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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Fractals using just modulo operation

Let us calculate the remainder after division of $27$ by $10$. $27 \equiv 7 \pmod{10}$ We have $7$. So let's calculate the remainder after divison of $27$ by $7$. $ 27 \equiv 6 \pmod{7}$ Ok, so let us continue with $6$ as the divisor... $ 27 \equiv…
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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 "does not use the properties of binomial…
Paulo Argolo
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Division of Factorials [binomal coefficients are integers]

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 $\frac{9!}{(2!3!4!)} = 1260$ (no remainder), where $9=2+3+4$. I can…
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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 $a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer? But…
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How to prove the divisibility rule for $3\, $ [casting out threes]

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 numbers are multiples of three, because we can…
Aj521
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How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?

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?
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Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$

As is the question in the title, I am wishing to find all positive integers $n$ such that $3^n + 5^n$ is divisible by $n^2 - 1$. I have so far shown both expressions are divisible by $8$ for odd $n\ge 3$ so trivially a solution is $n=3$. I'm not…
TI82
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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 $a_n=\dfrac{\phi(2^n-1)}{n}$, then $a_n$ is A011260,…
lsr314
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Why $\gcd(b,qb+r)=\gcd(b,r),\,$ so $\,\gcd(b,a) = \gcd(b,a\bmod b)$

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.stackexchange.com/a/4110/53259 and averting a…
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Is sum of digits of $3^{1000}$ divisible by $7$?

I'm working on a little exercise I found in my high school book (printed in 2007) which is pretty complicated. Is the sum of digits of $3^{1000}$ a multiple of $7$? Do you have any advice to solve this type of problem (without programming of…
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How to prove that all odd powers of two add one are multiples of three

For example \begin{align} 2^5 + 1 &= 33\\ 2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)} \end{align} I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
Ronikos
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What is vector division?

My question is: We have addition, subtraction and muliplication of vectors. Why cannot we define vector division? What is division of vectors?
Reader
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How many ways can I arrange the numbers $1$ to $N$ with this divisibility condition?

For the numbers $1, \ldots, N$, how many ways can I arrange them such that either: The number at $i$ is evenly divisible by $i$, or $i$ is evenly divisible by the number at $i$. Example: for $N = 2$, we have: $\{1, 2\}$ number at $i = 1$ is $1$…
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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 composite number. If you take any prime, reverse it,…
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Prove that there exist infinitely many integers $(n^{2015}+1)\mid n!$

I conjecture that there exist infinitely many integers $n$ such that $$(n^{2015}+1)\mid n!.$$ I have seen a simpler problem that there exist infinitely many integers $n$ such that $(n^2+1)\mid n!$. Alternatively, I considered the Pell…
user225250