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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
3 years, 3 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. · 3 years, 3 months ago

I feel dumb... Made a dumb arithmetic mistake on #17 [36+9=42] and ended up with 112.5 · 3 years, 3 months ago

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. · 3 years, 3 months ago

lol same here :P · 3 years, 3 months ago

On the AMC 10, I did my modular arithmetic wrong and got #2 wrong... · 3 years, 3 months ago

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 :) · 3 years, 3 months ago

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 · 3 years, 3 months ago

read my profile -- "I'm 17, not 23!" · 3 years, 3 months ago

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. $$\blacksquare$$ · 3 years, 3 months ago

The AIME cutoff's going to have to be 120 this year. · 3 years, 3 months ago

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) · 3 years, 3 months ago