Questions tagged [divisibility]

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

If $a$ and $b$ are integers, $a$ divides $b$ if $b=ca$ for some integer $c$. This is denoted $a\mid b$. It is usually studied in introductory courses in number theory, so add if appropriate.

A common notation used for the phrase "$a$ divides $b$" is $a|b$. It is also common to negate the notation by adding a slash like this: "$c$ does not divide $d$" written as $c\nmid d$. Note that the order is important: for example, $2|4$ but "$4\nmid 2$".

This notion can be generalized to any ring. The definition is the same: For two elements $a$ and $b$ of a commutative ring $R$, $a$ divides $b$ if $ac=b$ for some $c$ in $R$.

Divisibility in commutative rings corresponds exactly to containment the poset of principal ideals. That is, $a$ divides $b$ if and only if $aR\subseteq bR$. For commutative rings like principal ideal rings, this means that divisibility mirrors exactly the poset of all ideals of the ring.

The topics appropriate for this tag include, for example:

  • Questions about the relation $\mid$.
  • Questions about the GCD and LCM.

There are divisibility rule that is a shorthand way of determining whether a given integer is divisible by a fixed divisor without performing the division, usually by examining its digits.

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integer with all digits, multiple of 126

Find a positive integer containing all ten digits: $0,1,2,3,4,5,6,7,8,9$ that is a multiple of $126$ I don't really know where to start. I guess I could find the prime factorization of $126$, which is $2*3^2*7$, but I don't know how that helps.…
suomynonA
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A number is divisible by 13

I am studying divisibility and come across this rule. I think the rule is too complicated and hard to understand and remember. What is the best way to judge whether a number is divisible by 13 without having to remember this rule? Delete the last…
learning
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Prove that for every natural $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$

Prove that for every natural number $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$. I've noticed that $(n^2 + n) = n(n+1)$ so these are two successive numbers hence one of them can be divided by two. I suppose that I should prove that $(n^2 +…
tdudzik
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Brett has £135, Dustin has £70, Greg has £35.

Brett gives some money to Dustin & Greg. The ratio of the amount of money Brett, Dustin and Greg have now is 3:2:1 How much money did Brett give to Dustin? I considered saying Brett gets 3 parts of £240 but thats obviously wrong because what do the…
greg
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Simple Division Problem

I have the equation: $$(1-\frac{1}{2^2})...(1-\frac{1}{n^2}) = \frac{n+1}{2n}$$ for n ≥ 2 Trying to prove by induction and I get the following equation. $$\frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2} = \frac{k+2}{2(k+2)}$$ I can't to simplify it to the…
misheekoh
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check if large number $(9^{81}+6)$ is divisible by $11$

I would like to know if there is a mathematical way to check whether number $9^{81}+6$ is divisible by $11$, without actually calculating the whole number.
heky__
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Finding $n\in\mathbb{Z}$ such that $\frac{(n^2+3)(n^2-5)}{16n}\in\mathbb{Z}$

I'm trying to follow a step in a proof, which involves finding $n\in\mathbb{Z}$ such that $\frac{(n^2+3)(n^2-5)}{16n}\in\mathbb{Z}$. The proof then states that $\text{hcf}(n,n^2+3)$ divides 3, and $\text{hcf}(n,n^2-5)$ divides 5. Hence n divides…
maliky0_o
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For numbers divisible by three, why is the sum of their digits able to be divided by three?

When you add the digits of any number that is divisible by three, that sum of those digits also appears to be divisible by three (with no remainder). For example a number (which I randomly grab from the top of my head): 289752 whose digits sum to 33…
John
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If $a, b, c, d \in \mathbb Z$ and $p$ is a prime factor of $a - b$ and $c - d$, then $p$ is a prime factor of $(a + c) - (b + d)$

By hypothesis, $p \mid (a - b), (c - d)$. Then $p \mid [(a - b) + (c - d)]$. In other words, $p \mid [(a + c) - (b + d)]$. Does it work?
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Divisibility and GCD proof

I'm having trouble with this simple proof. Any help would be appreciated. I don't really know where to start to try to conquer this problem. Suppose $a|m$, $b|m$ and $\gcd(a,b) = 1$. Prove, without appealing to the fundamental theorem of…
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divisibility relations in sets.

How to draw an arrow diagram, a digraph and the matrix representation for the specified relation? The "divides" relation $|$ from the set $\{0,1,2\}$ to the set $\{0,3,6,9\}$
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Explain 'expressing a number using its digits'

While studying divisibilty and prime numbers in my maths book (IB Mathematic Higher Level Option 10: Discrete Mathematics), I came across an explanation of a way to '[express] a number using its digits'. It says: If $N$ is a $k$-digit number with…
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$k | x^{k} - x,$ for $k, x \in \mathbb{Z}$?

I seem to have found that: $$k | x^{k} - x, \ \text{for} \ k, x \in \mathbb{Z}.$$ I have tried it with a few values, and it seems to be true. I am sure that this has been discovered before.
Taylor
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Contest Problem - Divisibility

Find all ordered pairs (x, y) of positive integers x, y such that $x+y$ divides 2014 and (simultaneously) $x^yy^x$ divides $(x+y)^{(x+y)}$ . This is a contest problem from U Tenn, FERMAT contest. My try: Let S = x+y The prime factorization of 2014…
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GCD theory - gcd(x, y) = 1

Take $n + 1$ numbers out of $1, 2, ..., 2n$. Show that there will be two numbers $x, y$ so that $gcd(x, y) = 1$. What I've got is: Let $d=gcd(a,b)$; by definition there are integers $a′$ and $b′$ such that $a=a′d$ and $b=b′d$, so $a′dx+b′dy=d$.…