# 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.

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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.

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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 $$725278\equiv 78 \text{ and }67066\equiv 66 \pmod{100}.$$

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I don't quite understand why, could you elaborate further? – MathScratch Dec 5 '12 at 23:44

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 $y = 8$.

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thank you! This is perfect. – MathScratch Dec 5 '12 at 23:49

The last two digits of $725278\times 67066$ are the last two digits of $78\times 66=5148$. Hence $x+y=4+8=12$.

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66*78=5148

Therefore x=4 and y=8

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Why can only looking at 66*78 work to solve this problem? – MathScratch Dec 5 '12 at 23:43