This tag is for questions about divisibility, that is, determining when one thing is a multiple of another thing.

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62
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13answers
20k views

Dividing 100% by 3 without any left

In mathematics, as far as I know, you can't divide 100% by 3 without having 0,1...% left. Imagine an apple which was cloned two times, so the other 2 are completely equal in 'quality'. The totality ...
45
votes
4answers
7k views

Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

For all $a, m, n \in \mathbb{Z}^+$, $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
36
votes
5answers
6k views

If a prime number is reversed, and then appended to itself, why is the result always a composite number?

$2 \Rightarrow 22$ which is a composite number. $37 \Rightarrow 3773$ which is a composite number. $523 \Rightarrow 523325$ which is a composite number. $8123 \Rightarrow 81233218$ which is a ...
36
votes
4answers
1k views

How does the divisibility graphs work?

I came across this graphic method for checking divisibility by $7$. $\hskip1.5in$ Write down a number $n$. Start at the small white node at the bottom of the graph. For each digit $d$ in ...
33
votes
8answers
3k views

What is $\gcd(0,0)$?

What is the greatest common divisor of $0$ and $0$? On the one hand, Wolfram Alpha says that it is $0$; on the other hand, it also claims that $100$ divides $0$, so $100$ should be a greater common ...
23
votes
6answers
13k views

If $a^2$ divides $b^2$, then $a$ divides $b$

Let $a$ and $b$ be positive integers. Prove that: If $a^2$ divides $b^2$, then $a$ divides $b$. Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out ...
22
votes
10answers
1k views

Prove if $56x = 65y$ then $x + y$ is divisible by $11$

If $x$ and $y$ are natural numbers, and $56x = 65y$, prove that $x + y$ is divisible by $11$. I tried taking the $\gcd(56x,65y)$ using the Euclidean algorithm, but I got nowhere with it and do not ...
22
votes
7answers
4k views

Why $a^n - b^n$ is divisible by $a-b$?

I did some mathematical induction problems on divisibility $9^n$ $-$ $2^n$ is divisible by 7. $4^n$ $-$ $1$ is divisible by 3. $9^n$ $-$ $4^n$ is divisible by 5. Can these be generalized as $a^n$ ...
21
votes
3answers
636 views

Prove $n\mid \phi(2^n-1)$

If $2^p-1$ is a prime,(thus $p$ is a prime,too) then $p\mid 2^p-2=\phi(2^p-1).$ But I find $n\mid \phi(2^n-1)$ is always hold, no matter what $n$ is.Such as $4\mid \phi(2^4-1)=8.$ If we denote ...
20
votes
4answers
2k views

Is the number $333{,}333{,}333{,}333{,}333{,}333{,}333{,}333{,}334$ a perfect square?

I know that if the number is a perfect square then it will be congruent to $0$ or $1$ (mod $4$). Now since the number is even, I know that it is either $0$ or $2$ (mod $4$). How would I go about ...
20
votes
1answer
483 views

Is $\sum_{k=1}^{n} k^k / \sum_{k=1}^{n} k \in \mathbb{N}$ for some $n > 1$?

Let $ A = \sum_{k=1}^{n} k^k $ and $ B = \sum_{k=1}^{n} k$, where $n >1 $ is a positive integer. Is $A/B$ ever an integer?
19
votes
5answers
1k views

How does one show that two general numbers $n! + 1$ and $(n+1)! + 1$ are relatively prime?

How does one show that two general numbers $n! + 1$ and $(n+1)! + 1$ are relatively prime? I don't mind if someone uses a different example, I want to learn how to prove this class of problems. My ...
18
votes
3answers
843 views

My first induction proof (very simple number theory). Please mark/grade.

What do you think about my first induction proof? Please mark/grade. Theorem The sum of the cubes of three consecutive natural numbers is a multiple of 9. Proof First, introducing a predicate ...
17
votes
4answers
388 views

Prove that $\dfrac{(n^2)!}{(n!)^n}$ is an integer for every $n \in \mathbb{N}$

Prove that $\dfrac{(n^2)!}{(n!)^n}$ is an integer for every $n \in \mathbb{N}$ I know that there are tools in Number theory to proves the required but I want to use the tool that says that if you ...
17
votes
4answers
481 views

Do there exist infinitely many pairs of primes $(p,q)$ such that $pq$ divides $2^{p-1}+2^{q-1}-2$?

A mathematician friend gave me this question (partly as a joke) a few months ago and it has puzzled me for a long time:- Do there exist infinitely many pairs of primes $(p,q)$ such that ...
16
votes
2answers
859 views

Proving there are an infinite number of pairs of positive integers $(m,n)$ such that $\frac{m+1}{n}+\frac{n+1}{m}$ is a positive integer

The question is: Show that there are an infinite number of pairs $(m,n): m, n \in \mathbb{Z}^{+}$, such that: $$\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{Z}^{+}$$ I started off approaching this ...
15
votes
6answers
1k views

Divisibility criteria of 24. Why is this?

I am currently familiar with the method of checking if a number is divisible by $2, 3, 4, 5, 6, 8, 9, 10, 11$. While Checking for divisibility for $24$ (online). I found out that the number has to ...
14
votes
7answers
702 views

$n^5-n$ is divisible by $10$?

I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that ...
13
votes
4answers
1k views

Is there a sequence of 5 consecutive positive integers such that none are square free?

Is there a sequence of 5 consecutive positive integers such that none are square free? A number is square free if there is no prime number p such that $p^2 \mid n$ What I've tried doing so far is to ...
13
votes
2answers
291 views

Prove $6 \nmid [\left( \sqrt[3]{28} - 3 \right)^{-n}]$

Prove that: $$6 \not\left|\ \left\lfloor\frac 1 {(\sqrt[3]{28} - 3)^{n}}\right\rfloor \ (n \in Z^+)\right.$$ ($\lfloor x\rfloor$ = largest integer not exceeding $x$) I am very bad as English and ...
13
votes
1answer
590 views

My attempt to prove GCD exists

Please review my attempt to prove a theorem. Any mistakes you point would be highly appreciated by me. To prove the theorem, I'll be using the following properties which I'm assuming have already ...
13
votes
1answer
398 views

Do there exist two primes $p<q$ such that $p^n-1\mid q^n-1$ for infinitely many $n$?

We can prove that there is no integer $n>1$ such that $2^n-1\mid 3^n-1$. This leads to the following question: Is it true that for every pair of primes $p<q$ there are only finitely many ...
12
votes
5answers
4k views

Proof of the divisibility rule of 17.

Rule: Subtract 5 times the last digit from the rest of the number, if the result is divisible by 17 then the number is also divisible by 17. How does this rule work? Please give the proof. ...
12
votes
7answers
546 views

How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$

If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can´t find a way to use any of the elemental divisibility and gcd theorems to find them.
12
votes
7answers
7k views

The product of n consecutive integers is divisible by n factorial

How can we prove that the product of n consecutive integers is divisible by n factorial? Note: In this subsequent question and the comments here the OP has clarified that he seeks a proof that ...
12
votes
2answers
389 views

How rare are the primes $p$ such that $p$ divides the sum of all primes less than $p$?

This is just for fun! The title pretty much says it all. It's probably a very difficult question. Up to the $40,000^{th}$ prime $(479909)$, I have found only $5$, $71$ and $369119$ with this ...
12
votes
3answers
343 views

Mental Primality Testing

At a trivia night, the following question was posed: "What is the smallest 5 digit prime?" Teams (of 4) were given about a minute to write down their answer to the question. Obviously, the answer is ...
12
votes
2answers
1k views

Elementary proof of Zsigmondy's theorem

I've been writing a not-so-short introduction to elementary number theory, supplying proofs for all theorems. When coming across Zsigmondy's theorem, it seemed difficult to find a proof available on ...
12
votes
1answer
201 views

Family of GCDs all equal to $2$

Is it true that $$\gcd\left(5^{2^n} + 1, 13^{2^n} + 1\right) = 2$$ for all $n \in \mathbf{Z}_{\geq 0}$? I'm continually stumped with this and verifying it numerically is quite expensive very ...
11
votes
7answers
1k views

Proof that $2^{222}-1$ is divisible by 3

How can I prove that $2^{222}-1$ is divisible by three? I already have decomposed the following one: $(2^{111}-1)(2^{111}+1)$ and I understand I should just prove that $(2^{111}-1)$ is divisible by ...
11
votes
6answers
3k views

Show that $11^{n+1}+12^{2n-1}$ is divisible by $133$.

Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck. ...
11
votes
2answers
760 views

What is the remainder when $1! + 2! + 3! +\cdots+ 1000!$ is divided by $12$?

What is the remainder when $$1! + 2! + 3! +\cdots+ 1000!$$ is divided by $12$. I have tried to find the answer using the Binomial Theorem but that doesn't help. How will we do this? Please help.
11
votes
4answers
2k views

How many three digit numbers are not divisible by 3, 5 or 11?

How many three digit numbers are not divisible by 3, 5, or 11? How can I solve this? Should I look to the divisibility rule or should I use, for instance, $$ \frac{999-102}{3}+1 $$
11
votes
7answers
1k views

Prove by induction that an expression is divisible by 11

Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$. I am rather confused by this question. This is my attempt so far: For $n = 2$ $2^5 ...
11
votes
4answers
380 views

Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$

I can't crack this one. Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ (the only constraint is that $a,b,c,z \in \mathbb{Z}$).
11
votes
1answer
72 views

Does it follow that $(n!)^n$ divide $(n^2)!$

It is well known that $(n!)^2$ divides $(2n)!$. Does it follow that $(n!)^3$ divides $(3n)!$ and so on up to $(n!)^n$ dividing $(n^2)!$? If yes or no, could you provide the details behind the ...
11
votes
1answer
149 views

Find all $x,y,z\in\mathbb N$, $x,y,z>1$ such that satisfy $x\mid yz+1$, $y\mid xz+1$, and $z\mid xy+1$

Find all $x,y,z\in\mathbb N$, $x,y,z>1$ such that satisfy $$\begin{cases}x\mid yz+1\\y\mid xz+1\\z\mid xy+1\end{cases}$$ I've found out easily that $$\begin{cases}x\nmid yz\\y\nmid xz\\z\nmid ...
11
votes
1answer
170 views

Are there infinitely many pairs of primes where one divides one more than the square of the other?

I have the following question on number theory that is eating my head. Are there infinitely many primes $p,q$ such that $p | (q^2 + 1)$ and $q | (p^2 + 1)$? I can see $13,5$ and $2,5$ has the ...
11
votes
2answers
229 views

Show that $A[X]/(aX+b)$ is an integral domain

Let $A$ be an integral domain, $a$ and $b \in A-\{0\}$, and let $B = A[X]/(aX+b)$. Show that, if $Aa \cap Ab=Aab$, then $B$ is an integral domain. My attempt at proof (following a hint). Denote ...
11
votes
2answers
612 views

Does $n \mid 2^{2^n+1}+1$ imply $n \mid 2^{2^{2^n+1}+1}+1$?

There are two ways to try to prove this. One is in the title, the other is its de Morgan counterpart: $n \nmid 2^{2^{2^n+1}+1}+1 \implies n \nmid 2^{2^n+1}+1$. Disproving it requires only one example ...
11
votes
1answer
156 views

$ 0 < a < b\,\Rightarrow\, b\bmod p\, <\, a\bmod p\ $ for some prime $p$

If $\,a < b\,$ are natural numbers then a prime $\,p\,$ exists such that $\ a\bmod p\, >\, b\bmod p.$ The task seems understandable, but I have no idea how to prove this statement.
11
votes
1answer
237 views

Curious number theory problem

$k,m,n\in\mathbb{N}$ satisfy $k^{m+n}=nm^n$. How can I show that $m=k$ and $n=k^k\,?$
11
votes
0answers
259 views

My first proof that uses the well-ordering principle (very simple number theory). Please mark/grade.

What do you think about my first proof that uses the well-ordering principle? Please mark/grade. Theorem The sum of the cubes of three consecutive natural numbers is a multiple of 9. Proof ...
11
votes
2answers
481 views

How does one attack a divisibility problem like $(a+b)^2 \mid (2a^3+6a^2b+1)$?

In my current line of investigation, I am running into [many] divisibility questions like the one in the title, i.e. $$ (a+b)^2 \mid (2a^3+6a^2b+1), \qquad(\star) $$ where $a > b \ge 1$ are ...
10
votes
2answers
2k views

$\gcd(b^x - 1, b^y - 1, b^ z- 1,…) = b^{\gcd(x, y, z,…)} -1$ [duplicate]

Possible Duplicate: Number theory proving question? Dear friends, Since $b$, $x$, $y$, $z$, $\ldots$ are integers greater than 1, how can we prove that $$ \gcd (b ^ x - 1, b ^ y - 1, b ^ ...
10
votes
7answers
2k views

Proof for divisibility by $7$

One very classic story about divisibility is something like this. A number is divisible by $2^n$ if the last $n$-digit of the number is divisible by $2^n$. A number is divisible by 3 (resp., by ...
10
votes
5answers
2k views

Disproving the claim that the numbers 1+2+4, 1+2+4+8, 1+2+4+8+16… alternate between prime and composite

I am working through an elementary number theory book and I have come across the following problem. Show the following claims are wrong: Claim 1: The sequence 1+2+4, 1+2+4+8, 1+2+4+8+16, ...
10
votes
4answers
3k views

Why are all non-prime numbers divisible by a prime number?

In Euclid's infinite prime numbers proof, the logic is as follows: Assume a set $S$ of all prime numbers in existence is finite (there are a finite amount of primes) Then there must be a greatest ...
10
votes
3answers
1k views

For any $n$, is there a prime factor of $2^n-1$ which is not a factor of $2^m-1$ for $m < n$?

Is it guaranteed that there will be some $p$ such that $p\mid2^n-1$ but $p\nmid 2^m-1$ for any $m<n$? In other words, does each $2^x-1$ introduce a new prime factor?
10
votes
5answers
311 views

Show that $\gcd\left(\frac{a^n-b^n}{a-b},a-b\right)=\gcd(n d^{n-1},a-b)$

How to show that $$ \gcd\bigg( {a^n-b^n \over a-b} ,a-b\bigg )=\gcd(n d^{n-1},a-b ) $$ $a,b\in \mathbb Z$ where $d=\gcd(a,b)$? Note $\ $ Some of the answers below were merged from this ...