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Questions tagged [divisibility]

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

79
votes
9answers
4k views

Divisibility by 7 rule, and Congruence Arithmetic Laws

I have seen other criteria for divisibility by 7. Criterion described below present in the book Handbook of Mathematics for IN Bronshtein (p. 323) is interesting, but could not prove it. Let $n = (...
123
votes
7answers
23k views

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

For all $a, m, n \in \mathbb{Z}^+$, $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
35
votes
8answers
13k views

Why is $a^n - b^n$ 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 $a^n$ $-$...
29
votes
7answers
31k 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 "...
2
votes
3answers
637 views

Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = \gcd(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.

Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = hcf(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$. We know that if $a|c$ and $b|c$ then $a\cdot b\cdot s=c$ (for some ...
29
votes
5answers
8k views

Fibonacci modular results $\ F_n\mid F_{kn},\,$ $\, \gcd(F_n,F_m) = F_{\gcd(n,m)}$

Can any one give a generalization of the following properties in a single proof? I have checked the results, which I have given below by trial and error method. I am looking for a general proof, which ...
11
votes
3answers
2k views

Divisibility criteria for $7,11,13,17,19$

A number is divisible by $2$ if it ends in $0,2,4,6,8$. It is divisible by $3$ if sum of ciphers is divisible by $3$. It is divisible by $5$ if it ends $0$ or $5$. These are simple criteria for ...
30
votes
8answers
13k views

Why $\gcd(qb+r,b)=\gcd(b,r)$?

Given: $a = qb + r$. Then it holds that $\gcd(a,b)=\gcd(b,r)$. That doesn't sound logical to me. Why is this so? Addendum by LePressentiment on 11/29/2013: (in the interest of http://meta.math....
66
votes
6answers
10k views

Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)

I don't know if this is a well know fact but I have observed that every number no matter how large that is equally divided by $9$, will equal $9$ if you add all the numbers it is made from until there ...
72
votes
16answers
38k views

For any prime $p > 3$, why is $p^2-1$ always divisible by 24?

I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.
10
votes
4answers
14k views

The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)

How can we prove, without using the properties of binomial coefficients, the product of n consecutive integers is divisible by n factorial?
6
votes
2answers
3k views

How to prove that $z\gcd(a,b)=\gcd(za,zb)$

I need to prove that $z\gcd(a,b)=\gcd(za,zb)$. I tried a lot, for example, looking at set of common divisors of the two sides, but I can't conclude anything from that. Can you please give me some ...
11
votes
5answers
1k 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 ...
5
votes
7answers
20k views

If $\gcd(a,b)= 1$ and $a$ divides $bc$ then $a$ divides $c\ $ [Euclid's Lemma]

Well I thought this is obvious. since $\gcd(a,b)=1$, then we have that $a$ does not divide $b$ AND $a$ divides $bc$. this implies that $a$ divides $c$. done. but apparently this is wrong. help ...
7
votes
2answers
7k views

Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$.

Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ for $0\ne a,b,c\in \Bbb{Z}$. I tried solving it with sets but I sense there are some details I am missing. I would truly appreciate your reference.
4
votes
3answers
760 views

Proving $p\nmid \binom{p^rm}{p^r}$ where $p\nmid m$

A question from Advanced Modern Algebra by Joseph J.Rotman. Let $n=(p^r)m $ such that the prime $p\nmid m$.Prove that $p\nmid \dbinom{n}{p^r}$.HINT: Assume otherwise, cross multiply and apply ...
24
votes
6answers
14k views

If $n\ne 4$ is composite, then $n$ divides $(n-1)!$.

I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. ...
13
votes
2answers
5k views

Derive a formula to find the number of trailing zeroes in $n!$ [duplicate]

Possible Duplicate: How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? I know that I have to find the number of factors of $5$'s, $25$'s, $125$'s etc....
12
votes
4answers
1k 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 $a,b,c,z \in \mathbb{Z}$).
7
votes
2answers
29k views

If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$

Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$. This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was ...
8
votes
10answers
2k views

How to show that $2730\mid n^{13}-n\;\;\forall n\in\mathbb{N}$

Show that $2730\mid n^{13}-n,\;\;\forall n\in\mathbb{N}$ I tried, $2730=13\cdot5\cdot7\cdot3\cdot2$ We have $13\mid n^{13}-n$, by Fermat's Little Theorem. We have $2\mid n^{13}-n$, by if $n$ even ...
16
votes
7answers
38k views

If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$

How do I go about proving this? If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$. I'm very confused with gcd proofs.
35
votes
12answers
15k views

Division of Factorials

I have a partition of a positive integer $(p)$. How can I prove that the factorial of $p$ can always be divided by the product of the factorials of the parts? As a quick example $\frac{9!}{(2!3!4!)} =...
17
votes
5answers
7k views

Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$

Somewhere on Stack Exchange I saw the equation $$\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1.$$ I had never seen this before, so I started trying to prove it. Without success... Can anyone explain me (so ...
15
votes
3answers
830 views

Are associates unit multiples in a commutative ring with $1$?

My problem is: Are associates unit multiples in a commutative ring $R$ with $1$? Recall $a$ and $b$ are associates in $R$ if $\,a\mid b\mid a\,$ in $R$ or, equivalently $\,aR = bR,\,$ and $\,a,b\,...
15
votes
6answers
23k views

Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$

I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$ $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds. I think the first step should look something like this: $d = \gcd(a+b, a-b) = \gcd(...
27
votes
17answers
49k views

How to Prove the divisibility rule for $3$

The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three. This is quite helpful for determining if really large ...
15
votes
5answers
11k views

Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$

Prove if $p$ is a prime then $p \mid \binom pk$ for $k\in\{1,\ldots,p-1\}$ I don't really know where to begin with this one. I can see that I have to use the fact that $p$ is prime somewhere - the ...
15
votes
12answers
14k views

If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$

If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$ This seems clear, but I don't know how to prove this.. I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ ...
5
votes
9answers
845 views

If a prime $p\mid ab$, then $p\mid a$ or $p\mid b$

If a prime number $p$ is a divisor of a product $ab$, $p$ has to be a divisor of $b$ or $a$. How can I demonstrate this theorem? I demonstrated this theorem on one way using Bezout's theorem in an ...
2
votes
4answers
5k views

GCD Proof with Multiplication: gcd(ax,bx) = x$\cdot$gcd(a,b)

I was curious as to another method of proof for this: Given $a$, $b$, and $x$ are all natural numbers, $\gcd(ax,bx) = x \cdot \gcd(a,b)$ I'm confident I've found the method using a generic common ...
16
votes
2answers
5k 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 ^ z - 1 ,...
3
votes
4answers
3k views

Prove that $(ma, mb) = |m|(a, b)$

I'm trying to prove that $(ma, mb) = $|$m$|$(a, b)$ , where $(ma, mb)$ is the greatest common divisor between $ma$ and $mb$. My thoughts: If $(ma, mb) = d$ , then $d$|$ma$ and $d$|$mb$ → $d$|$max +...
14
votes
5answers
11k 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. ...
3
votes
2answers
741 views

If $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \cdot\gcd(b, c)$

How can I prove that if $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \times \gcd(b, c)$? By eea there exists $ax+by=1$ from $\gcd(a,b)=1$ so a and be are co-primes there also exists $dk=a$ and $...
4
votes
2answers
1k views

Extended Euclidean Algorithm problem

I'm confused about how to do the extended algorithm. For example, here's the gcd part gcd(8000,7001) $$\begin{align}8000 &= 7001\cdot1 + 999\\ 7001&=999\cdot 7+8\\ 999&=8\cdot 124+7\\ 8&...
44
votes
3answers
2k views

How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? [duplicate]

Possible Duplicate: Highest power of a prime $p$ dividing $N!$ How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
6
votes
1answer
550 views

$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$

Let $p$ be prime number ($p\gt2$) and $a,b\in\mathbb Z$ ,$a+b\neq0$ ,$\gcd(a,b)=1$ how to prove that $$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p$$ Thanks in advance .
4
votes
8answers
4k views

If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$.

Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$ I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case, $(a+b)x + ((a+b)^2 -3ab)y =1.$ I thought setting $x = (a+b)$ and $y = -1$...
3
votes
5answers
208 views

How to prove $\,x^a-1 \mid x^b-1 \iff a\mid b$

How to prove $x^a-1\mid x^b-1 \iff a \mid b$, where $x \ge 2$ and $a,b,x \in \Bbb Z$. I've tried the following in attempting to solve this: $$a\mid b \Rightarrow aq=b \Rightarrow x^{aq}=x^b \...
10
votes
5answers
7k views

Concise proof that every common divisor divides GCD without Bezout's identity?

In the integers, it follows almost immediately from the division theorem and the fact that $a | x,y \implies a | ux + vy$ for any $u, v \in \mathbb{Z}$ that the least common multiple of $a$ and $b$ ...
20
votes
7answers
19k views

How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$?

I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
12
votes
13answers
10k views

How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$

Intuitively it's true, but I just can't think of how to say it "properly". Take for example, my answer to the following question: Let $p$ denote an odd prime. It is conjectured that there are ...
6
votes
7answers
4k views

Prove that $6$ divides $n(n + 1)(n + 2)$

I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows: Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$. ...
6
votes
3answers
24k views

Show that $n^3-n$ is divisible by $6$ using induction

As homework, I have to prove that $\forall n \in \mathbb{N}: n^3-n$ is divisible by 6 I used induction 1) basis: $A(0): 0^3-0 = 6x$ , $x \in \mathbb{N}_0$ // the 6x states that the result is a ...
1
vote
2answers
675 views

Greatest common divisor is the smallest positive number that can be written as $sa+tb$

We know that $d = \gcd(a, b)$ can be written as $sa + tb$, where $s, t \in \mathbb{Z}$. Apparently, $d$ is the smallest positive number that can be written in this form. Why is this so?
64
votes
8answers
14k 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 ...
11
votes
3answers
530 views

Showing that $a^n - 1 \mid a^m - 1 \iff n \mid m$

Let $a\ge 2$ be an integer. Show that for positive integers $m,n$, we have $a^n - 1$ divides $a^m - 1$ if and only if $n$ divides $m$. I am having trouble showing this. I've seen a similar problem on ...
11
votes
4answers
4k views

Fibonacci sequence divisible by 7?

Make and prove a conjecture about when the Fibonacci sequence, $F_n$, is divisible by $7$. I've realized it's when $n$ is a multiple of $8$. I just don't know how to go about proving it.
6
votes
7answers
7k views

Prove: $\gcd(a,b) = \gcd(a, b + at)$. [duplicate]

I know that $\gcd(a,b)$ divides $a$ and $b$, and must also then divide $(a)(t)$ ($t$ being some integer). This makes sense to me, but how do I prove it? It seems that the addition of $(a)(t)$ is a ...