When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$? When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$?
Well given by two digit numbers in base $5$ I tried out the multiplication and tried to simplify.
$(ab_5)(cd_5)=4103_5$ or $(5a+b)(5c+d)=4*5^3+1*5^2+0*5^1+3$
Then get that $5^2ac+5ad+5cb+bd=5^3*4+5^2+3$ 
I notice that $bd=3$ so b=1,3 and $d=1,3$. Now this is where I'm stuck I don't see a clear way to get what the numbers should be without a lot of trial and error. 

Continuing along your route, where:
$$4103=(10a+b)(10c+d)$$
(notice all values are represented in base $5$ here)
All values will be in base $5$ throughout.
Notice that $40^2=3100$, which is less than $4103$, so $a$ and $c$ are both $4$:
$$4103=(40+b)(40+d)=3100+40(b+d)+bd$$
Now, notice that $\left|{b-d}\right|=2$ in order for the last digit in the product to be a $3$.  Since the order of factors is optional, and we know the first digits are the same, we will set $d=b+2$:
$$4103=3100+40(2b+2)+b(b+2)$$
$$4103=3100+(130b+130)+(b^2+2b)$$
$$4103=3230+132b+b^2$$
$$0=b^2+132b-323$$
$$0=(b-2)(b+134)$$
$$ b=\{2,-134\}$$
Since the digit must be between $0$ and $4$ inclusive, only $2$ can be valid.  For $d$:
$$d=b+2=2+2=4$$
Finally, since $a=4$; $b=2$; $c=4$; $d=4$, the factors are $\{42_5, 44_5\}$
