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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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up vote 7 down vote accepted

Remember how you learned to multiply by hand?


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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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

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