I made three case that $2$ is at extreme right then we have $9\times 10 = 90$ numbers, $2$ in middle has $81$ numbers,and with $2$ in beginning has $90$ numbers. This makes $90+90+81=261$ numbers and we add the numbers with no $2$ in them, that is $8\times 9 \times 9= 648$. Thus total will be $909$. What will be the common numbers counted multiple times and how to eliminate them?
122325
There is a number of three digits, 1 is never immediate right of 2 is? (A)771 (B) 791 (C) 881 (D)891.  I faced this Question in Aptitude Competitive exam and don't know what the question actually means. Please anyone help me answering this question.
55292
For how many positive integer $k$ does the ordinary decimal representation of the integer $k\text { ! }$ end in exactly $99$ zeros ?  By inspection I found that $400\text{ !}$ end in exactly $99$ zeros , but $399\text{ !}$ does NOT end in $99$ zeros ; using the formula $$\text{ number of zeros at the end }=\sum_{n=1}^{\infty}\left[\frac{k}{5^n}\right]$$where , $[x]$ denotes the greatest integer part not exceeding $x$. I also found that, for $k=401,402,403,404$ the number of zeros is same, but for $k=405$  the number of zeros increase by $1$ ; as $405$ is divisible by $5$ again , after $400$. Thus I got that there are only $5$ integers satisfying the condition which are $400,401,402,403,404$. The question is possible duplicate of  or  but my question is different from these two questions.  Does there exist any other rule  or easy formula from where I can get how many integers are there
111922
I came across the following problem and do not know how to tackle it.       Find the last two digits of $ 7^{81} ?$     Can someone point me in the right direction? Thanks in advance for your time.
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.
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
10400
Suppose I have $$ \large\sum_{i=1}^{10} 2. $$ Do I just add $2$ to itself $10$ times? I have worked on more complex ones with $n$ and such in the place where the $2$ is, but I have never done it when there is just a number there.
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.
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) = $9^k(n-k)$
51253
$6$ and $210$ share the property that both are the products of both two and three consecutive numbers. $6$ is $2\times3$ and $1\times2\times3$ and $210$ is $14\times15$ and $5\times6\times7$. It was easy enough to write a program to search for more numbers with this property, I found that there were no more up to at least $1{,}000{,}000{,}000{,}000$ ($1$ Trillion). But it is beyond me to prove that there are either no more numbers like this or to find the next one. Any ideas?
9540
Fix integer $n\geq 1$. Consider the number $$1-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$$where the sum continues as long as the lower number in the binomial is $\leq n$. Is there a way to simplify this sum?   The first few values are $1, 0, -2, -4, -4$.
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?
43985
Question: How many  zeroes will there be at the end of $(127)!$  Approach: Considering the fact that when two numbers ending in $x$ and $y$ zeroes are multiplied, the resulting number contains $x+y$ zeroes: The numbers to be multiplied that contain zeroes: $$120,110,100,90,80.....10$$ That comes out to be a total of 13 zeroes. However, that doesn't seem to be the correct answer. Rethinking it, it's obvious that I'm missing out on several cases. For example: $$25*22 = 550$$ This is just one of the cases which will add a zero. How would I account for all these cases?
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" ?
31861
Original problem  Show that there are exactly 16 pairs of integers $(x,y)$ such that $11x+8y+17=xy$.  My work  From case by case analysis I come to know that the equation will hold if and only if $x$ is odd and $y$ is even.  Also I found that $(8-x)|(11x+17)$ and $(11-y)|(8y+17)$  This is all what I have found.  Please see that my work are right or not.  This is a new kind of question which I have encountered so please help me is solving this problem.
95877
I have no clue how to begin these problems. How do I start? I don't think I should pound em out...Thanks.  Let P be the set of $42^{\text{nd}}$ roots of unity, and let Q be the set of $70^{\text{th}} $ roots of unity. How many elements do P and Q have in common?  Let P be the set of $42^{\text{nd}} $roots of unity, and let Q be the set of $70^{\text{th}} $roots of unity. What is the smallest positive integer n for which all the elements in P and all the elements in Q are $n^{\text{th}}$ roots of unity?
60872
I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$. Appreciate your help :)
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?