Online Course Discussion Forum
Math Challenge II-A (AMC 10) Spring 2017 (Number Theory): Solution to 8.7 & more
Hi, this is Lawrence.
For the in-class problem of 8.7, I don't quite get the part after it says "completing the rectangle". I'm asking this because of problem 8.16.
May you walk me through the solution so that I can understand the part where it factors -9?
Also, for problem 8.23, couldn't I just say that M = 60 because of the triple (3, 4, 5)?
For problem 8.28, do I have to use strategies from class or can I just do it guess-and-check style?
Finally, for problem 8.24, do I have to express the values of C in terms of A and B?
Thanks,
Lawrence
I'll do the homework questions first:
For 8.23: You should also at least mention why M=60 works, but then your mention of (3,4,5) shows that M can't be any larger.
For 8.24: Yes, you should give some sort of "formula" for the values of C that work. That is, given A, B, explain how to calculate/describe the values of C that work.
For 8.28: The methods from class may help, but you can use any method to solve them.
For 8.7: Just in case anyone else reads this I'll give a little more explanation of the whole solution. Clearing denominators we get $$3y-3x=xy \mbox{ or } xy + 3x - 3y = 0.$$ Subtracting $9$ on both sides we have $$xy+3x-3y - 9 = 0 \mbox{ which factors as } (x-3)(y+3) = -9.$$ This type of strategy to factor is what is sometimes called "completing the rectangle" (or sometimes SFFT: Simon's Favorite Factoring Trick). We want $x,y$ integers, so also $x-3, y+3$ are integers. Hence we must have integer times integer = -9, which gives us the six cases of (i) $1\times -9 = -9$, (ii) $-1\times 9 = -9$, (iii) $9\times -1 = -9$, etc. Each of these cases then gives us a pair (x,y) that is possibly a solution. For example, in case (i), $$x-3 = 1, y+3 = -9 \Rightarrow x = 4, y = -12.$$
Hope this helps!
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