How can I prove that every prime except 2 and 5 divide infinitely many of the following integers $11, 111, 1111, ...$ ?
110285
Prove that every number ending in a $3$ has a multiple which consists only of ones.      Eg. $3$ has $111$, $13$ has $111111$.   Also, is their any direct way (without repetitive multiplication and checking) of obtaining such  multiple of any given number( ending with $3$) ?
2872
This is just a curiosity. I have come across multiple proofs of the fact that there are infinitely many primes, some of them were quite trivial, but some others were really, really fancy. I'll show you what proofs I have and I'd like to know more because I think it's cool to see that something can be proved in so many different ways.  Proof 1 : Euclid's. If there are finitely many primes then $p_1 p_2 ... p_n + 1$ is coprime to all of these guys. This is the basic idea in most proofs : generate a number coprime to all previous primes.  Proof 2 : Consider the sequence $a_n = 2^{2^n} + 1$. We have that  $$ 2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1), $$  so that for $m &lt; n$, $(2^{2^m} + 1, 2^{2^n} + 1) \, | \, (2^{2^n}-1, 2^{2^n} +1) = 1$. Since we have an infinite sequence of numbers coprime in pairs, at least one prime number must divide each one of them and they are all distinct primes, thus giving an infinity of them.  Proof 3 : (Note : I particularly like this one.) Define a topology on $\mathbb Z$ in the following way : a set $\mathscr N$ of integers is said to be open if for every $n \in \mathscr N$ there is an arithmetic progression $\mathscr A$ such that $n \in \mathscr A \subseteq \mathscr N$. This can easily be proven to define a topology on $\mathbb Z$. Note that under this topology arithmetic progressions are open and closed. Supposing there are finitely many primes, notice that this means that the set $$ \mathscr U \,\,\,\, \overset{def}{=} \,\,\, \bigcup_{p} \,\, p \mathbb Z $$ should be open and closed, but by the fundamental theorem of arithmetic, its complement in $\mathbb Z$ is the set $\{ -1, 1 \}$, which is not open, thus giving a contradiction.   Proof 4 : Let $a,b$ be coprime integers and $c &gt; 0$. There exists $x$ such that $(a+bx, c) = 1$. To see this, choose $x$ such that $a+bx \not\equiv 0 \, \mathrm{mod}$ $p_i$ for all primes $p_i$ dividing $c$. If $a \equiv 0 \, \mathrm{mod}$ $p_i$, since $a$ and $b$ are coprime, $b$ has an inverse mod $p_i$, call it $\overline{b}$. Choosing $x \equiv \overline{b} \, \mathrm{mod}$ $p_i$, you are done. If $a \not\equiv 0 \, \mathrm{mod}$ $p_i$, then choosing $x \equiv 0 \, \mathrm{mod}$ $p_i$ works fine. Find $x$ using the Chinese Remainder Theorem.   Now assuming there are finitely many primes, let $c$ be the product of all of them. Our construction generates an integer coprime to $c$, giving a contradiction to the fundamental theorem of arithmetic.  Proof 5 : Dirichlet's theorem on arithmetic progressions (just so that you not bring it up as an example...)  Do you have any other nice proofs?
82022
Show that every prime $p&gt;3$ is either of the form $6n+1$ or of the form $6n+5$, where $n=0,1,2, \dots$
94048
How prove :         $$ {31}^{11}&lt;{17}^{14} $$       help please!
37470
To prove the following fact: given any sequence of digits in any base, eg 314159265358979323 base 10, there are infinitely many primes that start with these digits,eg when expressed in decimal they start with 314159265358979323. I think using a prime number theorem will be helpful . But I am still not getting a point of how to start the proof.
111147
Prove that there is no arithmetic progression that consists only of primes.  A question that I've been set; I'm guessing it makes use of primes being written in the form 4k+1 and 4k+3? Not sure where to start.  Thanks.
92895
I feel like there must exist a combinatoric proof of a theorem like: There is a prime between $n$ and $2n$, or $p$ and $p^2$ or anything similar to this stronger than there is a prime between $p$ and $(\prod_p p) + 1$ (Euclid's theorem).  I was trying to prove one by the Sieve on this grid      1     2     3     4 ... p   p+1   p+2   p+3   p+4 ... 2p  2p+1  2p+2  2p+3  2p+4 ... 3p ........................... ........................... ........................... p^2 p^2+1 p^2+2 p^2+3 p^2+4 ...   but it did not work.  Do any good arguments like this exist? I don't expect anything as strong as Prime Number Theorem or even Bertrand, but surely a direct combination proof can prove that there are lots of primes?
92419
Someone claimed that a number, multiplied by the number after it plus 17 is always prime, and showed several cases. I'm not a complete amateur in Number Theory, and I know that $17*18+17=17*19$, so it does not work for $n\equiv0(\mod17)$ but does it always work for other $n$ values? If not, can someone give me a counter example that I can show that person so they can correct their statement?
66947
There should be infinitely many primes of the form $5+6n$. How do you prove it? The same should be true for $7+6n$.
103535
Prove that if $p$ and $q$ are distinct primes such that $pq\mid n^2$ then $pq\mid n$.  My Attempt:  Assume that $pq\nmid n$. If $pq\nmid n$, then $p\nmid n$ or $q\nmid n$. Without loss of generality, assume that $p\nmid n$.  If $p\nmid n$, then $pq\nmid n^2$.
65396
I need something explained or corrected:  In my number theory class we proved that there are an infinite number of primes using Euler's Phi Totient. It went something like this:  Let $M = p_1 p_2 \dots p_n$ be the product of all primes. Consider $1 &lt; A \le M$:  Some prime must divide $A$, call it $q$. Since $q$ must be one of the primes, $q$ must divide $M.$  So the $gcd(A,M) &gt; 1.$ Thus $\phi(M) = 1$ ...?   Which is not even and contradicts the theorem that $\phi(N)$ is even for $N&gt;2.$ Therefore there exists an infinite number of primes.  I get confused by the statement "Thus $\phi(M) = 1$"....  Did i possibly copy this proof down wrong? or am I missing something? Thank you in advance.   Edit: by consider a such that...i meant consider an integer '$a$' I replaced it with $A$ to hopefully make it more clear.   Edit2: I'm sorry I am not familiar with using the equation editor. This is not a homework assignment, just studying for my exam. I just want to be able to understand this or clearify it.
40340
Prove that for any positive integers $m$ and $n$ , there exists a set of $n$ consecutive positive integers each of which is divisible by a number of the form $d^m$, where $d$ is some positive integer not equal to $1$.  I don't know how to approach this question.
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).
123401
Prove that $2730$ divides $n^{13} - n$ for all integers $n$.   What I attempted is breaking $2730$ into $2, 3, 5$, and $7, 13$. Thus if I prove each prime factor divides by $n^{13} - n$ for all integers $n$, it will prove the original problem.   I already proved $2, 3$, and $5$ work. Struggling with how to do $91$?  Other solutions are welcome, but they should be associated with mod and solutions of this kind, because my answer should pertain to what was learned in class.  (Edit: Fixed the 91 problem, being slow today...In any case, I've got this problem now, so no answers are necessary. Posting my own soon.)
98853
Prove: Given a prime $p$, there are infinitely many $n\in \mathbb{Z}^+$, for   which $pn+1$ is a prime.   This is a simplified version of , so is there any elementary solution to this problem?  A proof by contradiction is recommended. Just like the proof of the existence of infinitely many primes by Euclid.
30113
Take $p_1, p_2, \ldots, p_n, p_{n+1}$ be $n+1$ prime numbers in $\mathbb{P} \subseteq \mathbb{N}$. $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ seems to be quite clear, but still need a proof. I know some proofs are involved with Galois theory, which is not I want.
27077
Given any $10$ consecutive positive  integers , does there exist one integer which is relatively prime to the product of the rest ?