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A company is selling car batteries. Type A battery sells for $\$1200$ while type B sells for $\$700$ dollars. The company collects $\$32100$ for that month, but there was a critical error in their database and they could not see how many of each battery they sold. They come to you as a statistician to find out how can they find out how many of each battery type they sold in order to replenish their stock properly.

They want you to find out all the possible ways that the batteries could be bought. How would you go about completing your task if you are to present your solution to them in two different ways? They have specified that you are to use modular arithmetic in one solution since it provides an advantage in accuracy since it makes exponent problems easier.

What is the other method? I already know the modular arithmetic method? Any suggestions? Thanks

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    Odd job for a statistician.2012-11-16

5 Answers 5

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12x + 7y = 321 is equivalent to 12x = 321(mod 7) and x = 4(mod 7) or x = 4 + 7k.

Substituting this x into the original equation, y = 39 - 12k.

The complete solution is x = 4 + 7k and y = 39 - 12k.

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$\begin{eqnarray}\rm{\bf Hint}\quad 321\: &=& \ \ \:300 &+&\ \:21\\ &=&\: 12\cdot 25 &+&\: 7\cdot 3\\ &=&\: 12\cdot (25\!-\!7i) &+&\: 7\cdot(3+12i)\\ \end{eqnarray}$

Remark $\ $ See Wikipedia's Bezout Identity page for the theory of linear diophantine equations.

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The general solution to $12x+7y=321$ in integers is $x=963-7t$ and $y=-1605+12t$. The only $t$ producing nonnegative $x,y$ are $134 \le t \le 137$.

This gives the possible pairs $(x,y)$ as $(125,3),(18,15),(11,27),(4,39).$

Other $t$ don't work, e.g. $t=133$ gives $(32,-9)$, and $t=138$ gives $(-3,51)$.

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Look at the equation $12x+7y=321$. It is fairly easy to see that we can have $y=3$ and $x=300/12=25$.

Now trade in expensive batteries for cheap ones. For every $7$ expensive batteries, we can get $12$ cheap ones. So we can also have $x=18$, $y=15$ or $x=11$, $y=27$, or $x=4$, $y=39$, a total of $4$ possibilities.

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Hint: Take your equation $\pmod {700}$ and $\pmod {1200}$ and use the Chinese remainder theorem

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    Thank you very much, that is perfectly true. Thank you Ross Millikan.2012-11-16