Difference between revisions of "2021 April MIMC 10"

Latest revision as of 16:45, 19 September 2021

$\textit{This is a mock contest of the actual AMC competition. Feel free to add more solutions beneath the official solution.}$

Problem 1

What is the sum of $2^{3}-(-3^{4})-3^{4}+1$ ?

$\textbf{(A)} ~-155 \qquad\textbf{(B)} ~-153 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~9 \qquad\textbf{(E)} ~171$

Solution

Problem 2

Okestima is reading a $150$ page book. He reads a page every $\frac{2}{3}$ minutes, and he pauses $3$ minutes when he reaches the end of page 90 to take a break. He does not read at all during the break. After, he comes back with food and this slows down his reading speed. He reads one page in $2$ minutes. If he starts to read at $2:30$ , when does he finish the book?

$\textbf{(A)} ~4:33 \qquad\textbf{(B)} ~5:30 \qquad\textbf{(C)} ~5:33 \qquad\textbf{(D)} ~6:30 \qquad\textbf{(E)} ~7:33$

Solution

Problem 3

Find the number of real solutions that satisfy the equation $(x^2+2x+2)^{3x+2}=1$ .

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{(D)} ~3 \qquad\textbf{(E)} ~4$

Solution

Problem 4

Stiskwey wrote all the possible permutations of the letters $AABBCCCD$ ( $AABBCCCD$ is different from $AABBCCDC$ ). How many such permutations are there?

$\textbf{(A)} ~420 \qquad\textbf{(B)} ~630 \qquad\textbf{(C)} ~840 \qquad\textbf{(D)} ~1680 \qquad\textbf{(E)} ~5040$

Solution

Problem 5

Given $x:y=5:3, z:w=3:2, y:z=2:1$ , Find $x:w$ .

$\textbf{(A)} ~3:1 \qquad\textbf{(B)} ~10:3 \qquad\textbf{(C)} ~5:1 \qquad\textbf{(D)} ~20:3 \qquad\textbf{(E)} ~10:1$

Solution

Problem 6

A worker cuts a piece of wire into two pieces. The two pieces, $A$ and $B$ , enclose an equilateral triangle and a square with equal area, respectively. The ratio of the length of $B$ to the length of $A$ can be expressed as $a\sqrt[b]{c}:d$ in the simplest form. Find $a+b+c+d$ .

$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~12 \qquad\textbf{(D)} ~14 \qquad\textbf{(E)} ~15$

Solution

Problem 7

Find the least integer $k$ such that $838_k=238_k+1536$ where $a_k$ denotes $a$ in base- $k$ .

$\textbf{(A)} ~12 \qquad\textbf{(B)} ~13 \qquad\textbf{(C)} ~14 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$

Solution

Problem 8

In the morning, Mr.Gavin always uses his alarm to wake him up. The alarm is special. It always rings in a cycle of ten rings. The first ring lasts $1$ second, and each ring after lasts twice the time than the previous ring. Given that Mr.Gavin has an equal probability of waking up at any time, what is the probability that Mr.Gavin wakes up and end the alarm during the tenth ring?

$\textbf{(A)} ~\frac{511}{1023} \qquad\textbf{(B)} ~\frac{1}{2} \qquad\textbf{(C)} ~\frac{512}{1023} \qquad\textbf{(D)} ~\frac{257}{512} \qquad\textbf{(E)} ~\frac{129}{256}$

Solution

Problem 9

Find the largest number in the choices that divides $11^{11}+13^2+126$ .

$\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~16$

Solution

Problem 10

If $x+\frac{1}{x}=-2$ and $y=\frac{1}{x^{2}}$ , find $\frac{1}{x^{4}}+\frac{1}{y^{4}}$ .

$\textbf{(A)} ~-2 \qquad\textbf{(B)} ~-1 \qquad\textbf{(C)} ~0 \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution

Problem 11

How many factors of $16!$ is a perfect cube or a perfect square?

$\textbf{(A)} ~158 \qquad\textbf{(B)} ~164 \qquad\textbf{(C)} ~180 \qquad\textbf{(D)} ~1280 \qquad\textbf{(E)} ~3000$

Solution

Problem 12

Given that $x^2-\frac{1}{x^2}=2$ , what is $x^{16}-\frac{1}{x^{8}}+x^{8}-\frac{1}{x^{16}}$ ?

$\textbf{(A)} ~1120 \qquad\textbf{(B)} ~1180 \qquad\textbf{(C)} ~3780 \qquad\textbf{(D)} ~840\sqrt{2} \qquad\textbf{(E)} ~1260\sqrt{2}$

Solution

Problem 13

Given that Giant want to put $12$ green identical balls into $3$ different boxes such that each box contains at least two balls, and that no box can contain $7$ or more balls. Find the number of ways that Giant can accomplish this.

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~19$

Solution

Problem 14

James randomly choose an ordered pair $(x,y)$ which both $x$ and $y$ are elements in the set $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$ , $x$ and $y$ are not necessarily distinct, and all of the equations: $x+y$ $x^2+y^2$ $x^4+y^4$ are divisible by $5$ . Find the probability that James can do so.

$\textbf{(A)} ~\frac{1}{25} \qquad\textbf{(B)} ~\frac{2}{45} \qquad\textbf{(C)} ~\frac{11}{225} \qquad\textbf{(D)} ~\frac{4}{75} \qquad\textbf{(E)} ~\frac{13}{225}$

Solution

Problem 15

Paul wrote all positive integers that's less than $2021$ and wrote their base $4$ representation. He randomly choose a number out the list. Paul insist that he want to choose a number that had only $2$ and $3$ as its digits, otherwise he will be depressed and relinquishes to do homework. How many numbers can he choose so that he can finish his homework?

$\textbf{(A)} ~30 \qquad\textbf{(B)} ~62 \qquad\textbf{(C)} ~64 \qquad\textbf{(D)} ~84 \qquad\textbf{(E)} ~126$

Solution

Problem 16

Find the number of permutations of $AAABBC$ such that at exactly two $A$ s are adjacent, and the $B$ s are not adjacent.

$\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25$

Solution

Problem 17

The following expression $\sum_{k=1}^{60} {60 \choose k}+\sum_{k=1}^{59} {59 \choose k}+\sum_{k=1}^{58} {58 \choose k}+\sum_{k=1}^{57} {57 \choose k}+\sum_{k=1}^{56} {56 \choose k}+\sum_{k=1}^{55} {55 \choose k}+\sum_{k=1}^{54} {54 \choose k}+...+\sum_{k=1}^{3} {3 \choose k}-2^{10}$ can be expressed as $x^{y}-z$ which both $x$ and $y$ are relatively prime positive integers. Find $2^{x}(xy+2x+z)$ .

$\textbf{(A)} ~4632 \qquad\textbf{(B)} ~4844 \qquad\textbf{(C)} ~4860\qquad\textbf{(D)} ~4864 \qquad\textbf{(E)} ~8960$

Solution

Problem 18

What can be a description of the set of solutions for this: $x^{2}+y^{2}=|2x+|2y||$ ?

$\textbf{(A)}$ Two overlapping circles with each area $2\pi$ .

$\textbf{(B)}$ Four not overlapping circles with each area $4\pi$ .

$\textbf{(C)}$ There are two overlapping circles on the right of the $y$ -axis with each area $2\pi$ and the intersection area of two overlapping circles on the left of the $y$ -axis with each area $2\pi$ .

$\textbf{(D)}$ Four overlapping circles with each area $4\pi$ .

$\textbf{(E)}$ There are two overlapping circles on the right of the $y$ -axis with each area $4\pi$ and the intersection area of two overlapping circles on the left of the $y$ -axis with each area $4\pi$ .

Solution

Problem 19

$(0.51515151...)_n$ can be expressed as $(\frac{6}{n})$ in base $10$ which $n$ is a positive integer. Find the sum of the digits of $n^{3}$ .

$\textbf{(A)} ~6 \qquad\textbf{(B)} ~7 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~9 \qquad \textbf{(E)} ~\textrm{Does Not Exist}$

Solution

Problem 20

Given that $y=24\cdot 34\cdot 67\cdot 89$ . Given that the product of the even divisors is $a$ , and the product of the odd divisors is $b$ . Find $a \colon b^4$ .

$\textbf{(A)} ~512:1 \qquad\textbf{(B)} ~1024:1 \qquad\textbf{(C)} ~2^{64}:1 \qquad\textbf{(D)} ~2^{80}:1 \qquad\textbf{(E)} ~2^{160}:1 \qquad$

Solution

Problem 21

How many solutions are there for the equation $\left \lfloor{x}\right \rfloor^{2}-\left \lceil{x}\right \rceil=0$ . (Recall that $\left \lfloor{x}\right \rfloor$ is the largest integer less than $x$ , and $\left \lceil{x}\right \rceil$ is the smallest integer larger than $x$ .)

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{(D)} ~3 \qquad\textbf{(E)} ~4 \qquad$

Solution

Problem 22

In the diagram, $ABCD$ is a square with area $6+4\sqrt{2}$ . $AC$ is a diagonal of square $ABCD$ . Square $IGED$ has area $11-6\sqrt{2}$ . Given that point $J$ bisects line segment $HE$ , and $AE$ is a line segment. Extend $EG$ to meet diagonal $AC$ and mark the intersection point $H$ . In addition, $K$ is drawn so that $JK//EC$ . $FH^2$ can be represented as $\frac{a+b\sqrt{c}}{{d}}$ where $a,b,c,d$ are not necessarily distinct integers. Given that $gcd(a,b,d)=1$ , and $c$ does not have a perfect square factor. Find $a+b+c+d$ .

$\textbf{(A)} ~5 \qquad\textbf{(B)} ~15 \qquad\textbf{(C)} ~61 \qquad\textbf{(D)} ~349 \qquad\textbf{(E)} ~2009 \qquad$

Solution

Problem 23

On a coordinate plane, point $O$ denotes the origin which is the center of the diamond shape in the middle of the figure. Point $A$ has coordinate $(-12,12)$ , and point $C$ , $E$ , and $G$ are formed through $90^{\circ}$ , $180^{\circ}$ , and $270^{\circ}$ rotation about the origin $O$ , respectively. Quarter circle $BOH$ (formed by the arc $BH$ and line segments $BO$ and $GH$ ) has area $25\pi$ . Furthermore, another quarter circle $DOF$ formed by arc $DF$ and line segments $OF$ , $OD$ is formed through a reflection of sector $BOH$ across the line $y=x$ . The small diamond centered at $O$ is a square, and the area of the little square is $2$ . Let $x$ denote the area of the shaded region, and $y$ denote the sum of the area of the regions $ABH$ (formed by side $AB$ , arc $BH$ , and side $HA$ ), $DFE$ (formed by side $ED$ , arc $DF$ , and side $FE$ ) and sectors $FGH$ and $BCD$ . Find $\frac{x}{y}$ in the simplest radical form.

$\textbf{(A)} ~\frac{50\pi+1}{280} \qquad\textbf{(B)} ~\frac{50\pi\sqrt{2}+\sqrt{2}}{560} \qquad\textbf{(C)} ~\frac{50\pi+1}{140+100\pi} \qquad\textbf{(D)} ~\frac{50\pi+1}{280+100\pi} \qquad\textbf{(E)} ~\frac{50\pi^2+700\pi\sqrt{2}+3001\pi-70\sqrt{2}+60}{2\pi^2+240\pi+6920}\qquad$

Solution

Problem 24

One semicircle is constructed with diameter $AH=4$ and let the midpoint of $AH$ be $M$ . Construct a point $O$ on the side of segment $AH$ (closer to segment $AH$ than arc $AH$ ) such that the distance from $A$ to $O$ is $2\sqrt{5}$ , and that $OM$ is perpendicular to the diameter $AH$ . Three more such congruent semicircles are formed through multiple $90^{\circ}$ rotations around the point $O$ . Name the $6$ endpoints of the diameters $B$ , $C$ , $D$ , $E$ , $F$ , $G$ in a circular direction from $A$ to $H$ . Another four congruent semicircles are constructed with diameters $AB, CD, EF, GH$ , and that the distance from the diameters to the point $O$ are less than the distance from the arcs to the point $O$ . Connect $AC$ , $CD$ , $DO$ , $OG$ , and $GA$ . Find the ratio of the area of the pentagon $ACDOG$ to the total area of the shape formed by arcs $AB$ , $BC$ , $CD$ , $DE$ , $EF$ , $FG$ , $GH$ , $HA$ .

$\textbf{(A)} ~\frac{14+10\pi}{17} \qquad\textbf{(B)} ~\frac{13+\sqrt{2}}{28} \qquad\textbf{(C)} ~\frac{4+\sqrt{2}}{7+3\pi} \qquad\textbf{(D)} ~\frac{13}{28+6\pi} \qquad\textbf{(E)} ~\frac{13}{30\pi}\qquad$

Solution

Problem 25

Suppose that a researcher hosts an experiment. He tosses an equilateral triangle with area $\sqrt{3}$ $cm^2$ onto a plane that has a strip every $1$ $cm$ horizontally. Find the expected number of intersections of the strips and the sides of the equilateral triangle.

$\textbf{(A)} ~4 \qquad\textbf{(B)} ~\frac{12}{\pi} \qquad\textbf{(C)} ~\frac{2+3\sqrt{3}}{2} \qquad\textbf{(D)} ~\frac{4+\sqrt{3}}{2} \qquad\textbf{(E)} ~\frac{12+4\sqrt{2}-2\sqrt{3}}{4}\qquad$

Solution

Additional Information

1. The Committee on the Michael595 & Interstigation Math Contest (MIMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The MIMC also reserves the right to disqualify score from a test taker if it is determined that the required security procedures were not followed.

2. The publication, reproduction or communication of the problems or solutions of the MIMC 10 will result in disqualification. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules except the private discussion form.

Sincerely, the MIMC mock contest cannot come true without the contributions from the following testsolvers, problem writers and advisors: $\textrm{Interstigation (Problem Writer)}$ $\textrm{Michael595 (Problem Writer)}$ $\textrm{Fidgetboss\_4000 (Testsolver)}$ $\textrm{Skyscraper (Suggester)}$

@@ Line 1: / Line 1: @@
+<math>\textit{This is a mock contest of the actual AMC competition. Feel free to add more solutions beneath the official solution.}</math>
 ==Problem 1==
 What is the sum of <math>2^{3}-(-3^{4})-3^{4}+1</math>?
@@ Line 29: / Line 30: @@
 ==Problem 5==
-. Given <math>x:y=5:3, z:w=3:2, y:z=2:1</math>, Find <math>x:w</math>.
+Given <math>x:y=5:3, z:w=3:2, y:z=2:1</math>, Find <math>x:w</math>.
 <math>\textbf{(A)} ~3:1 \qquad\textbf{(B)} ~10:3 \qquad\textbf{(C)} ~5:1 \qquad\textbf{(D)} ~20:3 \qquad\textbf{(E)} ~10:1</math>
@@ Line 136: / Line 137: @@
 <math>\textbf{(E)}</math> There are two overlapping circles on the right of the <math>y</math>-axis with each area <math>4\pi</math> and the intersection area of two overlapping circles on the left of the <math>y</math>-axis with each area <math>4\pi</math>.
-[[2021 April MIMC Problems/Problem 18 |Solution]]
+[[2021 April MIMC 10 Problems/Problem 18 |Solution]]
 ==Problem 19==
@@ Line 143: / Line 144: @@
 <math>\textbf{(A)} ~6 \qquad\textbf{(B)} ~7 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~9 \qquad  \textbf{(E)} ~\textrm{Does Not Exist}</math>
-[[2021 April MIMC Problems/Problem 19 |Solution]]
+[[2021 April MIMC 10 Problems/Problem 19 |Solution]]
 ==Problem 20==
@@ Line 169: / Line 170: @@
 ==Problem 23==
-On a coordinate plane, point <math>O</math> denotes the origin which is the center of the diamond shape in the middle of the figure. Point <math>A</math> has coordinate <math>(-12,12)</math>, and point <math>C</math>, <math>E</math>, and <math>G</math> are formed through <math>90^{。}</math>, <math>180^{。}</math>, and <math>270^{。}</math> rotation about the origin <math>O</math>, respectively. Quarter circle <math>BOH</math> (formed by the arc <math>BH</math> and line segments <math>BO</math> and <math>GH</math>) has area <math>25\pi</math>. Furthermore, another quarter circle <math>DOF</math> formed by arc <math>DF</math> and line segments <math>OF</math>, <math>OD</math> is formed through a reflection of sector <math>BOH</math> across the line <math>y=x</math>. The small diamond centered at <math>O</math> is a square, and the area of the little square is <math>2</math>. Let <math>x</math> denote the area of the shaded region, and <math>y</math> denote the sum of the area of the regions <math>ABH</math> (formed by side <math>AB</math>, arc <math>BH</math>, and side <math>HA</math>), <math>DFE</math> (formed by side <math>ED</math>, arc <math>DF</math>, and side <math>FE</math>) and sectors <math>FGH</math> and <math>BCD</math>. Find <math>\frac{x}{y}</math> in the simplest radical form.
+On a coordinate plane, point <math>O</math> denotes the origin which is the center of the diamond shape in the middle of the figure. Point <math>A</math> has coordinate <math>(-12,12)</math>, and point <math>C</math>, <math>E</math>, and <math>G</math> are formed through <math>90^{\circ}</math>, <math>180^{\circ}</math>, and <math>270^{\circ}</math> rotation about the origin <math>O</math>, respectively. Quarter circle <math>BOH</math> (formed by the arc <math>BH</math> and line segments <math>BO</math> and <math>GH</math>) has area <math>25\pi</math>. Furthermore, another quarter circle <math>DOF</math> formed by arc <math>DF</math> and line segments <math>OF</math>, <math>OD</math> is formed through a reflection of sector <math>BOH</math> across the line <math>y=x</math>. The small diamond centered at <math>O</math> is a square, and the area of the little square is <math>2</math>. Let <math>x</math> denote the area of the shaded region, and <math>y</math> denote the sum of the area of the regions <math>ABH</math> (formed by side <math>AB</math>, arc <math>BH</math>, and side <math>HA</math>), <math>DFE</math> (formed by side <math>ED</math>, arc <math>DF</math>, and side <math>FE</math>) and sectors <math>FGH</math> and <math>BCD</math>. Find <math>\frac{x}{y}</math> in the simplest radical form.
-[[File:19(1).png]]
+[[File:19 (1).png|500px]]
 <math>\textbf{(A)} ~\frac{50\pi+1}{280} \qquad\textbf{(B)} ~\frac{50\pi\sqrt{2}+\sqrt{2}}{560} \qquad\textbf{(C)} ~\frac{50\pi+1}{140+100\pi} \qquad\textbf{(D)} ~\frac{50\pi+1}{280+100\pi} \qquad\textbf{(E)} ~\frac{50\pi^2+700\pi\sqrt{2}+3001\pi-70\sqrt{2}+60}{2\pi^2+240\pi+6920}\qquad</math>
 [[2021 April MIMC 10 Problems/Problem 23 |Solution]]
+==Problem 24==
+One semicircle is constructed with diameter <math>AH=4</math> and let the midpoint of <math>AH</math> be <math>M</math>. Construct a point <math>O</math> on the side of segment <math>AH</math>  (closer to segment <math>AH</math> than arc <math>AH</math>) such that the distance from <math>A</math> to <math>O</math> is <math>2\sqrt{5}</math>, and that <math>OM</math> is perpendicular to the diameter <math>AH</math>. Three more such congruent semicircles are formed through multiple <math>90^{\circ}</math>rotations around the point <math>O</math>. Name the <math>6</math> endpoints of the diameters <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math> in a circular direction from <math>A</math> to <math>H</math>. Another four congruent semicircles are constructed with diameters <math>AB, CD, EF, GH</math>, and that the distance from the diameters to the point <math>O</math> are less than the distance from the arcs to the point <math>O</math>. Connect <math>AC</math>, <math>CD</math>, <math>DO</math>, <math>OG</math>, and <math>GA</math>. Find the ratio of the area of the pentagon <math>ACDOG</math> to the total area of the shape formed by arcs <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, <math>FG</math>, <math>GH</math>, <math>HA</math>.
+<math>\textbf{(A)} ~\frac{14+10\pi}{17} \qquad\textbf{(B)} ~\frac{13+\sqrt{2}}{28} \qquad\textbf{(C)} ~\frac{4+\sqrt{2}}{7+3\pi} \qquad\textbf{(D)} ~\frac{13}{28+6\pi} \qquad\textbf{(E)} ~\frac{13}{30\pi}\qquad</math>
+[[2021 April MIMC 10 Problems/Problem 24 |Solution]]
+==Problem 25==
+Suppose that a researcher hosts an experiment. He tosses an equilateral triangle with area <math>\sqrt{3}</math> <math>cm^2</math> onto a plane that has a strip every <math>1</math> <math>cm</math> horizontally. Find the expected number of intersections of the strips and the sides of the equilateral triangle.
+[[File:25.png|500px]]
+<math>\textbf{(A)} ~4 \qquad\textbf{(B)} ~\frac{12}{\pi} \qquad\textbf{(C)} ~\frac{2+3\sqrt{3}}{2} \qquad\textbf{(D)} ~\frac{4+\sqrt{3}}{2} \qquad\textbf{(E)} ~\frac{12+4\sqrt{2}-2\sqrt{3}}{4}\qquad</math>
+[[2021 April MIMC 10 Problems/Problem 25 |Solution]]
+==Additional Information==
+. The Committee on the Michael595 & Interstigation Math Contest (MIMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The MIMC also reserves the right to disqualify score from a test taker if it is determined that the required security procedures were not followed.
+. The publication, reproduction or communication of the problems or solutions of the MIMC 10 will result in disqualification. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules except the private discussion form.
+Sincerely, the MIMC mock contest cannot come true without the contributions from the following testsolvers, problem writers and advisors:
+<cmath>\textrm{Interstigation (Problem Writer)}</cmath>
+<cmath>\textrm{Michael595 (Problem Writer)}</cmath>
+<cmath>\textrm{Fidgetboss\_4000 (Testsolver)}</cmath>
+<cmath>\textrm{Skyscraper (Suggester)}</cmath>

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Difference between revisions of "2021 April MIMC 10"

Latest revision as of 16:45, 19 September 2021

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Additional Information