<p>after reading diamond’s answer, i redid the problem and realized that i got it wrong.</p>
<p>the answer is 109, not 99.</p>
<p>whatever haha</p>
<p>My AIME ruined my chance to get into USAMO. Oh man i’m so mad at myself.
I could not believe i got a 4 on such a relatively easy AIME test while i got a 127.5 on the AMC 12. Dang</p>
<p>Way to go. Diamon 153!</p>
<p>BTW, seems 10 should be 202 and 11 should be 365.</p>
<p>i don’t know.
my friend got a different answer for both questions
maybe I should write java code to solve #10 right now.</p>
<p>number 10 is 202 I am sure of it. I did a stupid mistake to get 186. you are forgetting the 1 for first a, 0 for second a then numbers and 0 for first a and 1 for second a… lol long explanation but its 100 + 100 + 2 = 202. I was stupid. one of my carless errors.</p>
<p>I already did. And ended up with an answer of 202</p>
<p>i wrote the code, and got 202 !!!</p>
<p>i felt like a genius when I was solving #10.
i separated each a_n. For example, a0 = d1<em>10+d2
a1 = c1</em>10+c2
a2 = b1*10+b2
and a3 is just a3. Where all of these are a digit.
Then, 2010 = 1000(a3+b1)+100(b2+c1)+10(c2+d1)+d2 which is much easier to solve</p>
<p>What are #2 and #5 about again? I forgot already.</p>
<h1>2. Find the remainder when</h1>
<p>9<em>99</em>999* … *9999999999…9999
is divided by 1000</p>
<h1>5. positive integers a,b,c and d such that a+b+c+d = 2010, a^2-b^2+c^2-d^2=2010</h1>
<p>Thanks! And #6? (sorry, I was functioning on 3 hours of sleep when I was taking it >_<")</p>
<p>diamon, are you sure about 12? my friend and i both got 243 and for 80 it seems like it can be partioned into <3,4,5,6,7,8> and <9,10,11,…,80>.</p>
<ol>
<li>Let P(x) be a quadratic polynomial with real coefficients satisfying
x^2 - 2x + 2 <= P(x) <= 2x^2 - 4x + 3</li>
</ol>
<p>It’s a really easy problem once you figure out that both parabolas have vertices at 1,1.</p>
<p>
</p>
<p>That’s weird. I went to sleep at around 11 so I did not check the real-time updating (I blame the weird wordpress clock). It is definitely visible at the current moment though, so:</p>
<p>[28th</a> Annual Invitational Mathematics Examination (AIME I) The Uninteresting Chronicles of a High School Student](<a href=“http://excelexcel.■■■■■■■■■■■■■/2010/03/17/aime2010/]28th”>28th Annual Invitational Mathematics Examination (AIME I) | The Uninteresting Chronicles of a High School Student)</p>
<p>@UMTYMP student, as I said before, some of my answers are wrong</p>
<p>For number 12, I think 81 is the answer. That way, you have 3, 9, and 81, and since a,b and c need not be distinct, 3<em>3=9 and 9</em>9=81, so whether you put the nine with the three or the 81, the partition requires that one of the sets meets the condition.</p>
<p>I retract my earlier statement about 81 for question 12. 3 and 81 could be together and separate from nine. I can confirm 760 for number 7. Each minimally intersecting triple can be divided into equivalence classes each represented by an ordered triple of numbers 1 to 7. Each of the coordinates of the ordered triple represents the one value contained in the intersection of one of the three possible pairs of sets. Since these values cannot be the same, there are 7P3 possibilities, or 210 possibilities for ordered triples. Then, consider the other four numbers of the set. They can each be in exactly one of the three sets or not in any of them. This gives 4^4 combinations for each ordered triple equivalence class. Thus, we have 210*256=53760, which means that the answer is 760.</p>
<p>A guy from my school finished all 15 questions without guessing and still had 45 minutes to check them over… I’ll try to get the answers from him and post them here soon.</p>
<p>^ gg. god.</p>
<p>Well, the answer for Q12 is 243. 242 can be partitioned such that neither of the subsets contain numbers a,b,c such that ab = c as follows.</p>
<p>Set 1: 3,4,5,6,7,8 and 81,82,83,…,242
Set2: 9,10,11,…80</p>
<p>So m > 242. Now, note that 3 and 9 must be in different sets. (otherwise 3^2 = 9). Also, 81 must be with 3 (otherwise 9^2 = 81). Furthermore, 27 now must be with 9 (otherwise 27*3=81). So we have</p>
<p>Set 1: 3, 81
Set 2: 27,9</p>
<p>If we put 243 in set 1, then we have 3<em>81=243. If ewe put 243 in set 2 then 9</em>27 = 243. So we are done (and its not 81; nor is it 760.</p>
<p>I think the first Q answer is 94. there is 81 divider. 16 perfect square, the probility is
16*65/2 / C(81,2) , the sequence is not important</p>