2016 AMC 10A Narration

by shiningsunnyday, Feb 5, 2016, 12:40 PM

Chapter 2:

Q1:

A bit hard for a #1. Does this bode badly for the rest of the test? Whatever, factor out the $9!$. Wait, what if I just split this into a difference of two fractions? Whatever, I'm already behind, *circles $100$* GO GO GO!

Q2:

Gotta be a bit careful on this guy. The left hand side has $5x$ on top, the right has a $3*5$. The answer is $3$. Circle, go.

Q3:

*Reads question, a bit confused* Wait a sec, I've seen a problem of a similar vein. Does this mean that Ben has to pay a whole dollar before David gets a $0.25$ decrease? Or is this simply the ratio between the two? I'll assume the latter... Will come back to this if I have time. *Solves linear equation* C.

Q4:

What the heck? Gotta be really careful with my computation. Can't mess up this negative sign. Wait, floor of something that's slightly bigger than negative one is negative one right? Ahhh I used to get confused about this. Nice one, MAA. Can't trick me though! Wait wait, should I recheck my computation? (Debates for a sec) No time! *Circles B and moves on*

Q5:

Alright step one. Diagram. Wait a sec. What am I doing? This is just a trivial NT question. Dang, still no geometry questions yet? $x*3x*4x=12x^3$. Wait, how should I check which one's right. Oh right, I'll divide each answer choice by $12$. E doesn't work. Wait, D does. *Circles and moves on*

Q6:

*Stares for 20 secs* Oh no please don't tell me I got stuck on a question 6. Should I list everything out? No, can't afford to be so inefficient like last year. I'll consider the units digit and tens digit separately. The tens digit gets a difference of a hundred, the units digit? (Counts 02, 12, 22). Three differences. 103. Keep going!

Proctor: JUST A REMINDER. THOSE WHO'RE TAKING THE 10 GOT A 10 BOOKLET, NOT A 12 RIGHT? AND THOSE WHO'RE TAKING THE 12 GOT A 12 BOOKLET RIGHT? NOT THE 10, RIGHT? SO THE 10 AND THE 12S ARE NOT MIXED UP RIGHT?

Me: *Oh my god can you just stop your run on sentence? Stop distracting me!

Q7:

Dang I saw a similar question in the 2000 AMC or something like that. Do I have to consider a bunch of cases? The list has distinct numbers so $x$ has to be one of these numbers. The median can be two numbers. How about the mean? Ah... almost tricked me *sets up equation, solves. D* Go on.

Q8:

Trickster rabbit, foolish fox? I hate how MAA likes screwing up people by making up such tongue-twisting names. This sounds like one of those exponential models. Wait never mind, the fox changes from $x$ to $2x-40$ (realizes I haven't been dividing my work into boxes). Ok time to start dividing my work up. But oh wait, let me finish this guy first. *Does a triple nested parentheses and solves it* Ok, I'll do a quick computation check. Good. Next

Q9:

Oh my god, oh my god. I remember this user in the olympiads forum with a picture of Howard Wolowitz or something write that $2016$ is a triangular number. It was $60$something. Crap, why didn't I memorize this beforehand? $n(n+1)=4032$. So $n$ is like twenty-something? Wait, no, it's the thousands. Should be sixty-something? *Tries a bunch of numbers. $63$, (sure it's not 69?) go on.

Q10:

Ok. I've screwed up on these questions before. Wait, width, length, what's the difference? First write the area of the three rectangles. Subtract the second from the first and the third from the second. Shouldn't it be a quadratic since we're solving two equations? Wait, never mind, the difference is $8$ and the answer is $2$.

Q11:

Nothing is clicking. Hm... Shoelace? Am I crazy? Shoelace on a #11? Ok... gotta be smart... gotta be smart. Is this like a kite or something? I can multiply the two diagonals? Oh wait, I can drop two altitudes and make two differences of a trapezoid and a right triangle. *Messy calculations, whatever, move on!*

Q12:

Alright. Easy. Three odd. $\binom{1008}{3}$ and $\binom{2016}{3}$. Ok, they should cancel out. A, next.

Q13:

Why do you have to move Ada? What movie they watching? I watched A Brilliant Young Mind yesterday. Oh great movie. Wait don't get distracted. Why do these guys have such gaudy names? End seat... Oh man. There're two cases. Oh my god, Bea and Ceci keep getting in the way of Dee and Edie. And it's so hard representing these guys in one model! Ugh. Skip.

Q14:

Wait what? So something like $2+2+\cdots+3+3$ or something will work? This is endless! So it involves some combinations? No, this can't be this hard. Does it really just mean a linear diophantine equation? $2x+3j=2016$. How many solutions. Should I just list them? Nah I'll mod the equation base $3$. I'm dumb $x \equiv 0 \mod 3$. Alright. So it can be $0, 3, \cdots, 672$? Wait no $2*672 \neq 2016$ Oh I'm thinking it was $j$. It should be up to $1008$. How many numbers are there? *Adds 3 to each number, divides by 3, C, go go go.

Q15:

This might be hard. *Tries to draw a diagram, realizes hands are frozen.* I hate this test center. Alright, just write some areas. So $2\pi=2\pi*r$. Why isn't $r=1$? (Realizes I was being dumb, feeling lucky they didn't put it as a distraction. Yes answer is A. Next!

Q16:

*Takes out my graph paper, tries to fit it on my tiny desk.* Draws a tiny diagram. Should I restart? Nah. Alright. Gotta be careful. It goes here... then here... how should I get it back to the original? Rotation? No... Wait. Is it symmetrical across the $y=x$ line? Yes! D, Next!

Q17:

Wait, it approaches a value? Does this involve a limit argument? Skip!

Q18:

Oh I remember a similar problem in an old AMC. $12=1+\cdots$ I'm out of ideas. Do I have to list everything? Wait, the sum on all the faces should be... $36$ cause each vertex is shared by $3$ faces. Right? How does this help. Well the face that $8$ is on can't have a $7$ cause otherwise the other faces' sum won't be big enough. Can a $6$ and a $8$ be on the same face though? Crap, this is taking forever. Skip!

Q19:

This looks intimidating.

Q13:

Something has to work. Come on... Ada! Where can you be?? Alright, the problem is that the two switchers keep getting in the way of Bea and Ceci. *Plays with some arrangements for like 3 mins* Ok I think I got it. Ada is at seat 1! Wait no, seat 2! Yea!

Q20:

Ok. I'm actually good at counting. I got this... wait no. Ugh the $1$ term is so annoying. Don't waste time, skip!

Q21:

Should I go back and do the ones I didn't do? This is geo. Nah.

Q17:

Ok, well where can the red ball be? It has to be in front of a pack of 3n green balls (where N=5n) or in the back. So there're 4n+2 total spots. Ok, don't screw up solving the inequality. Wait, I'm getting a negative answer. Stupid me, It's greater than 479. 480? Yup, go go go!

Q18:

I think I'm going to put a guess for now. D has a lot of factors. I'll guess this. Wait, C looks promising cause $3!=6$ and it's a factor of the two greater choices. Nah, D. Next

Q19:

Gotta face my fear. What can I do? Oh I remember a MC mini on this. Richard, help me! Ugh. Coordinates. (Hands are frozen by now). Proctor murders something unintelligible. Someone yells outside in the hall. Michael. Trust your bashing skills. Don't get lost in the computation. *Gets lost immediately.* Ok. Write the equations of the lines. Omg, which is which again. I'll have to draw a diagram. *Takes out ruler for the first time.* MOVE, HAND, MOVE! *Manages to draw a rectangle* *Does some computation, getting an answer with a radical.* I hate myself.

Q20:

Wait, if a term involves each of the four variables the remaining $N-4$ parentheses can be anything. In fact it's the sum of $5$ nonnegative numbers. So... That's $\binom{N}{4}$. How am I ever going to calculate this? Well let me do some approximating. Time is running out! Stop moving, clock! Ok. So it's approx the fourth root of this huge number. Oh wait, prime factorize this baby. $1001$ is divisible by $7$ and $11$ and $13$ (thanks 2002 AMC). Ugh. Four consecutive numbers. $77$? Too big. What's the fat $13$ doing there. $11*12*13*14$. Works! Yes! Answer is B.

Q21:

*Shakes my hands like it's the end of the world, managing to finally spin my paper to draw some circles after a while. Wait the triangle is degenerate. That's impossible. Where's the triangle going to be? *After a while, realizes* Ohhhh it's a skinny obtuse triangle. How in the world can I compute this? Wait I can subtract some trapezoids and stuff. *Gets area of $\sqrt{2}-\sqrt{6}$. Where did I screw up? This is negative! *Sees grimacing snare from the proctor.* Ok, D, move on.

Q19: *Redos bash*. Ok, so the root fives will cancel out. Pray to god I didn't mess up. Yes. An answer of $20$. *Circles*

Q22: Divisor problem? Should've did this earlier! Wait, Q25 is so easy! Just realized. Ok, no time. *Factorizes $110$*. Ugh. $n$ could include a $2$, $5$ and/or $11$. Messy casework... Ok, trial and error. $110=2*5*11$ factors? The exponents of the factors must be $1, 4$ and $10$. Perhaps throw in an $11^3$? Yes, that gives us the $10$ exponent. Wait, we can't include a $5$. I'll try a $2$ and put it in $n$. Wait... This works! $325$ (multiplies carefully), Next!

Q24: This has got to be Ptlomey's. But I can't find a way to apply it! Wait, is this guy a square? Oh, how about some 45 degree angles cause of the root 2? Ugh, not working. The diagonals seem to be leading no where. Oh, wait, this is trig bash!

Proctor: Get ready to turn in your test! The test is over!

Then I made two mistakes I'll regret.

$\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$... I don't think this is going to work. *Gives up*

Ugh. This isn't enough for a 132! All the work I poured in during the entire year, all in vain? I might've made some sillies too! Everything I did. I. NEED. A. BETTER. SCORE. *Throws a tantrum*

Proctor: HAND IT IN!

Q24: *Guesses B for some reason*

Q18: *Begins to change my guess to D, then somehow thinks this is one of those MAA trick questions. Aha! The answer must be $1$. All the complicated casework boils down to one possibility, I think. A!*

*Begins to feel really sad*

The test is over.
This post has been edited 5 times. Last edited by shiningsunnyday, Feb 5, 2016, 12:46 PM

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2 Comments

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This is quite amusing. :-D

XD
This post has been edited 1 time. Last edited by shiningsunnyday, Feb 8, 2016, 1:45 AM

by 15Pandabears, Feb 7, 2016, 3:03 PM

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You have the same name as my little brother :O

That's so cool :)
This post has been edited 1 time. Last edited by shiningsunnyday, Feb 9, 2016, 1:27 AM

by Sun13, Feb 9, 2016, 12:22 AM

The ones who are crazy enough to think they can change the world are the ones who do.

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  • Yea, doing a bit of cleaning before migrating to new site

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    I ask because I was purging my thunderbird inbox and I found emails indicating new blog posts of yours.

    email do not lie

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  • @below sorry not accepting contribs

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    first shout of december

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