Question

$\operatorname{lcm}(n,m,p)\times \gcd(m,n) \times \gcd(n,p) \times \gcd(n,p)= nmp \times \gcd(n,m,p)$, solve for $n,m,p$?

score 279 · Answer 1 · 2026-03-27 14:03:14

279

Views

$\operatorname{lcm}(n,m,p)\times \gcd(m,n) \times \gcd(n,p) \times \gcd(n,p)= nmp \times \gcd(n,m,p)$, solve for $n,m,p$?

Published on 27 Mar 2026 - 14:03

score 881 · Answer 2 · 2026-03-22 22:59:51

881

Views

How to show that $\displaystyle [a,b,c] = \frac{abc}{(ab,bc,ca)}$ without prime factorization?

Published on 22 Mar 2026 - 22:59

score 1.4k · Answer 3 · 2014-05-23 01:44:49

1.4k

Views

Let $a,m,n \in \mathbf{N}$. Show that if $\gcd(m,n)=1$, then $\gcd(a,mn)=\gcd(a,m)\cdot\gcd(a,n)$.

Published on 23 May 2014 - 1:44

score 161 · Answer 4 · 2014-05-26 22:12:41

161

Views

Let $a$ and $b$ be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)= 1$ or $5$

Published on 26 May 2014 - 22:12

score 186 · Answer 5 · 2014-05-27 00:02:08

186

Views

Prove that $\mathrm{lcm}(n,m)= \mathrm{lcm}(|n|,|m|)$

Published on 27 May 2014 - 0:02

score 810 · Answer 6 · 2014-05-29 02:23:12

810

Views

Prove $\gcd(k, l) = d \Rightarrow \gcd(2^k - 1, 2^l - 1) = 2^d - 1$

Published on 29 May 2014 - 2:23

score 67 · Answer 7 · 2014-06-09 13:39:47

67

Views

$a,b,c\in Z, (x,y) \in Z$ is a solution of $ax+by=c$, proof that c is divisible by $\gcd(a,b)$.

Published on 09 Jun 2014 - 13:39

score 120 · Answer 8 · 2014-06-12 04:17:58

120

Views

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

Published on 12 Jun 2014 - 4:17

score 2.9k · Answer 9 · 2014-06-25 01:25:38

2.9k

Views

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

Published on 25 Jun 2014 - 1:25

score 444 · Answer 10 · 2014-07-01 23:09:27

444

Views

Identity involving LCM and GCD: $\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$

Published on 01 Jul 2014 - 23:09

score 3.6k · Answer 11 · 2014-07-01 23:18:37

3.6k

Views

Notation for the least common multiple and greatest common divisor

Published on 01 Jul 2014 - 23:18

score 2.1k · Answer 12 · 2014-07-23 06:38:31

2.1k

Views

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

Published on 23 Jul 2014 - 6:38

score 2.7k · Answer 13 · 2014-07-23 14:27:34

2.7k

Views

Show $\,m$ odd $\Rightarrow \gcd(a^m-1,a^n+1)=1$ or $2$

Published on 23 Jul 2014 - 14:27

score 162 · Answer 14 · 2014-07-30 00:25:43

162

Views

Show that $\gcd(a,b) = \gcd(a+b,b)$

Published on 30 Jul 2014 - 0:25

score 1k · Answer 15 · 2014-08-01 09:33:13

1k

Views

$\gcd( x+y, x^2 - xy + y^2) $ is 1 or 3 for coprime $x$, $y$

Published on 01 Aug 2014 - 9:33

List Question

$\operatorname{lcm}(n,m,p)\times \gcd(m,n) \times \gcd(n,p) \times \gcd(n,p)= nmp \times \gcd(n,m,p)$, solve for $n,m,p$?

How to show that $\displaystyle [a,b,c] = \frac{abc}{(ab,bc,ca)}$ without prime factorization?

Let $a,m,n \in \mathbf{N}$. Show that if $\gcd(m,n)=1$, then $\gcd(a,mn)=\gcd(a,m)\cdot\gcd(a,n)$.

Let $a$ and $b$ be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)= 1$ or $5$

Prove that $\mathrm{lcm}(n,m)= \mathrm{lcm}(|n|,|m|)$

Prove $\gcd(k, l) = d \Rightarrow \gcd(2^k - 1, 2^l - 1) = 2^d - 1$

$a,b,c\in Z, (x,y) \in Z$ is a solution of $ax+by=c$, proof that c is divisible by $\gcd(a,b)$.

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

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

Identity involving LCM and GCD: $\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$

Notation for the least common multiple and greatest common divisor

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

Show $\,m$ odd $\Rightarrow \gcd(a^m-1,a^n+1)=1$ or $2$

Show that $\gcd(a,b) = \gcd(a+b,b)$

$\gcd( x+y, x^2 - xy + y^2) $ is 1 or 3 for coprime $x$, $y$

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