$10000=abc$ 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?
 A: Recognize that the numbers cannot both contain a multiple of $5$ and a multiple of $2$; otherwise they would include a zero. Since $10000=2^4\cdot 5^4$, we know that each of $a,b,c$ is of the form $2^r5^s,$ where $0\leq r,s\leq 4$. Using the above fact we only have three possibilities to consider, though two of them can be ruled out by intuition. $1$) $5^4$ is a term. Then the minimum sum must be greater than $625$. $2$) $5^3$ is a term. Then the minimum sum must be greater than $125$. $3$) $5^2$ is a term. Then another term must be $5^2$ and the remaining term is $2^4$. So the minimum sum is $5^2+5^2+2^4=66.$
A: Recognize that you cannot have a factor that includes both a power of $2$ and $5$ at the same time, and recognize that (obviously) $5^2+5^2<5^4$.  We would try to cover all the $2$s required, then split up the $5$s as evenly as possible. 
A: Clearly one of $a,b,c$ is $2^4$
We need to minimize $b+c$ with $bc=5^4$
Now $(b+c)^2-4bc=?\ge0$
So $(b+c)^2\ge4bc=?$ the equality occurs if $b=c$
Fortunately, that condition keeps $b,c$ natural
Else we need to resort to check for possible combinations
$$5^k,5^{4-k},0\le k\le4-k$$
A: 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 $\{a,b,c\} = \{25,25, 16\}$.
