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AMC10/12A

Who took the AMC10/12??? Did you meet your expectations?

Scores are not yet out, so please tell me if they are out :)

Thank You

Note by Bob Yang
2 years, 11 months ago

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I got 130.5 on AMC 10. Made a stupid mistake on number 4 (multiplied by 2). Finished the first 22, missed one. I heard from my friend that #23 was easy, but otherwise, I think I did well. The AMC answers are all on AoPS, so you can go on there to check. Well, I'm taking the AMC 12B next week. Hopefully, I can get at least a 110. Alan Hu · 2 years, 11 months ago

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I feel dumb... Made a dumb arithmetic mistake on #17 [36+9=42] and ended up with 112.5 Jerry Lee · 2 years, 11 months ago

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I got a 132, but I made a stupid mistake on number 4. It was really easy this time; previously, I had scored 1.5 less than the cutoff almost every time. Trevor B. · 2 years, 11 months ago

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@Trevor B. lol same here :P Bob Yang · 2 years, 11 months ago

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@Trevor B. On the AMC 10, I did my modular arithmetic wrong and got #2 wrong... Tristan Shin · 2 years, 11 months ago

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AMC12....expecting a 118.5, enough to make the AIME. During the test, I only focused on the first 20 questions(as I mentioned in my strategy), I skipped number 8 on due to the confusing wording of the problem, later I found out all of other participants in my school was able to do that problem. Next there was number 18, which twisted up my brain......

Other than these, I feel satisfied since I didn't leave any counting and geometry problems blank :) Xuming Liang · 2 years, 11 months ago

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i tooko the 10A, and i got a 114. Not as good as I wished it could be, but I am overall satisfied

@Michael tong- you're 23, how you take the amc12 Jeremy Lu · 2 years, 11 months ago

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@Jeremy Lu read my profile -- "I'm 17, not 23!" Michael Tong · 2 years, 11 months ago

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I'm mad at myself for making FOUR stupid mistakes:

  1. Forgot to add the original 35 miles (#11)

  2. For the palindrome question, I added all \(abc\) rather than \(abcba\)

  3. For the arith/geo sequence question, forgot \(a<b<c\)

  4. For question 24 with the absolute values (which was stupidly easy and didn't deserve to be question 24) I thought \(f_0 (x)\) was \(f_1 (x)\)!

But it's okay. I'm taking the 12B next week too. (I got a predicted 102 on this one, sadly)

Favorite problems: Question 21 and 23.

Question 21 because it's one of those "looks hard but isn't" questions

Question 23 because it's a great question.

Solution 23: Note that \(\frac{1}{99} = .\overline{01} = (\frac{1}{100} + \frac{1}{10000} + \dots\), so \(\frac{1}{99^2} = (\frac{1}{100} + \frac{1}{10000} + \frac{1}{1000000} + \dots) (\frac{1}{100} + \frac{1}{10000} + \frac{1}{1000000} + \dots)\).

Multiplying the first term, \(\frac{1}{100}\), by the whole of \( (\frac{1}{100} + \frac{1}{10000} + \frac{1}{1000000} + \dots)\). ) yields the repeating decimal \(0.00\overline{01}\)

Multiplying the second term yields \(0.0000\overline{01}\), and so forth. Thus, the decimal looks like so:

\(0.000010203040506070809101112131415\dots969799\) -- note that \(98 + 99 + 100\) at the end turns into \(990000\).

Adding it up, the sum of the first \(9\) natural numbers occurs \(10\) times, so that's \(450\). Then the digit \(1-9\) appears as the "tens place" \(10\) times each, so that's another \(450\). Subtract \(8 + 9 = 17\) since \(98\) isn't counted to get \(883\) as the answer. \(\blacksquare\) Michael Tong · 2 years, 11 months ago

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The AIME cutoff's going to have to be 120 this year. Tristan Shin · 2 years, 11 months ago

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You can find the answer key online with a google search. I exceeded my expectations! The test had beautiful problems, like #25, (which I skipped) Daniel Wang · 2 years, 11 months ago

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@Daniel Wang Oh Thanks :) Bob Yang · 2 years, 11 months ago

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