Higher values of $b$ will not work since $a$ cannot exceed 3. This is an explanation we obtained from a mathematics group.It works well if the square is a three digit number.What will happen if square is a 4 digit number... Or if we try to extend this problem for numbers having more than two digits ?
95227
I have a question and got strucked on this.. To find the total no. of six digit numbers that can be formed having property that every succeeding digit is greater than preceding digit. Please guide me to solve this. Thanks in advance.
122378
I stumbled upon the following in a book:     till 400 all even numbers will be divisible by 2 ( 200 even numbers)   remaining 200 odd numbers  1 3 5 7 9 ..... 399 200/3 = 67 will be   divisible by 3, as we have discarded all even numbers and out of   remaining numbers 2nd number is 3 then 1 + 198/3 = 67 numbers are   divisible by 3      now we have discarded all multiples of 2 and 3 remaining numbers will   be  200 - 67 = 133;   1 5 7 11 13 ... so on out of these 133 numbers 2nd   number is 5 so numbers divisible by 5 will be 1 + 131/5 = 1+26 = 27      now remaining numbers = 133 - 27 = 106; 1 7 11 13 .. so on again 2nd   number is 7 , so numbers divisible by 7 = 1 + 104/7 = 1+14 = 15      total numbers from 1 - 400 divisible by any of 2 3 5 7 = 200+67+27+15   = 309   But I can't seem to understand how the following is derived? :     as we have discarded all even numbers and out of remaining numbers 2nd   number is 3 then 1 + 198/3 = 67 numbers are divisible by 3   How does the result come out to be "1 + 198/3" ?
45198
Clearly an even number $n$ cannot divide $2^{n}-1$, but about odd ones ? If $n$ is an odd prime this cannot happen neither since for an odd prime $p$ we have $2^p\equiv 2\pmod p$ and so $p$ cannot divide $2^p-1$ but what about the general case ($i.e~n$ an odd number larger than 1) ?
121952
My question is this : how many distinct two digit numbers can be produced from numbers $4, 3, 3, 1$? When applying the formula $$\frac{4!}{(4-2)!2!}$$ you come up with $6$, yet when doing the problem manually, I come up with $7$ numbers, namely $41, 43, 14, 13, 33, 31, 34$. Can anybody explain this discrepancy?  Thanks.
120131
The sum of the digits in $9 k$ (where $k$ is an integer) is a multiple of $9$: for example   $$9\cdot 1=9$$ $$9\cdot 7=63 \qquad  \text{and } 6+3=9\cdot 1$$ $$9\cdot 11=99 \qquad \text{and }  9+9=9\cdot 2$$  But why?
82023
Prove the divisibility test by $7,11,13$ for numbers more than six digits    Attempt:  We know that $7\cdot 11 \cdot 13 = 1001$. The for a six-digit number, for example, $120544$, we write it as $$ 120544 = 120120 + 424 = 120\cdot1001 + 424 $$ thus we just check the divisibility of $424$ by $7,11,13$.  Know for a number with more than six digits,  for example: $270060340$,  $$270060340 = 270270270 - 209930$$ $$ = 270 \cdot (1001001) - 209930 $$ $$ = 270 \cdot (1001000) + (270 - 209930) =270 \cdot (1001000) - 209660$$  so we check the divisibility of $209660 = 209209 + 451$, or just $451$.  But the test states that: for $270060340$, we group three digits from the right: $$ 270, 60, 340$$ then check divisibility of $340+270 - (60)$.  How to prove this?
53938
Find a recurrence relation for the number of $n$-digit binary sequences with no pair of consecutive $1$s.  I know my base case:$$a_1 = 2$$ for either $1$ or $0$. Normally to construct the recurrent relation I would say fill in $n-1$ spaces then add a $1$ or $0$, which is simply:$$a_n = 2\cdot a_{n-1}$$However, I don't believe this is right, because this doesn't take into account that there can't be any consecutive $1$s.  Even when I use the base case of $2$ digits:$$a_2 = 3$$for either $01$, $10$, or $00$, I still get the recurrence relation of:$$a_n = 3\cdot a_{n-2}$$which also does not account for $01$ and $10$ occurring consecutively.   So how would I express this?
79254
I have to figure out a way to count how many sequences are there s.t:   My alphabet is $1$ up to $49$. Each number that is chosen, is chosen only once. The sequence is $6$ digits long. The sequence is in increasing order.  e.g that is a valid sequence: $(1, 2, 40, 42, 43, 49)$  and this is not: $(1, 2, 40, 25, 25, 12)$   I know that for the first 3 rules the answer is: $\binom{49}{6}\cdot 6!$  How do I deal with the fourth?
81118
How many 6 digit numbers are possible with at most three digits repeated?  My attempt:  The possibilities are:   A)(3,2,1) One set of three repeated digit, another set of two repeated digit and another digit (Like, 353325, 126161)  B)(3,1,1,1) One set of three repeated digit, and three different digits.(Like 446394, 888764)  C)(2,2,1,1) Two sets of two repeated digits and two different digits (Like, 363615, 445598)  D)(2,2,2) Three sets of two repeated digits (Like, 223344, 547547)  E)(2,1,1,1,1,1) One set of two repeated digit and four different digits (Like 317653, 770986)  F)(1,1,1,1,1,1) Six Different digits (like 457326, 912568)  G)(3,3) Two pairs of three repeated digits. Let's try to calculate each possibilities separately.    F) is the easiest calculate.   Let us try to workout  Case E)  Let's divide the case into two parts:  Case E(1) Zero is not one of the digit   We can choose any $5$ numbers form $9$ numbers $(1,2,3,\cdot, 9)$ in $\binom{9}{5}$ ways , the digit which one is repeated can be chosen in 5 ways, and you can permute the digits in $\frac{6!}{2!}$ ways. The total number of ways$=\binom{9}{5}\times 5\times \frac{6!}{2!} $  Case E(2) Zero is one of the digit.  Case E(2)(a) Zero is the repeated digit We need to choose four other numbers which can be done in $\binom{9}{4}$ ways, the digits can be permuted in $\frac{6!}{2!}$ ways, but we need to exclude the once which starts with zero ($5!$ many). The total number of ways =$=\binom{9}{4}\times  (\frac{6!}{2!} -5!)$.  Case E(2)(b) Zero is not the repeated digit We need to choose four other numbers which can be done in $\binom{9}{4}$ ways, the repeated digit can be chosen in 4 ways,  the digits can be permuted in $\frac{6!}{2!}$ ways, but we need to exclude the once which starts with zero ($5!$ many). The total number of ways =$=\binom{9}{4}\times 4\times  (\frac{6!}{2!} -5!)$.  Before, proceeding to workout the other cases, I want to know   Is my attempt correct? If it is correct, it is too lengthy, is there any other way to solve this?