as my last question  this is the same problem but with 2 repeated digits,   here is my approach:   To solve this problem we need to devide it into 5 cases:  1 The integer contains one 8 and one 6 : 4 x 4 x 3 x 2 x 1 = 96  2- The integer contains one 8 and two 6s or one 6 and two 8s and no 0: (5C2) x 3 x 2 x 1 = 60  3- As same as 2 but containing 0: (4C1) x (4C2) x (3C2) x 2 = 144  4- The integer contains two 8’s and two 6’s with no 0: (5C2) x (3C2) x (2C1) x 2 = 120  5- Same as 4 with 0: (4C1) x (4C2) x (2C2)  = 24  The final result would be : 96 + 2(60) + 2(144) + 120 + 24 = 648  but when I used an online tool I got 588 as result.  (ii)Find the number of integers in (i) which satisfies “If there is 8, then it is followed by 6”
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.
97166
Problem description:  Suppose we use a set of digits $\{0,1,2\}$ to form a sequence, for example \begin{equation} 120210120102012102010210120212010\cdots  \end{equation} The length can be finite or infinite.  The sequence is invalid if there are two finite subsequences appears consecutively. For example, the following sequences are invalid  \begin{equation} 0{\bf 11}\cdots, \quad 010{\bf 201\, 201 }\cdots  \end{equation} where the consecutive subsequences are highlighted.   Question: Does there exist a valid sequence of infinite length? (a non-constructive proof is also acceptable; but a constructive one would be better.)  One way to think about it:  We can transform the sequence to a (base 3) real number,  \begin{equation} 120\cdots \rightarrow 0.120\cdots  \end{equation} Then the only possible infinite sequence should be a irrational number. Irrational number has no period, which means the pattern will not repeat infinite many times, but may repeat finite many times. A famous example is the ,      The Feynman point is a sequence of six 9s that begins at the 762nd   decimal place of the decimal representation of $\pi$.   So not all irrational number($\le 1$) are valid sequences.   I use a short program to do enumeration, but it quickly falls in exponential growth of candidates. I failed to come up with a contraction.   Generalization:  Can we take a larger set like $\{0,1,2,3, \cdots, n\}$? Can we relax the constraint to be three consecutive repetition or more?  Motivation  This problem origins from  in chess,     In chess and some other abstract strategy games, the threefold   repetition rule (also known as repetition of position) states that a   player can claim a draw if the same position occurs three times, or   will occur after their next move, with the same player to move. The   repeated positions do not need to occur in succession. The idea behind   the rule is that if the position occurs three times, no progress is   being made.   It reduces to the proposed problem if we modify the this rule to be "the same position occurs two times with the same sequences of moves".   Thanks for your help.  Edit: I find a related question:
101268
I was asked this question in a recent interview. You are given two digits $0$ and $9$. Using only these two digits you must form the least positive number that is a multiple of a given '$n$' . The value of $n$ will be provided before hand. For example if we are given $n$ as $7$ then the answer will be $9009$ .  I tried using some basic divisibility principles but was not able to come up with a correct solution and code it up. Probably I need some help in number theory
3924
Can anyone help me with this?     If 738 consecutive integers are added together, where the 178th number in the sequence is 4,256,815, what is the remainder when this sum is divided by 6?
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" ?
38963
How many 4 digit numbers are there which contain not more than two different digits?  As usual, I will highlight my attempt:   I can have 9 numbers using only 1 digit, e.g. $1111, 2222, \dots 9999$ = 9 digits Using two digits out of 9 (excluding 0) (selected in $\,^{8}C_{2}$ ways), I can arrange them in $4!$ ways. Furthermore, I can use one digit out of the selected 2 once or thrice (e.g. $2111$ or $2221$)  or simply both digits 2 times (e.g. $2288$, $6677$). Hence I get: $$\,^{8}C_{2} \times \left(\frac{4!}{2!\times2!} + \frac{4!}{3! \times 1!}\times 2\right) = 392$$ Finally, I consider numbers with zeroes. With zero selected as a digit, I have 9 options for other digits. And using the same logic as in 2) I can have either 1 zero ($8880$), 2 zeroes (e.g. $8800$) or 3 zeroes in a number ($7000$). This gives me $$9\times \left( \frac{4!}{2!\times2!} + \frac{4!}{3!\times1!} + 1\right) = 99$$   the above 3 steps yield a total number of $9+392+99=500$  However, the answer i have with me is $576$.
55623
The question is to determine the number of positive integers up to $2000$ that are not divisible by $2,3$ or $5$ but are divisible by $7$. The answer is supposed to be $76$ but not sure how it was derived  I know that if the question was how many integers are not divisible by $2,3,5$ or $7$ then the answer would be $458$ and I know how to derive this.
66500
I am currently self-studying introductory combinatorics by reading Introduction to combinatorial mathematics. I am currently in the first chapter, and I have a question regarding one of the examples. The question was asking to count the number of n-bit strings with an even number of zeros. The answer is of course $2^{n-1}$. The author gave 2 solutions. I however didn't completely understand what I think is the straightforward one. The solution I got was that he took out 1 bit, leaving $(n-1)$ bits, if the number of zeros is even in the $(n-1)$-bit number, then he will just append a 1, if not then he will append a zero. So in the end we just needed to count the number of $(n-1)$-bit strings. The other solution (the straightforward one) that I didn't understand examined the symmetry that half of the $2^n$ must have an even number of zeros, and the other half will have an odd number of zeros. I just don't get why this property must hold. I can understand that half of the $2^n$ numbers will have even parity, but I can't see how it holds for the parity of the number of zero or one bits. If anyone can show me how that property holds, I'd be very grateful. I'd also be interested to see different explanations and proofs if possible.  Thank you.
61582
Question: Take any $37$ integers from the set {${1,2,…,112}$}. Show that there will always exist two integers $x,y$ out of those $37$ integers such that $ \mid x-y \mid \in \{9,10,19\}$. Approach: What have I done so far: Let $S=$ {${1,2,…,112}$} We partition the set into subsets $S_i$ of the form $\{x, x+9, x+19\}$. Then the difference of any two elements is in $\{9, 10, 19 \}$, which is easy to see. The following partitions are made (for $x=1, 2, \cdots , 9$): $$\{1, 10, 20 \}, \{2, 11, 21 \}, \cdots, \{9, 18, 28\}$$ Then next we choose $x=29$ to maintain the disjoint sets essential for our proof. This gives the disjoint sets $$ \{29, 38, 48\}, \cdots, \{37, 46, 56 \} $$ and similarly $$\{85, 94, 104 \}, \cdots, \{93, 102, 112 \}$$ However as the quick reader may see that the numbers $\{19, 47, 75, 103\}$ do not occur in any set. However, I decide to do a case by case analysis of the situations. Name this set as $X$. Suppose, initially, that out of the $37$ integers in our selection, none belong to $X$. Then it follows from Dirichlet's box principle that at least two elements are from the same set (i.e. from one of the subsets that we partitioned $S$ into) then the claim is valid. Now suppose that one element from $X$ is in our selection, say $19$ then if at least one of $9, 28, 29, 38$ is in our selection, then the claim is valid. Suppose none of these are present, and we are deficit in four integers. We  can now select any four integers from the remaining numbers and the box principle guarantees that at least two are from the same set. For, even if all $3$ other elements of $X$ are now present, there is one empty slot which can be filled using the other subsets from where we already have at least one element in our choice of $37$ numbers. I hope this is comprehensible because my combinatorics proof writing abilities are horrible. I am preparing for Olympiads and also for being a good Mathematician in general. Thank you for the help :) Edit: It seems I didn't enter my main question, apologies. Is my solution correct?
93203
How many bit strings containing exactly eight $0$s and twelve $1$s have either all the $0$s consecutive, or all the $1$s consecutive?  My working :  There are $20$ positions $(8 + 12)$   Since there are $2$ possibilities for each position   Thus $2^{20}$
8059
I've been asked the following question:  What five odd integers from the set  $\{1, 3, 5, 7, 9, 11, 13, 15\}$ that when summed together equals to $30$? Note that any integer can be used more than once.   If my limited knowledge of maths is correct, there should be no answer, as no odd number of odd integers summed together can give an even number.
35774
SEND + MORE = MONEY   Each letter represents a single digit number.  No two letters represent the same number. (Ex: if M=1 the no other letter in the problem can equal 1)  So far we've figured out that M=1, O=zero, and S is either 8 or 9  This is a very interesting problem, but we're a bit stuck and don't know where to go from here.
16530
During self-study, I ran across the question of how many ways six numbers can be chosen from the numbers 1 - 49 without replacement, stipulating that no two of the numbers be consecutive.  I can obtain a simple lower bound by saying that, in the worst-case scenario, when you choose a particular number, there are now three numbers that cannot be chosen next.  For example, if I first pick 17, then I can't choose 16, 17, or 18 for the remaining choices.  This gives me the lower bound  $$\frac{49*46*43*40*37*34}{6!} = 6,773,770 \frac{8}{9}$$  This is about 48% of ${}_{49}C_6 =  13,983,816$.  The real answer must be bigger (and an integer).  I haven't found a way to calculate it, though.  The original problem asked to show that the probability of having non-consecutive integers when you choose six is greater than 50%, so if the problem is complicated to count exactly, better approximations that show the answer is above 50% would also be appreciated.   Of course, I can use a computer to do the counting, but I'm interested in learning what methods I'm missing.
41147
Problem :   Solve in $\mathbb N$ :   $8n+9\equiv 0\pmod{1163}$  The mathematics give me : $n=435+1163k$ , $k\in\mathbb N$   $1163$ is a prime number   $8n\equiv -9\pmod{1163}$   $8n\equiv 1154\pmod{1163}$   I don't know any ideas to complete my work ?   I have to see your hints
80007
Anyone have any ideas on this question? I think you have to use the pigeon hole principle..but I am not sure about that?     The numbers $1,2,3,\ldots,101$ are written down in a row in some order. Is it always possible to cross out $90$ numbers in a way such that all $11$ numbers left will stay either in increasing or decreasing order?
20105
$T(n) = 2$ if $n=0$  $T(n) = 9T(n-1)-56n +63$ if $n&gt;=1$  Repeated substitution  $k=1$  $T(n) = 9T(n-1)-56n+63 $  $k =2$  $T(n) = 81T(n-2) -560n + 1134$   $k =3$  $T(n) = 729T(n-3) -5096n + 15309$  I cant find the pattern for the n term and the integer  For now i just have  T(n) = $9^k(n-k)$
47058
How many 10-digit numbers can be made by using the digits {5,6,7} (all of them) and with the additional constraints that no two consecutive digits must be the same and also that the first and last digits of the number must be the same?   I am trying to find a solution by using combinatorics. I start from the 1st leftmost digit, which can have any value from {5,6,7} (3 possibilities).  Then we move to the 2nd digit, which can have 2 values (since it can't be the same with the 1st) and so on, and for the last digit we only have 1 option. But this is not correct, because for the 9th digit we have the restriction that it must be different from the 8th and also different from the 10th, which, in turn, is equal to the 1st.  I don't know how to express this.  I therefore tried to find a recursive relation.  I found that the general relation is $a(n) = 2*a(n-1)$ if n odd and $2*a(n-1) + 6$ if n is even.   For n=4, we have 6 such numbers (4-digit numbers, but with the given restrictions). Then if we add one more digit to the right, we remove the rightmost (4th) digit, which had to be the same with the first, and now for the 3rd digit we have 2 options instead of 1 (we can also add the options that were rejected because they were neighboring with the 4th digit). So in total we now have $2 x 6 = 12$ options. Therefore, $a(4)=6$ and $a(5)=12$. I don't understand, however, where this $+6$ (in the recursive relation) comes from!  By the way, the correct answer is 510.  Many thanks in anticipation.