$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?

Attempt:
First write as prime factors: $10000 = 2^{4} 5^{4}$. The possible triples are:
$$ 2, 2^{3}, 5^{4} $$
$$ 2^{2}, 2^{2}, 5^{4} $$
$$ 2^{3}, 2, 5^{4} $$
$$ 2^{4}, 2^{4}, 5^{4} $$
$$ 5, 5^{3}, 2^{4}$$
$$ 5^{2}, 5^{2}, 2^{4}$$
$$ 5^{3}, 5, 2^{4}$$
The smallest sum is $5^{2} + 5^{2} + 2^{4}$. Are there better approaches?

Well.  "no zero" means none of the $a,b,c$ have both factors of $2$ and $5$ so you either have two numbers that are powers of $2$ and one that is a power of $5$ or you have $2$ that are powers of $5$ and one that is a power of $5$.
CASE 1:  $\{a,b,c\} = \{2^k, 2^{4-k}, 5^4\}$.
So $a + b + c = 2^k + 2^{4-k} + 625$. 
We have to minimize $2^k + 2^{4-k}$.
By AM-GM $2^k + 2^{4-k} \ge 2\sqrt{2^k2^{4-k}} = 2\sqrt{2^4} = 2*4=8$ with equality holding if and only if $2^k = 2^{4-k}$.
So $\{a,b,c\} = \{4,4,625\}$ and $a + b +c =633$. Is the minimum sum for this case.
CASE 2:  $\{a,b,c\} = \{2^4, 5^{4-k},5^{k}\}$
Same argument of minimizing $a+b+c$ via $AM.GM$ $a=5^2, b=5^2, c=16$ and $a + b + c = 66$.
So the min is $\{a,b,c\} = \{25,25, 16\}$.
