Find all triples of positive integers a,b and c satisfying the equations: $ (a,b,c)=10$ and$ [a,b,c]=100$ simultaneously. This method below worked for gcd and lcm of two integers a,b, but I'm having trouble using the same argument.  Perhaps that is the reason, but then I'm stuck with how to move forward.
$(a,b,c)=10,$ Let $a=10a',b=10b',c=10c'$.  Then 
$$(a,b,c)=(10a',10b',10c')=10.$$  But,
$$(10a',10b',10c')=10(a',b',c')=10  ⇒  (a',b',c')=1$$
  By the product of the greatest common factor and least common multiple, 
$$[a,b,c]=\frac{abc}{(a,b,c)}   ⇒  100= \frac{1000a'b'c}{10}   ⇒  1=a'b'c'$$
This leads to an erroneous conclusion.  Without rigor I can see that solutions include
$(10,20,50), (20,20,50), (10,10,100), (10,100,100), (20,50,50), (20,50,100), (10,20,100), (10,50,100)$, but how do I show all solutions?
 A: The only mistake you've got is to assume that $[a, b, c] = \frac{a b c}{ ( a, b, c) }$, which is not true. You shurely inspired in the old formula for 2 numbers $[a, b] = \frac{a b } { (a , b) }$.
If you want to derive a formula for 3 numbers you should thing as follows, and I warn you by now that you should try this by yourself, the following is not so nice to read :/ -
 $ [ a, b, c ] = [ a' d, b'd , c' d ]  = d [ a', b', c']$ where $d = (a, b, c),  \, \, \, (a', b', c')  = 1$, you want to know what is $[ a', b', c']$ and at this point you can see that it's not always $ a' b' c'$.
Set $a' = \prod_{p \mid a' } p^{ \alpha_p}$, $b' = \prod_{p \mid b' } p^{ \beta_p}$, $c' = \prod_{p \mid c' } p^{ \gamma_p}$. Looking at the prime factorisation, $[a', b', c'] = \prod_{p \mid a' b' c' } p ^{max ( \alpha_p, \beta_p, \gamma_p )}$ and you know by the fact $( a' , b' , c' ) = 1 $ that for each $p$ prime, one of $\alpha_p, \beta_p, \gamma_p$ is zero, so $p^{max(\alpha_p, \beta_p, \gamma_p)} = \frac{p^{\alpha_p} p^{\beta_p} p^{\gamma_p}}{p^{max_2(\alpha_p, \beta_p, \gamma_p)} p^{max_3(\alpha_p, \beta_p, \gamma_p)}p^{max_3(\alpha_p, \beta_p, \gamma_p)}}$, where $max_2, \, \,max_3$ stands for the second maximum and third maximum.
Multiplying for each prime we get $[a', b', c'] = \frac{ a' b' c'} {(a' , b') ( b', c') (c', a')} =  \frac{ a b c} {(a , b) ( b, c) (c, a)}$ so $[a, b, c] =  (a, b, c) \frac{ a b c} {(a , b) ( b, c) (c, a)}$
You should anyways keep the job without using this formula, as $[a, b, c] = 100 \Rightarrow [a', b', c' ] = 10$ is small enough to brute-force all the cases: assume $a\leq b \leq c$ then for $a=1, \, b=1$ we get $c=10$, for $a=1, \, b=2$ we get $c=10$ or $c=5$, etc...
A: If $a = p_1^{e_1} \cdots p_k^{e_k}$ and $b = p_1^{f_1} \cdots p_k^{f_k}$ are prime factorizations, then $\gcd(a,b) = p_1^{\min\{e_1,f_1\}} \cdots p_k^{\min\{e_k,f_k\}}$. Similarly for $[a,b]$ you look for the max of the exponents in the prime factorization. This holds also for three numbers $a,b,$ and $c$. So our numbers can only have prime factors 2 and 5 and we need minimum and maximum exponents to statisfy $(a,b,c) = 10 = 2^1 \cdot 5^1$ and $[a,b,c] = 2^2 \cdot 5^2$. This doesn't leave many options.
A: Every one of $a,b,c$ 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 $a \le b \le c$ and permute.  This means we can have $(10,10,100),(10,20,100),(10,50,100),(10,100,100), (10,20,50),(20,20,50),(20,50,50),(20,50,100)$ and all permutations of these.
A: Since LCM is $10$ All numbers are divisible by $2$ and $5$, there is at least one number exactly divisible by $2$ and at least one number exactly divisible by $5$. Also no numbers are divisible by other primes. Since $GCD$ is 100 there is at least 1 number exactly  divisible by $4$ and at least 1 number exactly  divisible by $25$. No number is divisible by powers of $3$ or higher.
We look at possibilities of powers of $2$ in prime factorization:
$2,4,4$ or $2,2,4$ ($6$ ways counting permutations)
and now we look at the powers of $5$
$5,25,25$ or $5,5,25$ ($6$ ways counting permutations).
Thus there are $36$ triples.
