Sum of some place digits in a product If $x$ is the tens place digit and $y$ is the ones place digit of the product $725278\times 67066$, what is $x+y$? I have no idea how to even approach this.
 A: It is enough to use the last $2$ digits:
Write $A=100a+b$ and $C=100c+d$, then $AC=10000ac+100(ad+bc)+bd$, all the rest is dividable by $100$, so would not affect the last $2$ digits.
In other words, using the notation $a\equiv b \pmod{100}$ for giving the same residue mod $100$, i.e. $100|a-b$, we have that $a\equiv b$ and $c\equiv d$ implies $ac\equiv bd$. In the giving example we have
$$ 725278\equiv 78 \text{ and }67066\equiv 66 \pmod{100}.$$
A: Remember how you learned to multiply by hand?
       725278  
        67066  
       ------  
      .....68  
     ......8  
   .......      
  .......       
  -----------  
  .........48

This may help you see why, as others have pointed out, it’s enough to look at the last two digits of the two numbers.
A: So the product is $xy = 48641494348$, so $x = 4$ and $y = 8$.
However, as you see in the other answer, you don't actually have to compute the product.
Note that 
$$
x = 725200 + 78\quad\text{and}\quad y = 67000 + 66
$$
so the product is
$$\begin{align}
(725200 + 78)(67000 + 66) &= 725200\times 67000 + 725200\times 66 + 67000\times 78 + 78\times 68 \\ &= Z + 78\times68.
\end{align}
$$
Here $Z$ is a number that has $0$ in the tens and ones places. So for the digits that you are looking for you need just consider the product
$$
78 \times 68 = 5148.
$$
Again $x = 4$ and $y = 8$.
A: The last two digits of $725278\times 67066$ are the last two digits of $78\times 66=5148$. Hence $x+y=4+8=12$.
A: 66*78=5148
Therefore x=4 and y=8
