What is the smallest possible value of $n$? Is an easy way to solve the problem? 

The product of the digits of positive integer $n$ is $20$, and the sum of the digits is $13$. What is the smallest possible value of $n$?

The way I did is to list equations $$n_1+n_2+n_3+...+n_i=13$$ $$n_1*n_2*n_3*...*n_i=20$$ Then I tried $n_s$. I don't think this is not a very good method.
 A: First, let's list out all of the ways $20$ can be factored with digits:
$$20=2*2*5$$
$$20=4*5$$
If we take $225$, we can add four $1$s at the beginning to get a number with a sum of $13$, so we have $1111225$.
If we take $45$, we can add four $1$s at the beginning to get a number with a sum of $13$, so we have $111145$.
Clearly, $111145$ is the smaller number, so that is our answer.
A: Hint: If the product is $20=2 \cdot 2 \cdot 5 = 4 \cdot 5$, then the only possible digits are $1,2,4,5$.
A: We see that $20$ can be factored as $2\times2\times5$.So,we can use only these three numbers or add $1$'s to our number since multiplying by $1$ gives same result.
So,we see,13 can be written as $2+2+5+1+1+1+1$.We add $4$ ones without changing the product.
So,least number possible with this is $111145$.
Note that we multiply the two $2$'s to form $4$ to give the least possible number. 
A: Since $n_1,\dots,n_i$ are positive integers and $20$ is primarily decomposed as $20=2^2\cdot 5$, we quickly find the largest $n$, that is $5221111$. 
But before finding the smallest, lets note since $20$ is a multiple of $5$ and no digits other than $0$ and $5$ aren't divisible by $5$. and since product of the digits is non-zero, one and only one of the digits is $5$, but by similar but not not same reasoning we can say either, one and only one digit is $4$, or, two and exactly two digits are $2$. in either ways the some of non-units is $9$. So, we have 4 other digits, each of which are unit. so the numbers we can write with the properties asserted, have either, 6, or 7 digits. of course the smallest among them should have 6 digit. and is $111145$ 
