Find two different pairs of integers $a,b$ satisfying $(a,b)=10$ and $[a,b]=100$ Since $(a,b)=10$, we can assume $a=10m$, $b=10n$ and $(m,n)=1$.
Want $[a,b]=100$. But $a$ and $b$ have gcd, which is $10$. So $[a,b]=100$ is amount to $[m,n]=10$. Hence the two paris are $(1,10)$ and $(2,5)$.
Actually, I came up with this solution when I tried to ask. I'm not sure whether it is fully correct. If I'm correct, how can I organise the language? Like the pairs $(1,10)$, $(2,5)$ may be ambiguous with gcd. Is there any other way to solve this question?
 A: The product of two positive integers is the product of their gcd and lcm, that is
$$ab=(a,b)[a,b]$$
The proof is easy. Think on what exponent has in each side every prime that divides $a$ or $b$ (or both).
So we have $ab=1000$.
Also,
$$\frac{a}{(a,b)}\frac{b}{(a,b)}=\frac{1000}{100}=10$$
Since $a/(a,b)$ and $b/(a,b)$ are coprime, then $a/(a,b)$ is $1,2,5$ or $10$.
A: Yes, it is correct. It is a special case of a general technique of reduccing to the coprime case when we have distributive operations such as gcd and lcm. Let's do it that way.
Cancel $\ (a,b)=10\,$ from both reduces them to $\ (\bar a,\bar b)=1,\  \overbrace{[\bar a,\bar b]}^{\large  \bar a\,\bar b}= 10,\  $ for $\ \smash[b]{\begin{align} \bar a=a/10\\ \bar b = b/10\end{align}}$
Thus the problem reduces to splitting $10=\bar a \bar b\,$ into coprime factors two ways, which is easy, namely $\ 1\cdot 10 = 10 = 2\cdot 5.$

Note that said cancellation implicitly uses the distributive law for gcd and lcm to distribute out the common factor $\, c= 10\,$ that we seek to cancel, i.e. it uses the properties.
$$\begin{align} (ac,bc) &= (a,b)c\\ [ac,bc] &= [a,b]\,c\end{align} $$
So these distributive laws are one way to make rigorous your "amounts to" claim. If you write out the details implicit in your proof you will see that it uses them (so some equivalent properties).
A: Every one of $a,b$ is a multiple of $10$ and divides into $100$, so they can each only be $10,20,50,100$  To have the GCD be exactly $10$, we either need one to be $10$ or to have one be $20$ and another be $50$.  To have the LCM be exactly $100$ we either need one to be $100$ or one be $20$ and another $50$.  We might as well assume $a \le b $ and permute.  This means we can have $(10,10),(10,20),(10,50),(10,100),(20,50)$. Out of these, all are accepted except $(10,10)$ due to the condition of the numbers being distinct.
