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Math Challenge II-A (AMC 10) Spring 2017 (Number Theory): Solution to 8.7 & more

 
 
ZhangLawrence的头像
Math Challenge II-A (AMC 10) Spring 2017 (Number Theory): Solution to 8.7 & more
ZhangLawrence - 2017年04月30日 Sunday 21:10
 

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

 
LensmireJohn的头像
Re: Math Challenge II-A (AMC 10) Spring 2017 (Number Theory): Solution to 8.7 & more
LensmireJohn - 2017年05月1日 Monday 09:08
 

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!

ZhangLawrence的头像
Re: Math Challenge II-A (AMC 10) Spring 2017 (Number Theory): Solution to 8.7 & more
ZhangLawrence - 2017年05月1日 Monday 20:36
 

Thanks a lot!