Questions tagged [gcd-and-lcm]

For questions related to gcd (greatest common divisor, also known as the hcf, the highest common factor), lcm (least common multiple), and related concepts from integer and ring theory.

The greatest common divisor (also known as highest common factor) of two or more integers is the largest integer that divides all of them. It may be computed using the Euclidean algorithm.

Bézout's identity states that for non-zero integers $a$ and $b$ there exist integers $x$ and $y$ with $ax+by=\gcd(a,b)$.


If $a, b \in \mathbb{N}$, write $a \mid b$ if $a$ divides $b$, i.e. there is $k \in \mathbb{N}$ such that $b = ka$.

The least (or lowest) common multiple of $a_1, \dots, a_k \in \mathbb{N}$ is the smallest positive integer $N$ such that $a_i \mid N$ for $i = 1, \dots, k$. We usually denote $N$ by $\operatorname{lcm}(a_1, \dots, a_k)$. Note that $\operatorname{lcm}(a_1, \dots, a_k)$ can be defined recursively from a binary definition. That is,

$$\operatorname{lcm}(a_1, \dots, a_k) = \operatorname{lcm}(\operatorname{lcm}(\dots\operatorname{lcm}(\operatorname{lcm}(a_1, a_2), a_3), \dots, a_{k-1}), a_k).$$

If $a, b \in \mathbb{N}$ and $a = p_1^{r_1}\dots p_m^{r_m}$, $b = p_1^{s_1}\dots p_m^{s_n}$ are their prime decompositions (where some of the $r_i$ and $s_j$ can be zero), we have

$$\operatorname{lcm}(a, b) = p_1^{\max(r_1, s_1)}\dots\ p_m^{\max(r_m, s_m)}.$$

Note that $\operatorname{lcm}(a, b)\operatorname{gcd}(a, b) = ab$.


These notions can be generalised to any commutative ring; the above is just the particular case of (positive elements of) the ring $\mathbb{Z}$.

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Prove if given L|M and gcd(M, D) = 1, then gdc (D, L) = 1 is true

I want to prove if given $L|M$ and $gcd(M, D) = 1,$ then $gdc (D, L) = 1$ is true (As stated in the title) This seems very easy, yet I am having trouble on where to really start? I feel like Bezout's Identity should be used here, however I don't…
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Can you prove this implication between $\gcd$ equalities?

Given six positive integers $\{a_1,b_1,a_2,b_2,a_3,b_3\}$ such that $$ \gcd(a_1,b_1) = \gcd(a_2,b_2) = \gcd(a_3,b_3) $$ then the following equality also holds, $$ \gcd(d_{31},d_{32}) = \gcd(d_{31},d_{21}) = \gcd(d_{32},d_{21}) $$ being…
rtomas
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What is $\gcd(a^2-1, a^3-1)$?

What is $\gcd(a^2-1, a^3-1)$? Is it $1$? The exponents seem to follow the pattern of $\gcd(a, a+1)$.
user838817
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Question about gcd (greatest-common-divisor)

I have this question about gcd and I have doubts: "The gcd between two numbers is 48 and the largest number is 384. What is the other number?" I thought that the missing number must be a multiple of 48 as well, but I can't find a way to find that…
Marina
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Proof of The GCD Distributive Indentity

$$\gcd(\text {lcm}(x, y), \text {lcm} (x, z)) = \text {lcm}(x,\gcd(y, z))$$ where $x,y,Z$ are three integers. I came across this property on GCD but was not able to prove this.Can anyone suggest me a method.
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GCD( LCM(a,b) , LCM(a,c) ) = LCM(a, GCD(b,c)) , How?

I actually asked this question already once here, but I marked it answered by mistake and it was also titled wrong. I wanted to ask where I went wrong in the following lines and what should be the correct way to arrive at the conclusion. I suspect…
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How to prove that total numbers which are divisible by $a$ and $b$ and are less than $N$ are always $\lfloor N/LCM(a,b) \rfloor$?

I am trying to solve this question : find total numbers which are divisible by $a$ and $b$ and are less than $N$ are always $\lfloor N/LCM(a,b) \rfloor$, by intuition I first find all numbers which are divisible by $a$ and out of these numbers I…
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Prove gcd equality

How to prove in Euclidean ring $Z$ with $a\in Z$ $$gcd(a^2+3, 3a+5)=gcd(a^2+3, 3a^2+5a, 3a+5)$$ ?
sbond
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let $a,b$, and $c$ be integers with $b \ne 0$, and $b|c$. Prove that $\gcd(a,b) =\gcd(a+c, b)$

since $( b|c)$, $c = xb$ So, $\gcd(a,b) = \gcd(a + xb, b)$ Euclidean algorithm shows shows that for $a=bq + r$, $\gcd(a,b) = \gcd(b,r)$ This is where im stuck, I've approached this quesiton so many ways but can't figure out how to apply the…
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The GCF of all "bad" numbers less than $2020^{2020}+1$

A number is considered bad if it can be written in the form $2020^n+1$ for some positive integer $n.$ What is the greatest common divisor of all bad numbers less than $2020^{2020}+1?$ Can somebody please guide me on how to solve this?
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Finiteness of GCD

Binary GCD algorithm is described in the following link: https://en.wikipedia.org/wiki/Binary_GCD_algorithm I wonder if there exists any mathematics proof of the finiteness of this algorithm,that is,how to prove that the GCD can be achieved within…
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GCD: If $d\mid a, d\mid b$, and $d\mid c$, then $d|\mid \gcd(a,b,c)$

How do I prove that If $d\mid a, d\mid b$, and $d\mid c$, then $d\mid \gcd(a,b,c)$ without using gcd(a,b,c)=gcd(gcd(a,b),c)? I understand that the Common Divisor Divides GCD is only defined for two variables.
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Find number of pairs $(i, j)$ such that (gcd(i, j), lcm(i, j)) is the same for all pairs.

Suppose we have a pair (p, q) and a pair (gcd(p, q), lcm(p, q)). I need to find a number of pairs (i, j) such that (gcd(i, j), lcm(i, j)) = (gcd(p, q), lcm(p, q)). How can i do it?
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find $\gcd(2n+1,n)$ assuming that $n$ is non negative integer

Should I use the rule $$n=(2n+1)\cdot q+r$$ I am not sure how to find the gcd while $n$ is unknown or should I assume that the numbers will be odd so the gcd will be $1$?
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Find the greatest $x$ that divides 14, 19, 25, 52 and leaves remainders 4, 1, 5 and 2, respectively

Given a question as follows. Find the greatest $x$ that divides 14, 19, 25, 52 and leaves remainders 4, 1, 5 and 2, respectively. For me this question does not make sense. Because the $\text{HCF}$ of $14-4$, $19-1$, $25-5$, and $52-2$ is…
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