AMC 10A

     Well, last week I took the AMC 10 A. In my school, Magnet High School, there were ~40 students invited to take the AMC 10A, and we did it in the middle of the school day. Luckily, I was allowed to skip Spanish, easily my least favorite class. However, I wouldn't celebrate that fact for much longer.

     When I got to the test room, I discovered a bunch of tables lined up in rows, similar to how it was arranged when I took the AMTNJ, a different math competition from December. On the AMTNJ, I got 10/15, an incredibly disappointing score for someone who can normally score 13-15 on practice tests. As I sat down and took out my pencil pouch, I heard the test proctor say, "No compasses allowed". Instantly, I turned my head. How could they have us take this test without compasses? It literally says in the test booklet that we were allowed to use them! Luckily, this mistake was quickly rectified, and we were able to get started with our tests quickly.

   As I started my test, I raced through #1 and #2 in ~40 seconds, as I knew that they would be straightforward problems with little to no tricks. However, I instantly ran into a roadblock with #3.

A unit of blood expires after $10!=10\cdot 9 \cdot 8 \cdots 1$ seconds. Yasin donates a unit of blood at noon of January 1. On what day does his unit of blood expire?
$\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}$

   This problem seems really simple, and it is. However, my mind was like, "I have a great idea. Let's calculate all of this in a tiny 1"x1" space and then mark our answer!" And I did so, forgetting to multiply $10!$ by 10. Additionally, I did my long division wrong, and got a fraction the first 3 times I attempted the problem. After finally switching to the scrap paper and wasting 3 minutes, I finally managed to get the correct answer, E.
   Then, we have the next problem, #4.

How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

     To solve this, first, I first laid out the schedule block with three classes and two classes separating them, as the bare minimum. And then, I knew that there would still be another class that could go in any of the 6 slots between the classes. Therefore, I put B as my answer. However, there were two problems with my work. First of all, the statement says that the courses the student takes during the other three periods don't matter. Therefore, there weren't 6 possible places, there were only 4! (Not 4 factorial). Next, you do have to consider the order of the 3 classes, meaning that you have to multiply the final answer by 3! (Yes, 3 factorial). Therefore, the answer is 24, or E.

     The next two problems were simple word problems that could very easily be broken down algebraically. At this point, I was already pretty frustrated from #3, which would prove to be my downfall. I hit #7.

For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

     At this point, I saw this problem and was trying to go as fast as possible. Therefore, I completely ignored the "(not necessarily positive)" part of the question and went straight towards prime factorizing 4000 to $2^{5}5^{3}$ . Then, I saw that there were 3 factors of 5, and marked A. I completely forgot that 0 through -5 would also, work, meaning that there were 6 other solutions, making the answer E.

     Problem #8 was another algebraic bash, and problem #9 was a simple problem once realizing that all of the triangles were similar with a visually intuitive scale factor.

All of the triangles in the diagram below are similar to iscoceles triangle $ABC$ , in which $AB=AC$ . Each of the 7 smallest triangles has area 1, and $\triangle ABC$ has area 40. What is the area of trapezoid $DBCE$ ?
$[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]$
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$

Essentially, you have to realize that since the upper isosceles triangle is similar to the smaller ones, the scale factor of the small ones to the large one is 3, since three small bases make up the big base. Then, by squaring the scale factor, the area of the upper triangle is 9. From them, its simple subtraction to get 24, E.

    The next problem is #10, where my contest experience fell apart.

Suppose that real number $x$ satisfies $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ . What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$ ?
$\textbf{(A) }8 \qquad \textbf{(B) }\sqrt{33}+8\qquad \textbf{(C) }9 \qquad \textbf{(D) }2\sqrt{10}+4 \qquad \textbf{(E) }12 \qquad$

   I saw this problem, and my first thought was:

must be a simple number, let's bash it! I spent the next 3 minutes bashing it, before giving up. Then, I tried another approach. I set

, making the equation

. I then tried to bash for

, and got a huge fraction with absolutely no hope of simplifying in time. Then, after skipping the problem and coming back later, I found that if I instead did

, the equation became much simpler, since I didn't have to deal with any subtraction. This gave me a much more friendly value for

, and I solved the problem, after ~10 minutes, to get 8, A. However, obviously, there was a much more elegant solution.

   If you multiply both sides of the equation by its conjugate, AKA, what you are solving by, the left-hand side turns into $(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24$ .
Since this equals 3 times what you are solving for, the answer is 8, A.

     Next is problem #11.

When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as $\frac{n}{6^7},$ where $n$ is a positive integer. What is $n$ ?
$\textbf{(A) } 42 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 84$

   The way I solved this problem was by bashing every possible combination of dice rolls, and then finding out how many ways they could occur. However, I forgot to include a certain case, and I got the incorrect answer, 63, D. That makes 2 wrong, so far, and it isn't going to get any better :(

     Then we have #12:

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? $x+3y=3$ $\big||x|-|y|\big|=1$

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$

     I bashed the problem algebraically and somehow missed a solution. I got only 2 solutions, and the answer was 3. Someone else in my school also algebra bashed it and got 2, making me believe that this problem is somehow resistant to algebra bashing. However, the solution becomes apparent once you graph the system:

$[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy]$

The answer is now very obviously 3, C.

   Problem #13 is a simple problem once you realize that the crease will form similar triangles, allowing you to get the length pretty easily. I won't put the problem here, but you can go look at it at it by going to the link at the very bottom of the page.


What is the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?$

$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$

I think almost everybody in our school got this wrong. The clever solution is to realize that the powers of 2 are completely outclassed by the powers of 3, making this every so slightly smaller than

. This means that the answer is A, 80. Unfortunately, I didn't see this, and left this one blank, as I couldn't figure out how to solve it. The funny thing I realized after the test is that the answer is so close to 81 that Google calculator actually rounds it up to 81. In fact, if you plug it into WolframAlpha, you can see that there are ~15 9s after the 80.

   Next up, problem #15.

Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$ , as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?
$[asy] draw(circle((0,0),13)); draw(circle((5,-6.2),5)); draw(circle((-5,-6.2),5)); label("$B$", (9.5,-9.5), S); label("$A$", (-9.5,-9.5), S); [/asy]$

$\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93$

     This problem was also pretty easy, especially once you drew out the triangles.

$[asy] draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$X$", (0,0), N); label("$Y$", (-5,-6.25),NW); label("$Z$", (5,-6.25),NE); [/asy]$

All of the triangles were similar, and you knew a lot of side lengths, making it pretty easy to get 69, D, as the answer.

   Afterwards, we have the most frustrating problem on the test, #16.

Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$ . Including $\overline{AB}$ and $\overline{BC}$ , how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$ ?
$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }15 \qquad$

   Half of the students I know literally took out their graph paper and their compass and bashed the problem by drawing all of the possible circles. I didn't do this, and I couldn't solve the problem, so I left it blank. However, one of my friends told me that the way you do it is by taking the altitude from B, and then using geometric continuity to find the number of possible solutions. So I didn't solve this one :/

     Next up, we have #17.

Let $S$ be a set of 6 integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S?$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$

     For this problem, I put 2, A, since I misread the problem and thought that we had to take a set of 5 integers :( This was a really trivial #17, as you can just bash all of the answer choices. The answer was 4, C.

     Next, #18.

How many nonnegative integers can be written in the form $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,$ where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$ ?

$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$

     For this, I saw that there would be

possible numbers, and then I thought that half of them would be negative, except for 0. Therefore, the answer should be

, or 3281. However, I thought that that would be too simple for a #18, and I decided to leave it blank. Unfortunately, the answer was actually 3281, meaning I gave up free points.

The last problem I attempted was #19.

A number $m$ is randomly selected from the set $\{11,13,15,17,19\}$ , and a number $n$ is randomly selected from $\{1999,2000,2001,\ldots,2018\}$ . What is the probability that $m^n$ has a units digit of $1$ ?
$\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5}$

I realized that only a few cases actually resulted in an answer that was 1 mod 10. Therefore, I bashed every case, and added them all up to get E,

.

Overall, I answered 16 problems, and got 5 incorrect :( This means that I got a 79.5, which I am really disappointed about. I think my biggest problem was overthinking the actual meaning of the contest, and I put a lot of pressure/stress on myself. Luckily, my school offers us the opportunity to take the AMC 12B, and I took it yesterday. During yesterday's test, I felt a lot better, meaning that I am hopefully improving. Obviously, I can't qualify with my abysmal AMC 10A score, but I think I still have a chance of qualifying through the AMC 12B, since I answered 14 questions there. I learned a lot from this test, and I hope that I can continue to get better.

All problems are owned by the MAA (Mathematical Association of America)'s AMC (American Mathematics Competitions).
All formatting/diagrams were borrowed from the Art of Problem Solving wiki:
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems

This work is licensed under a Creative Commons Attribution 4.0 International License.

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