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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Divisibility by a prime number and multiplicity

We want to determine the number of times a prime number $p$ divides a number $n$ or, in other words, determine the greatest integer $k$ such that $p^k\ |\ n$. Is there a better method than trying naive division successively?
Adam54
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Given two numbers a and b , find Nth number which is divisible by a or b.

Input: a=2 b=3 N=10 The numbers which are divisible by 2 or 3 are: 2,3,4,6,8,9,10,12,14,15 and the 10th number is 15.
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A question related to divisibility

Given that $2n + 1$ is divisible by $3$, and $n$ is a natural number, what form can $n$ take? As per the solution I am referring to, $n$ must in the form $3k+1$, where $k$ is a whole number. The solution further adds that it is equivalent to $6p +…
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Let n and d be positive integers. How many positive integers not exceeding n are divisible by d?

There is a discussion in my book that, despite my best efforts, I was unable to understand. The problem asks: Let $n$ and $d$ be positive integers. How many positive integers not exceeding $n$ are divisible by $d$? I understood everything…
Aleksandr Hovhannisyan
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How do I prove this divisibility by 17 method?

Take any number. Drop the final two digits. Subtract from it nine times the number made by the two digits you dropped. The original number and the new number are either both divisible by 17 or both not divisible by 17.
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The smallest integer that produces remainders of $2$, $4$,$ 6$ and $1$ when divided by$ 3$, $5$,$ 7$ and $11$ respectively is?

I am finding the Chinese Remainder theorem very complicated, Any other wa to solve this will be really helpful, Thanks in advance!
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The GCD of Certain Multiples of Coprimes

Let $m,n\in \mathbb{Z}$, such that $\gcd(m,n)=1$, and let $p,q\in \mathbb{Z}$. Is there anything that we can conclude about $\gcd(pn,qnm)$? I am asking this because it might help me answer the truth of the statement: if $\gamma | \gcd(pn,qnm)$ (with…
Kurome
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Solve equation with Euclidean Division

I can't solve the following problem: Find $n$, $d$ and $q$ that satisfies that: The division of $n+5$ and $d$ (I mean $(n+5)/d$) produces a quotient of $q$ and a remainder of $2$. The division of $4n-4$ and $d+1$ produces a quotient of $2q+2$ and a…
gobaldia
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For all positive integers, If $a | b, b | c$ and $c |a$ then we must have $a = b = c$

I have a hard time understanding why this statement is true, I have used many examples and it holds true if $a, b$ and $c$ are the same integer. Could someone help me understand why?
Globey163
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Every integer of the form $(n^{3}-n)(n^{2}-4)$ (for n = 3,4,....K) is divisible by?

Every integer of the form $(n^{3}-n)(n^{2}-4)$ $(for n = 3,4,....K)$ is (A) divisible by $6$ but not always divisible by $12$; (B) divisible by 12 but not always divisible by 24; (C) divisible by 24 but not always divisible by 120; (D) divisible by…
user405925
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Prove if $a^n|b^n$ then $a|b$

So assume that $a^n|b^n$ then $b^n=a^n(x)$ for some integer $x$ taking the nth root of both sides you get $b=a(x^{\frac{1}{n}})$. Now I'm not sure if this is ok since $x^{\frac{1}{n}}$ is not always an integer. But it seems like it should always…
HighSchool15
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Prove that if $d\mid a$ and $d\mid b$ then $d\mid a\pm b$

My attempt so far is that if $d\mid a$ and $d\mid b$ then $a=dx$ and $b=dy$ for some integers $x$ and $y$. Then squaring both $a$ and $b$ and taking there difference I get $a^2-b^2=d(dx^2-dy^2)$ so $(a+b)(a-b)=d(dx^2-dy^2)$ So I know that…
HighSchool15
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How many such numbers are there?

Using only the digits $2$, $3$ and $9$, how many six digit numbers can be formed which are divisible by $6$? My attempt : To be divisible by $6$, the number has to be divisible by $2$ and $3$. Thus, the one's digit has to be $2$ only. Now we…
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When $p \mid a^2+ab+b^2$?

Let $p$ be an odd prime; and $a$ and $b$ are integers with $1\leq a, b \leq \dfrac{p-1}{2}$. Is it true that $p \mid a^2+ab+b^2 \text{ (for some choice of } a \text{ and } b) \iff p \not \equiv 2 \pmod{3},$ or equivalently, $p \not\mid a^2+ab+b^2…
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KS2 long division problem

Just seeking a little help with one of my daughter's Yr6 math questions, as the answer I get on a calculator is different from what she (and I) get by doing it the long division way. The question is 2912 ÷ 52 Now by calculator it's 56, but by the…