Recently, I have just registered for the American Mathematics Competition, or AMC for short. The AMC is split into two types- AMC 12, which requires pre-calculus and trig, and AMC 10, which tests regular 9th grade and 10th grade math. Yet, this test is anything but regular. In fact very challenging. (I registered for AMC 10).

I was just practicing this test, and I got really frustrated. This test is 25 multiple-choice under 75 minutes. You think that’s enough time? It’s anything but that; it’s definitely time-consuming. The beginning problems are plain simple, but as the test progresses more and more, so does the more and more difficult the tests become. For example, the first problem took me just 1 minute. Later, another problem took me 15 minutes. Anyway, here’s one problem you might want to try out. This is perhaps a medium-level problem:

Let a and b be relatively prime integers with a>b>zero and (a³-b³)/(a-b)³ = 73/3. Find a-b.

Choices: (a) 1 (b)2 (c)3 (d)4 (e)5

So, try out this problem by yourself. Once you’re done, here’s the solution:

Factor a^3 – b^3.

a^3 – b^3 = (a-b)(a^2 + ab + b^2)

……..

Thus, (a^3 – b^3)/((a-b)^3) = (a^2 + ab + b^2)/[(a-b)^2]

……..

Cross-multiply.

……..

73(a-b)^2 = 3(a^2 + ab + b^2)

73(a^2 – 2ab + b^2) = 3a^2 + 3ab + 3b^2

73a^2 – 146ab + 73b^2 = 3a^2 + 3ab + 3b^2

………

0 = 70a^2 – 149ab + 70b^2

0 = 70a^2 – 100ab – 49ab + 70b^2

0 = 10a(7a – 10b) – 7b(7a – 10b)

0 = (10a – 7b)(7a – 10b)

………

a = (7/10)*b OR a = (10/7)*b

………

Since a > b, a cannot be (7/10)*b.

………

Therefore, a = 10b/7, or a/b = 10/7.

Since a and b are relatively prime, a = 10, and b = 7. Thus, (a – b) = (10 – 7) = 3. Thus (c).

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