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. 
 A: Notice that $\;4103_5 = x^2 - 4x + 3\;$ where $\;x = 5^2.\;$ Now factor it as $\;(x-3)(x-1).$
A: Maybe this is cheating, but ... $4103_5 = 528 = 2^4 \cdot 3 \cdot 11$.
The pair of numbers greater than or equal to $5$ and less than $25$ that we can make with these factors is $22=42_5$ and $24=44_5$.
A: 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 $\{42_5, 44_5\}$
A: So $n_1 = 5a + b; n_2 = 5c + d$ and $(5a+b)(5c+d) = 4103_5 = 4*5^3 + 1*5^2 + 3$
$(5a + b)(5c+d) = ac*5^2 + 5(bc + ad) + bd = 4*5^3 + 1*5^2 + 3 = 21*5^2 + 0 + 3$
So $ac \approx 4*5$ which is a very large value considering $a,c \le 4$.
Note $40_5*100_5 = 4000_5 < 4103_5$ so as $(5a+c) < 100_5$ we now $(5b+d) > 40_5$ and vice versa.  So $a = c = 4$.
$16*5^2 + 5(4b + 4d) + db = 4*5^3 + 1*5^2 + 3 = 21*5^2 +3$
$20(b+d) + db = 5^3 + 3$
So $20(b+d) \approx 125$ so $b + d \approx 6$.
$bd \equiv 3\mod 5$ and $b,d \le 4$ so $bd \le 13$. so $bd=3,8,13$.  
$13$ is prime so that is out.  $bd = 3\implies b,d = 1,3$ and $b + d = 4\not \approx 6$.  $bd = 8\implies b,d =2,4$ and $b + d = 6$.
So if an such number exist they are $42_5$ and $44_5$.
And $(4*5 + 2)(4*5+4) = 16*5^2 + 5(2*4 + 4*4) + 8 =$
$3*5^3 + 5^2 + 5(24) + 5 + 3 = $
$3*5^3 + 5^2 + 5(25 - 1) + 5 + 3=$
$3*5^3 + 5^2 + 5^3 +3 =$
$4*5^3 + 5^2 + 3 = 4103_5$.
And $42_5*44_5 = 4103_5$ or in base 10 $22*24 = 528$.
