I want to find a number $m$ such that $$\frac{7(m-6)}{10}$$ is an integer multiple of $100$ and such that $$\frac{3(m-6)}{10}$$ is also an integer multiple of $100$. If possible, I would like the second to be a smaller multiple of the first.
How is it possible?
I tried doing $$\frac{7(m-6)}{10} = 100p$$ and then finding $q$ such that the other is equal to $100q$ but it didn't bring me anywhere
On
Since $7(m-6)/10$ and $3(m-6)/10$ are multiples of $100$, $7(m-6)$ and $3(m-6)$ are multiples of $1000$. And given that $3$ and $7$ are coprimes with $1000$, this happens only if $m-6$ is a multiple of $1000$.
Since you want $7(m-6)/10>3(m-6)/10$, you need $m-6$ to be positive. The minimum value for $m$ is $1006$.
On
If $\frac {7(m-6)}{10}$ is a multiple of $100$ then $7(m-6)$ is a multiple of $1000$. As the $100$ and $7$ have no factors in common, this means $m-6$ is a multiple of $1000$ and $m = 1000k + 6$ for some integer $k$.
So if $m = 1006, 2006, 3006....$ then $\frac {7(m-6)}{10} = 700,1400,2100....$.
The same goes for $\frac {3(m-6)}{10}$. If $m = 1006, 2006, 3006,.... $ then $\frac {7(m-6)}{10} = 300,600,600....$
You say you want $\frac {3(m-6)}{10}$ to be a multiple of $\frac {7(m-6)}{10}$. But that would mean, if $m\ne 6$ that $\frac {\frac{3(m-6)}{10}}{\frac {7(m-6)}{10}} = \frac 37$ is an integer. Never happens.
.... unless $m -6$. In that case $0$ is a multiple of $0$.
(I don't know what you mean by "smaller" mutliptle.)
You want $$7(m-6)=1000p \tag1$$ and $$3(m-6)=1000q \tag 2$$
Dividing $(1)$ by $(2)$$$\implies \frac pq=\frac 73$$
$$\implies 3p=7q$$ $$\implies p|7 \; \text{and} \; q|3$$
So, $p=7n ;\; n \in \Bbb N$.
from $(1)$, we get
$\color{blue}{m=1000n+6} \;$ is general solution for this problem.