If 2C8$\cdot$3C1$=$90C58, what is C? I recently came across this question:

If C is a digit such that the product of the three-digit numbers $2C8$ and $3C1$ is the five-digit number $90C58$, what is the value of $C$?


I start by multiplying out the two three digit numbers and simplifying:
$$2C8\cdot 3C1=60000+5000C+2000+100C^2+600+90C+8=90000+100C+50+8$$
$$\implies6000+500C+200+10C^2+60+9C=9000+10C+5$$
$$\implies10C^2+499C-2745=0$$
$$C=5$$
This quadratic was a pain of solving, due to the extremely large numbers. Is there any way I can avoid getting a quadratic but still arrive at my answer?
Thanks!
Max0815
 A: We may check each possible value of $C$, brute force, quick answer.
Well, it is simpler maybe to check the relation modulo nine, i.e. find possible values for $C$ so that
$$
(2+C+8)(3+C+1) = (9+0+C+5+8)\text{ modulo nine.}
$$
We get the simpler equation $(C+1)(C+4)=(C+4)$ modulo nine, so $C(C+4)=0$ modulo nine. If one factor is divisible by $3$, the other is not. So we have only two cases, $C=0,9$ or $C=5$, so that the one or the other factor is divisible by $9$. We check the $5$ first, 
sage: 258*351
90558

and of course,  $298\cdot 391=300\cdot 391-2\cdot 391> 117300-1000$ has too many digits. Later edit: The $0$ is ruled out because (working modulo $100$, we get $1\times 8=8$) the last two digits of $208\cdot 301$ are $08$, not $58$.
A: Here is an easy way to finish the problem once you got the quadratic:
$$499 C = 2745-  10C^2$$
Now since the RHS is a multiple of $5$ so is the LHS, and hence $5|C$. 
Since $C$ is a digit, $C=0$ or $C=5$, but $0$ is not a solution to your equation.
All you have to do is check if $5$ is a solution.
P.S. The solution by dan_fulea becomes much shorter if you look $\pmod{11}$ instead of $\pmod{9}$:
$$(2-C+8)(3-C+1) = (9-0+C-5+8) \pmod{11} \\ (-1-C)(4-C)=(1+C) \pmod{11}\\(1+C)(5-C) = 0 \pmod{11}$$
This means that 
$$C =10 \pmod{11} \mbox{ or } C=5 \pmod{11}$$ and since $C$ is a digit....
A: Once you've built that quadratic, you can avoid factoring it by taking it modulo 10, since you know the answer you want is a single digit integer.
$$10C^2+499C−2745\equiv5-C\equiv0\pmod {10}$$
From which C=5 quickly.
A: Just render the tens digit of the product.
Since $1×8<10$, there is no carry from the product of units digits into the tens place.  Then the tens digit of the product must satisfy
$8C+1C=9C\equiv 5\bmod 10$
where the $5$ comes from the given tens digit on the right side.  The only digit that could possibly work for $C$ is then $5$.
A: If you keep it simple, I realized that 
2C8
 x 3C1=
90C58 where C^2= has a 5 in the ones place meaning that it has to be 5 because it is the only number that squares to make a 5 in the one's place.
