<p>I got the email…and I got in!!! I’m so happy =) Thanks for your help.</p>
<p>My stats: I took the AMC 12 and qualified for AIME. I’m a sophomore and I’m in Capstone Category Theory. I solved all of the entrance problems. I took 6 ap’s last year, 5 this year with a capstone, I got all 5’s and I have a 4.0 GPA. I take 8 extracurriculars (AMC is my only math extracurricular though). I got a 234 on the PSAT, haven’t taken the SAT. Hope this helped in some way.</p>
<p>Congrats!</p>
<p>Now I see why I didn’t get accepted…</p>
<p>soyoung, did you make USAMO? Just curious at what level SUMaC is looking for.</p>
<p>asdf1253 - Thanks! And you were wait-listed right? So you still have a chance =)
NevadaKid- Yeah I did. Did you get your results yet?</p>
<p>Not yet, IDK why though D:</p>
<p>I didn’t make USAMO though, and I think SUMaC people probably check the USAMO lists. </p>
<p>Anyways, congrats soyoung! I’ll post again when I get notified I guess… wish me luck XD</p>
<p>Thanks and, of course, I wish you the best of luck and hope to see you at SUMaC! And not making USAMO probably won’t affect your acceptance so don’t worry about it. Let me know what happens!</p>
<p>Congrats… </p>
<p>BTW, what is Capstone Category Theory?</p>
<p>I’m not soyoung, but this link explains what a capstone course is:
[Capstone</a> Courses](<a href=“http://users.etown.edu/m/moorerc/capstone.html]Capstone”>http://users.etown.edu/m/moorerc/capstone.html)</p>
<p>So I assume it’s some kind of in-depth study of category theory.</p>
<p>I finished AP Calculus BC in my freshmen year, so Capstone is basically a post-AP class. It’s an in-depth look at Category Theory, which is a college level math course. It’s lots of fun!</p>
<p>By the way, how did you guys solve the last part of the last problem on the entrance test?</p>
<p>“What is the smallest number of tickets you need to purchase to guarantee that you win in 5-bit binary lottery?”</p>
<p>I got 7, but my “proof” was a convoluted mess that didn’t make any sense haha.</p>
<p>I got 7 two. I just prove that 6 is not ok and just make a example of 7. My example is(10000 01100 00011 11010 11001 10111 00111)</p>
<p>Rejected. Good luck to everyone else.</p>
<p>This was my proof. In 5-bit lottery, the maximum number of tickets is 32 and the number of winning tickets is 6 (the exact winning ticket, and the 5 other tickets with a variation in 1 digit each). Then we know that there are 26 non-winning tickets. So, if you bought all of the non-winning tickets (26) then you only need to buy one more to guarantee that you win. Therefore, the minimum number of tickets you need to buy to guarantee you win is 27.</p>
<p>That’s what I thought, but I don’t know if I’m right. How did you guys get 7?</p>
<p>I interpreted the problem to mean that we are allowed to choose the tickets that we buy. Here’s what I did:</p>
<p>For every possible winning ticket, list out all its variations and the winning ticket itself. Then buy the tickets so that you have at least one ticket from each group that you listed. Show that you can’t do this with six tickets, but that you can with seven (give an example).</p>
<p>The problem with my solution though was that my explanation as to why you can’t have a group of six was not logical… lol</p>
<p>I did what soyoungchung did; I thought it was a trivial pidgenhole principle problem. However, I see your approach, asdf1253. In retrospect, that was almost definitely the right interpretation, since my interpretation took literally 2 sentences to explain. How did you guys do the one about integer roots? That was the only one I didn’t (couldn’t) do.</p>
<p>The x^2+4xy+y^2=1 problem?</p>
<p>I solved for x using the quadratic formula and got -2y +/- sqrt(3y^2+1). Then I guess and checked some solutions and got (-4,1) and (-15,4). I had nothing else to try (I had been working on the problem for a long time) so I guessed that another solution might be (x,15) because in the first two solutions I found, there were 4s in each one (x = -4 for the first one and y = 4 for the second). So by extension, the next solution should have y = 15.</p>
<p>Surprisingly, plugging in y = 15 produced an integer value for x. Then I proved by induction that for a solution (x<em>n, y</em>n), there’s also a solution (x<em>n+1, y</em>n+1) = (x<em>n+1, -x</em>n).</p>
<p>Just to be on the safe side, I showed that y<em>n (could’ve been x</em>n; doesn’t matter) isn’t a periodic sequence (so that there are indeed an infinite number of solutions).</p>
<p>Hopefully someone can post a more elegant solution?</p>
<p>I agree with the solution to #8 d of systematically checking every case for the set of 6 tickets, and showing that they cannot yield a win for any chosen ticket. Then, you can find easily find one that does work for 7 tickets.</p>
<p>On 4, you notice that a^2+4ab+b^2=(a+2b)^2-3b^2. Let c=a+2b. This forms a bijective (1 to 1) mapping, so we can look at this as c^2-3b^2 (a pell-like equation). Take two of these: c^2-3b^2 and f^2-3d^2. When we multiply these (and simplify), we get (cf+3bd)^2-3(cd+bf)^2, which is of the form we want. (btw I will use a<em>b to mean a subscript b) After we find c</em>1=2, b<em>1=1 as a solution to c^2-3b^2, we can generate infinitely many solutions, where c</em>n=c<em>(n-1)c</em>1+3b<em>(n-1)b</em>1 and b<em>n=c</em>(n-1)b<em>1+c</em>1b_(n-1). Furthermore, we can show that these are strictly increasing as n increases. Since c=a+2b is bijective, this implies infinitely many distinct (a,b) that satisfy the equation.</p>
<p>It should say c^2-3b^2=1: “After we find c<em>1=2, b</em>1=1 as a solution to c^2-3b^2=1…”</p>
<p>I proved the 6is not ok in my admission paper. But I did not find the example for 7. Then I made a addition to the office with the example for 7 almost half a month later. I am waiting listed. A little disappointed. How about you gays solve the last question of the 7th question?</p>