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.
Sum of some place digits in a product
5 Answers
66*78=5148
Therefore x=4 and y=8
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0Why can only looking at 66*78 work to solve this problem? β 2012-12-05
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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2I don't quite understand why, could you elaborate further? β 2012-12-05
Remember how you learned to multiply by hand?
725278 67066 ------ .....68 ......8 ....... ....... ----------- .........48
This may help you see why, as others have pointed out, itβs enough to look at the last two digits of the two numbers.
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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1th$a$nk $y$ou! This is perfect. β 2012-12-05
The last two digits of $725278\times 67066$ are the last two digits of $78\times 66=5148$. Hence $x+y=4+8=12$.