$\gcd(n,m)=10$, and $\text{lcm}(n,m)=100$

Question

I'm having some trouble with the following question:

Determine all natural numbers $n,m$ such that: $\gcd(n,m)=10$, and $\text{lcm}(n,m)=100$.

Using this I arrived at the following conditions:

But I don't know how to proceed. How can I do this?

Maybe try to assgin the prime factors of 1000 to each one of them? — justadzr, Dec 22 '20 at 19:59
I did that and found that $n=10$ and $m=100$ is a solution, but how do I know if that is the only solution? @Yourong'DZR'Zang — Eduardo Magalhães, Dec 22 '20 at 20:03
Both $n$ and $m$ have to be divisible by $10$ and have to divide $100$. There are very few such numbers, so you can just check all cases. — Mark, Dec 22 '20 at 20:05
@EduardoMagalhães you need to move the factors around to get all solutions. — justadzr, Dec 22 '20 at 20:06

score 3 · Accepted Answer · answered Dec 22 '20 at 20:07

3

Let $n = 2^a\cdot 5^b, m=2^c\cdot5^d$

Since $\gcd(n, m)=10$, we know that $\min(a,c)=1$ and $\min(b,d)=1$.

Since $lcm(n, m) = 100=10^2,$ we know that $\max(a,c)=2$ and $\max(b,d)=2$.

Hence, $(a, c), (b,d) \in \{(1,2), (2,1)\}$.

answered Dec 22 '20 at 20:07

Siong Thye Goh

score 1 · Answer 2 · answered Dec 22 '20 at 20:04

1

Hint:$$mn = 1000, \ \ 10 \le m,n \le 100$$ So what you need to do is just to check $10$ numbers that are divisible by $10$ and not exceeding $100$.

answered Dec 22 '20 at 20:04

VIVID

score 0 · Answer 3 · answered Dec 22 '20 at 20:36

0

HintL Let $m=10m'$ and $n=10n'$. Then $\gcd(m',n')=1$ and $m'n'=10$. There are only a few solutions now.

answered Dec 22 '20 at 20:36

lhf

3 Answers3