The number of numbers lying between 1 and 200 which are divisible by either of two , three or five?
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" ?
75666
Let $S={1, 2, 3, 4, ....., 2070}$ find the number of subsets of $S$ whose sum of elements in $S$ is divisible by 9     In the expression $$f(x)=(1+x)(1+x^2)......(1+x^{2070})$$    Terms of the form $x^{9k}$ will be the subsets divisible by 9. I am not able to proceed from here.
76503
Find a number $n$ which has      $12$ divisors $(1 = d_1 &lt; d_2 &lt; \cdots &lt;d_{12}=n )$ and    $d_{d_4-1}=(d_1+d_2+d_4)d_8$.       Note: This is a problem from Russian Mathematical Olympiad 1989. I think this is related with  one. But this seems harder to me.
50871
Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
105964
Recently I've had my 24th birthday, and a friend commented that it was a very boring number, going from 23 which is prime, 25 which is the first number that can be written as the sum of 2 different pairs of squared integers $3^2+4^2 =0^2+5^2 =25$, 24 seems like a very boring number  however, it seems to have a very special property  Theorem: product of 4 positive consecutive numbers is divisible by 24.  I managed to prove this via long and dry induction, not very interesting. I wonder if anyone can propose a different more elegant and witty proof, rather than dry algebra like me.
63095
I'm solving a problem, and the solution makes the following statement: "The common difference of the arithmetic sequence 106, 116, 126, ..., 996 is relatively prime to 3. Therefore, given any three consecutive terms, exactly one of them is divisible by 3." Why is this statement true? Where does it come from? Is it generalizable to other numbers?
4694
I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, $3,1,5,2,4$ is an acceptable permutation where $3,1,2,4,5$ is not because 5 is in position 5.  I know that the number of total permutations is $n!$. Is there a formula for how many are acceptable given the case that no position holds its own number?
83951
I want to prove that if you choose $101$ numbers from the set $\{1,2,3,4,\dots ,200\}$, there are always two numbers such that one divides the other with no remainder. The prove should involve the "pigeonhole principle".  I am not sure how to define the pigeonholes and how to define the pigeons. Any assistance with the proof will be most appreciated.  Thank you.
32627
Suppose we have a natural number $N$ with decimal representation $A_kA_{k-1}\ldots A_0$. How do I prove that if the $\sum\limits_{i=0}^kA_i$ is divisible by $9$ then $N$ is divisible by $9$ too?
95408
So the question is  Let $x,y, z$ Be digits so that the 9-digit number  5$x$36$y$98$z$2 is divisible by 792.  1) Without calculating $x$ and $y$, explain why z = 3 or 7.  2) WIthout calculating $x$, show that y = 0 or 9.  3) FInd all combinations of $x, y, z$
29331
Find the number of three element subsets $A$ of $\{1,2,...,10\}$ such that $A$ contains at least one even and one odd integer.  I think in this way: First we distribute one even and one odd number to 3 sets.Then there are 4 digits left of which 2 are even and 2 are odd.But how to distribute this 4 digits into 3 sets?
114180
I want to find the sum of all combination of numbers whose sum is x, for e.g.  when x = 3 f(x) = countOf(111,12,21,3) = 4
111588
Simple question.  How many numbers between 1 and 10000 can't be written as $n^2$ or $n^3$ when $n \in \mathbb N$?  I know the way to solve this is with inclusion-exclusion. but for that I need to find the cardinality of these sets:  1) The set of all numbers between 1 to 10000 that can be written as $n^2$  2) The set of all numbers between 1 to 10000 that can be written as $n^3$  3) The set of all numbers between 1 to 10000 that can be written as $n^6$  How do I find out how many elements are in those sets?
99761
How many numbers of 6 digits, that can be formed with digits 2,3,9. And also divided by 3?  $$$$ I was trying to to add 2^6 (when there is no 2)+ C(6,2)2^2 (when 2 can be in two places)+C(6,3)2^3 (when 2 can be in three places)...
2521
Let's have 2 numbers, N and K, where K divides N. The number of K-combinations from a given set S of N elements is a well known formula. Let's concatenate N/K groups (resulting in N elements) such as the resulting set is N. How many possibilities are there, i.e. what's the formula?  For instance:    N=4, K=2, C(4, 2) = 6 {1,2},{1,3},{1,4},{2,3},{2,4},{3,4}   The 3 possibilities are:     {1,2},{3,4}   {1,3},{2,4}   {1,4},{2,3}   I generated these combinations and I think the number goes like this:  (4, 2): 3 (6, 3): 10 (6, 2): 15 (10, 5): 126 (9, 3): 280 (10, 2): 945 (14, 7): 1716 (12, 4): 5775 (15, 5): 126126 (15, 3): 1401400   Apparently, the result always divides with (N-1).
59103
Find n such that $n^5 - n$ is divisible by $120$. (*Here $n$ belongs to the natural number set)  Seems a trivial problem, but the answer mentioned in the text book seemed incomplete to me so I'm here just to ensure whether approach to this problem is correct or not.  So my approach is as follows: We know that $n^5 - n$ can be written as $n^5-n = n(n-1)(n+1)(n^2 + 1)$  Now since it's product of three consecutive natural numbers so it's divisible by $3! = 6$. Now using Fermat's Little theorem we can prove that $n^5= n \mod5$ hence $n^5 - n$ is divisible by $5$ as well. So $n^5 - n$ is divisible by $30$. Now to ensure their is another factor of $4$ in the term we need to ensure that $n^5 - n$ is divisible by $8$. I found that all odd natural numbers and multiples of $8$ makes $n^5 - n$ divisible by $8$, hence $n^5 - n$ is divisible by $120$. But in the text book they have only mentioned odd natural numbers as answer, which seemed incomplete to me.  Thanking everyone in advance:P
29626
How do I prove that among any $5$ integers, you are able to find $3$ such that their sum is divisible by $3?$ I realize that this is a number theory question and we use modular arithmetic, but I'm unsure of where to begin with this specific situation.