### 2022 AMC 10B Exam Problems

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All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

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1.

Define $x~\diamondsuit~ y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of $(1~\diamondsuit~(2~\diamondsuit~3))-((1~\diamondsuit~2)~\diamondsuit~3)?$

$-2$

$-1$

$0$

$1$

$2$

###### Answer: A

###### Solution:

$(1~\diamondsuit~(2~\diamondsuit~3))-((1~\diamondsuit~2)~\diamondsuit~3) =$ $|1-|2-3|| - ||1-2|-3| =$ $|1-1| - |1-3| = 0-2 = -2.$

Thus, the answer is **A**.

2.

In rhombus $ABCD,$ point $P$ lies on segment $\overline{AD}$ so that $\overline{BP} \perp \overline{AD},$ $AP = 3,$ and $PD = 2.$ What is the area of $ABCD?$ (Note: The figure is not drawn to scale.)

$3\sqrt 5$

$10$

$6\sqrt 5$

$20$

$25$

###### Answer: D

###### Solution:

Since we have a rhombus, we know $AB = AD = AP + PD = 5.$ And by the Pythagorean Theorem, $AP^2 + BP^2 = AB^2$We know that $9+ BP^2 = 25.$ $BP = 4.$ The area of a rhombus is $bh,$ so the area is $5\cdot 4 = 20.$

Thus, the answer is **D**.

3.

How many three-digit positive integers have an odd number of even digits?

$150$

$250$

$350$

$450$

$550$

###### Answer: D

###### Solution:

First, we can choose any combination for the first two digits. This would have $9\cdot 10 = 90$ choices.

Then, if there are an odd number of even digits among them, I make the units digit odd, which can be done in $5$ ways. Otherwise, I make the units digit even, which can be done in $5$ ways. Regardless of my choice of the first two digits, I have $5$ ways to choose the units digit.

Therefore, there are $90$ ways to choose the first two digits, and $5$ ways to choose the last two, so the total number of ways is $90\cdot 5 = 450.$

Thus, the answer is **D**.

4.

A donkey suffers an attack of hiccups and the first hiccup happens at $4:00$ one afternoon. Suppose that the donkey hiccups regularly every $5$ seconds. At what time does the donkey’s $700$th hiccup occur?

$15 \text{ seconds after } 4:58$

$20 \text{ seconds after } 4:58$

$25 \text{ seconds after } 4:58$

$30 \text{ seconds after } 4:58$

$35 \text{ seconds after } 4:58$

###### Answer: A

###### Solution:

Since we want to look at the $700$th hiccup, we need to look at time that is $699$ hiccups after the first one.

This would be $699\cdot 5 = 3495$ seconds. Note that $3495 = 60\cdot 58+15,$ so the time would be $58$ minutes and $15$ seconds after the first hiccup. This would therefore be $4:58$ and $15$ seconds.

Thus, the answer is **A**.

5.

What is the value of $\frac{\left(1+\frac{1}{3}\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?$

$\sqrt3$

$2$

$\sqrt{15}$

$4$

$\sqrt{105}$

###### Answer: B

###### Solution:

Lets work with the denominator first. By using difference of two squares on each term, we get the denominator as $\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}$ $=\sqrt{(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})}$ $\cdot\sqrt{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}$ $=\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }.$

Furthermore, we simplify the numerator as follows: $(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})$$= (\frac 43)(\frac 65)(\frac 87) .$

Dividing the numerator and denominator yields $\begin{align*}&\frac{(\frac 43)(\frac 65)(\frac 87)}{\sqrt{ (\frac 23)(\frac 45)(\frac 67) } \sqrt{ (\frac 43)(\frac 65)(\frac 87) }} \\&=\frac{\sqrt{ (\frac 43)(\frac 65)(\frac 87) }} {\sqrt{ (\frac 23)(\frac 45)(\frac 67) }} \\&= \frac{\sqrt 8 } { \sqrt 2} \\&= 2.\end{align*}$

Thus, the answer is **B**.

6.

How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?

$0$

$1$

$2$

$3$

$4$

###### Answer: A

###### Solution:

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the $n$th number is $\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$ $\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$ $= (10^n+1)(\sum_{k=0}^{n} 10^k).$ This shows that the number can be written as the product of two numbers greater than $1,$ so there are no primes.

Thus, the answer is **A**.

7.

For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?

$\ 6$

$\ 8$

$\ 9$

$\ 14$

$\ 16$

###### Answer: B

###### Solution:

Let the roots be $r,s.$ Then: $\begin{align*}&x^2+kx+36 \\ &= (x-r)(x-s) \\&= x^2-(r+s)x+rs \\&=0 \end{align*}$ And so, $rs = 36$ and $r + s = -k.$

Therefore, we need $r$ and $s$ distinct such that $rs = 36.$ All the possible factor pairs are $\pm\{1,36\},\pm\{2,18\},\pm\{3,12\}$ $\text{and }\pm\{4,9\}.$

Each of these unordered pairs produces a unique value for $k,$ so there are $8$ possible values for $k.$

Thus, **B** is the correct answer.

8.

Consider the following $100$ sets of $10$ elements each: $\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}$ How many of these sets contain exactly two multiples of $7?$

$\ 40$

$\ 42$

$\ 43$

$\ 49$

$\ 50$

###### Answer: B

###### Solution:

We can analyze the units digit of the first multiple of $7$ in each set.

If the last digit is $4,5,6,7,$ then adding $7$ yields numbers outside the set, so the sets with a multiple of $7$ such that its units digit is $4,5,6,7$ would have only one multiple.

If the last digit is $1,2,3$ then adding $7$ would yield a units digit of $8,9,0$ in the same set.

Out of the first $98$ sets, there are an equal number of occurrences of sets such that the first multiple of $7$ has each of the units digit of $1,2,3,4,5,6,7$ since they cycle every $7$ sets. These yield $42$ sets whose first multiple of $7$ contains two multiples of $7.$

The $99$th and $100$th set wouldn't work since they yield a set whose first multiples are $987$ and $994$ respectively, which can't have $2$ multiples of $7.$ This means we have $42$ sets that work.

Thus, the answer is **B**.

9.

The sum $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{2021}{2022!}$can be expressed as $a-\dfrac{1}{b!},$ where $a$ and $b$ are positive integers. What is $a+b?$

$\ 2020$

$\ 2021$

$\ 2022$

$\ 2023$

$\ 2024$

###### Answer: D

###### Solution:

We claim $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$$= 1- \dfrac 1{n!}.$

To prove this, we can use induction.

If $n=2,$ then the sum is $\dfrac 1{2!} = 1-\dfrac 1{2!} .$

If it works for $n-1,$ then $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\dots+\dfrac{n-1}{n!}$ $= 1- \dfrac 1{(n-1)!}+\dfrac{n-1}{n!}$$= 1- \dfrac n{n!}+\dfrac{n-1}{n!} = 1 - \dfrac 1{n!}.$

This means our formula is proven, so we can get our answer by plugging in $n=2022.$ Our answer is $1- \dfrac 1{2022!},$ yielding $a=1,b=2022,$ making our answer $2023.$

Thus, our answer is **D**.

10.

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

$\ 5$

$\ 7$

$\ 9$

$\ 11$

$\ 13$

###### Answer: D

###### Solution:

Let the integers in order be $a,b,c,d,e.$

The median of this list is $c.$ Since the mode is greater than the median, both $d$ and $e$ are equal to the mode, so we can write the list as $a,b,c,c+2,c+2.$

The mean is now $\dfrac{a+b+c+c++c+2}5$$= c-2,$ which yields $a+b+3c+4 = 5c-10.$ This means $a+b=2c-14.$ This means $a+b$ is even, and $a,b$ are different positive integers since we have a unique mode. This means $a+b \geq 4,$ since the uniqueness eliminates $a=1,b=1,$ so $a+b \neq 2.$ This means $a+b=4$ yielding a median of $c=9,$ making the mode $11.$

Thus, the answer is **D**.

11.

All the high schools in a large school district are involved in a fundraiser selling T-shirts. Which of the choices below is logically equivalent to the statement "No school bigger than Euclid HS sold more T-shirts than Euclid HS"?

All schools smaller than Euclid HS sold fewer T-shirts than Euclid HS.

No school that sold more T-shirts than Euclid HS is bigger than Euclid HS.

All schools bigger than Euclid HS sold fewer T-shirts than Euclid HS.

All schools that sold fewer T-shirts than Euclid HS are smaller than Euclid HS.

All schools smaller than Euclid HS sold more T-shirts than Euclid HS.

###### Answer: B

###### Solution:

First, we have no information about schools that are smaller than Euclid HS, so we can eliminate all the choices that mention smaller schools. This leaves just **B** and **C** to look at.

Now, given our statement, we know that if a school is bigger than Euclid, then it couldn't have sold more than Euclid. This means if a school sold more, it couldn't have been bigger, which corresponds to choice **B**.

Thus, the answer is **B**.

12.

A pair of fair $6$-sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\dfrac{1}{2}?$

$2$

$3$

$4$

$5$

$6$

###### Answer: C

###### Solution:

To compute this, we can also find the least $n$ such that the probability of not rolling a $7$ is less than $\dfrac 12.$ Each roll has an independent probability of $\dfrac 16$ of getting $7,$ so it has a $\dfrac 56$ probability of not landing on $7.$

Thus, the probability of none of the rolls being $7$ is $\left(\dfrac 56\right)^n.$ We must find the least $n$ such that $\left(\dfrac 56\right)^n < \dfrac 12.$

If $n=3,$ then the probability is $\dfrac{125}{216},$ which is greater than $\dfrac 12.$

If $n=4,$ then the probability is $\dfrac{625}{1296},$ which is less than $\dfrac 12.$ This makes the answer $4.$

Thus, the answer is **C**.

13.

The positive difference between a pair of primes is equal to $2,$ and the positive difference between the cubes of the two primes is $31106.$ What is the sum of the digits of the least prime that is greater than those two primes?

$\ 8$

$\ 10$

$\ 11$

$\ 13$

$\ 16$

###### Answer: E

###### Solution:

Since the primes are $2$ away from each other, we can make them equal to $m-1,m+1,$ where $m$ is their average.

Then, $(m+1)^3-(m-1)^3 = 31106 ,$ making $m^3+3m^2+3m+1$$-(m^3-3m^2+3m-1)$$= 6m^2+2 = 31106.$

Therefore, $m^2= 5184,$ so $m=72.$

The primes are therefore $71,73.$ The least prime greater than both of those is $79,$ and its digit sum is $16.$

Thus, the answer is **E**.

14.

Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?

$\ 12$

$\ 13$

$\ 14$

$\ 15$

$\ 16$

###### Answer: B

###### Solution:

First, note that we can make a set with size $13$ using $S = \{13,14 \cdots ,25\}.$

Now, we prove no arbitrary set of size greater than $13$ work. Let $m$ be the maximum element of $S.$ Then, for all $i$ in $S,$ we know $m-i$ isn't in $S.$

This would eliminate $\lceil{\dfrac{m-1}{2}} \rceil$ of the numbers below $m.$ This means the maximum number of elements below $m$ is $\lfloor \dfrac {m-1}2 \rfloor ,$ making the maximum number of elements $\lfloor \dfrac{m-1}2 \rfloor +1.$

The maximum value of this has $m=25,$ yielding $13.$

Thus, the answer is **B**.

15.

Let $S_n$ be the sum of the first $n$ term of an arithmetic sequence that has a common difference of $2.$ The quotient $\dfrac{S_{3n}}{S_n}$ does not depend on $n.$ What is $S_{20}?$

$340$

$360$

$380$

$400$

$420$

###### Answer: D

###### Solution:

Let the sequence be $a_1, a_2, \cdots.$ Create $a$ by making it the term before $a_1$ in the sequence. This would make $a_n = a+2n.$

This would make $S_n = \sum_{i=1}^n (a+2i) =$ $a\cdot n + 2\sum_{i=1}^ni = an + n^2+n.$

This makes $\dfrac{S_{3n}}{S_n} = \dfrac{3an + 9n^2+3n}{an + n^2+n} =$ $\dfrac{3a + 9n+3}{a + n+1}.$ Thus, we must find $a$ such that this value is constant.

If our given value is constant, than the given value minus $9$ is constant, so $\dfrac{3a + 9n+3}{a + n+1}-9 = \dfrac{-6a-6}{a+n+1}$ is constant. As $n$ increases, the numerator is constant and the denominator is increasing. Therefore, if the number is constant, the numerator must be $0.$

Since $-6a-6 = 0,$ we have $a=-1.$

With our formula from before, we have $S_{20} = -1(20)+20^2+20 = 400.$

Thus, the answer is **D**.

16.

The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5.$ Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

$15\dfrac{1}{8}$

$15\dfrac{3}{8}$

$15\dfrac{1}{2}$

$15\dfrac{5}{8}$

$15\dfrac{7}{8}$

###### Answer: D

###### Solution:

Firstly, let's label the points as follows:

Since we have a rectangle, $AB = 4.$ By the Pythagorean Theorem, we have $AC = 3.$ Then, since $\angle ACB$ and $\angle DCE$ are complementary, $\angle BAC = \angle CDE = 90^\circ,$ and $BC = CE$ we know $ABC \cong CDE.$ Therefore, $ED = 3$ and $EF = 1.$

Since $\angle CED$ and $\angle GEF$ are complementary and $\angle GEF = \angle CDE = 90^\circ,$ we know $EFG$ and $CDE$ are similar. This means $\dfrac{EC}{CD} = \dfrac{EG}{EF},$ so $EG = 1.25.$ Since the shaded region is a trapezoid, we can get the area as $\dfrac{(GE+BC)EC}2 = \dfrac{ 5(5+1.25)}2$$= \dfrac{5\cdot 6.25}{2} = 15.625.$

This is equal to $15 \dfrac 58.$

Thus, the answer is **D**.

17.

One of the following numbers is not divisible by any prime number less than $10.$ Which is it?

$2^{606}-1$

$2^{606}+1$

$2^{607}-1$

$2^{607}+1$

$2^{607}+3^{607}$

###### Answer: C

###### Solution:

Note that $a^n-b^n$ is divisible by $a-b.$

For **A**, let $a = 4,b=1,n=303.$ Then we get that $4^{30}3-1 = 2^{606}-1$ is divisible by $3.$

For **B**, let $a = 4,b=-1,n=303.$ Then we get that $4^{303}-(-1)^{303} = 2^{606}-1$ is divisible by $5.$

For **D**, since $2^{606}-1$ is divisible by $3,$ we have $2^{607} -2$ is divisible by $3,$ leaving $2^{607} +1$ being divisible by $3.$

For **E**, let $a = 3,b=-2,n=607.$ Then we get that $3^{607}-(-2)^{607} = 3^{607}+2^{607}$ is divisible by $5.$

For **C**, we know $2^{607} +1$ is divisible by $3,$ so our value isn't divisible by $3.$ We also know $2^{607} +1$ is divisible by $5$ from choice D so our value isn't divisible by $5.$

Also, $64^{101}-1 = 2^{606} -1$ is divisible by $7,$ so $2^{607} -2$ is divisible by $7.$ This means our value isn't divisible by $7.$ Since it isn't divisible by $2,$ it isn't divisible by a prime under $10.$

Thus, our answer is **C**.

18.

Consider systems of three linear equations with unknowns $x,$ $y,$ and $z,$ $\begin{cases} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{cases}$ where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0.$ For example, one such system is $\begin{cases} 1 x + 1 y + 0 z & = 0 \\ 0 x + 1 y + 1 z & = 0 \\ 0 x + 0 y + 0 z & = 0 \end{cases}$ with a nonzero solution of $(x,y,z) = (1, -1, 1).$ How many such systems of equations are there? (The equations in a system need not be distinct, and two systems containing the same equations in a different order are considered different.)

$\ 302$

$\ 338$

$\ 340$

$\ 343$

$\ 344$

###### Answer: B

###### Solution:

There are $2^9 =512$ total configurations. Now, we can use complementary counting to determine how many have more than one solution.

If a configuration has $3$ equations which don't contain redundant information, then it has only one solution.

This means every equation has to be different. Also, if any equation has $a,b,c =0,$ then it doesn't provide any information, making it redundant. This means we have $7$ choices for the first equation, $6$ choices for the second, and $5$ choices for the third.

This yields $210$ configurations. However, some configurations may still yield redundant information. If two equations add to the other equation, then there is a redundancy.

There are two cases for this to happen.

Case $1:$ $1$ of the equations has $a,b,c=1,$ another equation has $1$ of the variables being $1$ and the other equation has $2$ variables being $1.$ There are $3$ ways to choose which equation has every variable as $1.$ Then, there are $2$ ways to choose which variables have one variable being $1,$ and this equation has $3$ ways to choose which variable is $1.$ This case has $3\cdot 2\cdot 3=18$ configurations to exclude.

Case $2:$ $1$ of the equations has $2$ variables being $1,$ and the other two equations have only one variable being $1,$ with those variables being different from each other, but one of the variables chosen in the first equation. There are $3$ ways to choose the equation with $2$ variables being $1,$ there are $3$ ways to choose which variables are $1,$ and $2$ ways to choose the order of the other equations. This case has $3\cdot 2\cdot 3=18$ configurations to exclude.

There are a total of $210-18-18=174$ cases which have only one solution. This means $512-174=338$ configurations have multiple solutions, making at least one nonzero.

Thus, the answer is **B**.

19.

Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

• Any filled square with two or three filled neighbors remains filled.

• Any empty square with exactly three filled neighbors becomes a filled square.

• All other squares remain empty or become empty. A sample transformation is shown in the figure below.

Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

$\ 14$

$\ 18$

$\ 22$

$\ 26$

$\ 30$

###### Answer: C

###### Solution:

Suppose the center is initially filled. Then, there are either either $2$ or $3$ other filled squares, each of which can't have $2$ or $3$ filled neighbors.

This means that there are at most $4$ filled squares, so each square has at most $3$ neighbors. Since they don't have $2$ or $3$ neighbors, they must have at most $1$ neighbor. The center square is a neighbor, so they can't have any other neighbor.

Suppose I have a filled square on an edge. Since there is some filled square that isn't a neighbor of the square, we can examine the two edges which are neighbors of the filled edge. If I have a filled edge on the corner, the edge on the same side as the corner would have three neighbors. If I choose the opposite edge, the adjacent edges would have three neighbors.

Suppose I choose a corner. Then, I need to choose another corner. If I choose the adjacent corner, then the edge between would have three neighbors, making it filled. Therefore, it must be the adjacent corner. This has $2$ configurations.

Suppose the center is initially empty. Then, there are $3$ filled neighbors of the center, each with at most $1$ neighbors. This means no square has two filled neighbors. This makes it only possible to do in the following ways:

The first three can rotated making $4$ configurations, and the last one can be rotated and reflected making $8$ configurations. There are $20$ configurations with the center being empty. This means there are $20+2=22$ different configurations.

Thus, the answer is **C**.

20.

Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ.$ Let $E$ be the midpoint of $\overline{CD},$ and let $F$ be the point on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}.$ What is the degree measure of $\angle BFC?$

$\ 110$

$\ 111$

$\ 112$

$\ 113$

$\ 114$

###### Answer: D

###### Solution:

First, we extend $BE$ and $AD$ such that they meet at $G.$ Since $\angle GDE = \angle ECB,$ $\angle GED = \angle BEC \text{ and } DE = EC,$ we know $GDE \cong BCE.$ Therefore, $DG =BC = AD.$ This means that if we construct a circle with center $D$ that includes $A,$ $C,G$ are also on it.

Also, since $AFG$ is a right triangle, the drawn circle would be its circumcircle, placing $F$ on the circle.

Since $\angle GDC = 134^\circ,$ we can get $\angle CFE = \angle CFG = \dfrac {\overset{\Large\frown}{CG}} 2$$= \dfrac{134^\circ}{2} = 67^\circ.$ Therefore, $\angle BFC = 180^\circ - 67^\circ = 113^\circ.$

Thus, the answer is **D**.

21.

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1,$ the remainder is $x+2,$ and when $P(x)$ is divided by the polynomial $x^2+1,$ the remainder is $2x+1.$ There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

$\ 10$

$\ 13$

$\ 19$

$\ 20$

$\ 23$

###### Answer: E

###### Solution:

Since $P(x)$ has a remainder of $x+2$ when divided by $x^2+x+1,$ it must be able to be written as $P(x) =$$(x^2+x+1)Q(x)+x+2$ for some polynomial $Q(x).$ Note that the remainder of $P(x)$ when divided by $x^2+1$ is equal to the remainder when $xQ(x) + x+2.$

If $Q(x)=c$ for some constant, then the remainder when $P(x)$ is $(c+1)x+2,$ which can't happen since we need a $1$ for our constant.

If $P(x) = ax+b,$ then the remainder when divided by $x^1$ is equal to the remainder when $(ax+b)x+x+2 =$$ax^2 + (b+1)x+2$ is divided by $x^2+1.$ We can get the remainder by subtracting $a(x^2+1),$ yielding $(b+1)x +(2-a).$ This means $b+1=2$ and $2-a=1,$ so $a=b=1.$

This means $P(x) =$$(x+1)(x^2+x+1)+x+2 =$$x^3+2x^2+3x+3.$ The sum of the squares of the coefficients is $23.$

Thus, the answer is **E**.

22.

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $\begin{align*}x^{2}+y^{2}&=4,\\ x^{2}+y^{2}&=64, \\ (x-5)^{2}+y^{2}&=3.\end{align*}$ What is the sum of the areas of all circles in $S?$

$\ 48 \pi$

$\ 68 \pi$

$\ 96 \pi$

$\ 102 \pi$

$\ 136 \pi$

###### Answer: E

###### Solution:

Let $x^2 + y^2 = 64$ be circle $O,$ $x^2 + y^2 = 5$ be circle $P,$ and $(x - 5)^2 + y^2 = 3$ be circle $Q.$

First note that every circle, $R,$ in $S$ is internally tangent to $O.$ Then we case on the tangency of $R$ with $P$ and $Q.$

Case $1:$ This corresponds to the pink circle. This is where $P$ and $Q$ are internally tangent to $R.$

Case $2:$ This corresponds to the bluish circle. This is where $P$ and $Q$ are externally tangent to $R.$

Case $3:$ This corresponds to the green circle. This is where $P$ is externally and $Q$ is internally tangent to $R.$

Case $4:$ This corresponds to the red circle. This is where $P$ is internally and $Q$ is externally tangent to $R.$

We can consider cases $1$ and $4$ together. Note that $O$ and $P$ have the same center. This means that the line connecting the center of $R$ and $O$ passes through the tangency point of both $S$ and $O$ and $S$ and $P.$

This line is the diameter of $R,$ and it has length $r_P + r_O = 2 + 8 = 10.$ Therefore, the radius of $R$ is $5.$

Consider cases $2$ and $3$ together. Similarly to above, the line connecting the center of $R$ and $O$ will pass through the tangency points.

This time, however, the diameter of $R$ is $r_P - r_O = 8 - 2 = 6.$ This makes the radius of $R$ $3.$

$S$ contains $8$ circles: $4$ of which have radius $5$ and $4$ of which have radius $3$ (this is because we can flip all the circles in the diagram over the x-axis to get $4$ more circles).

The total area of the circles in $S$ is therefore $4 (5^2 \pi + 3^2 \pi) = 136 \pi.$ Thus, **E** is the correct answer.

23.

Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1?$

$\dfrac 13$

$\dfrac 12$

$\dfrac 23$

$\dfrac 34$

$\dfrac 56$

###### Answer: C

###### Solution:

We begin by breaking the question into two smaller problems:

• The probability such that if $x,y$ were random numbers in the interval $(0,1),$ that $x+y < 1.$

• The probability such that if $x,y,z$ were random numbers in the interval $(0,1),$ that $x+y+z < 1.$$$$$

To solve both of these, we use geometric probability. That means, for a 2-D or 3-D coordinate system, that $x+y < 1$ and $x+y+z < 1.$

To solve the first one, we take the area of the region where $x+ y < 1,$ and $x,y > 0.$ This would be $\frac 12$ since it is a triangle with base and height of 1. Since the total area is just $1^2 = 1,$ the probability is also $\frac{1}{2}.$

To solve the second one, we take the area of the region where $x+ y + z < 1,$ and $x,y > 0.$ This would be $\frac 16$ since it is a volume of a pyramid with base of area $\frac 12$ and height of $1.$ Again, the volume is $1^3 = 1,$ so the probability is also $\frac{1}{2}.$

Now, we can look at the cases of the problem. Annie can either go until $n=2$ if $t_1 + t_2 > 1$ or until $n=3$ otherwise.

Case $1:$ The probability that $t_1+t_2 > 1$ is $\frac 12$ since the probability that $t_1+t_2 < 1$ is $\frac 12.$ Similarly, the probability that $x_1+x_2 > 1$ is $\frac 12.$ The total probability with this case is $\frac 12 \cdot \frac 12 = \frac 14.$

Case $2:$ The probability that $t_1+t_2 < 1$ is $\frac 12.$ Also, the probability that $x_1+x_2+x_3 > 1$ is $\frac 56$ since the probability that $x_1+x_2+x_3 < 1$ is $\frac 16.$The total probability with this case is $\frac 12 \cdot \frac 56 = \frac 5{12}.$

The total probability therefore is $\frac 14 + \frac 5{12} = \frac 23.$

Thus, the answer is **C**.

24.

Consider functions $f$ that satisfy $|f(x)-f(y)|\leq \dfrac{1}{2}|x-y|$ for all real numbers $x$ and $y.$ Of all such functions that also satisfy the equation $f(300) = f(900),$ what is the greatest possible value of $f(f(800))-f(f(400))?$

$25$

$50$

$100$

$150$

$200$

###### Answer: B

###### Solution:

Note that $|f(f(400))-f(f(300))|$ $\leq \dfrac 12 | f(400) - f(300)|$ $\leq \dfrac 14 |400-300| = 25,$ and $|f(f(900))-f(f(800))|$ $\leq \dfrac 12 | f(900) - f(800)|$ $\leq \dfrac 14 |900-800| = 25.$

Since $f(900) = f(300),$ by the triangle inequality, we know $|f(f(800)) - f(f(400))| =$ $|(f(f(800)) - f(f(900))) -$ $(f(f(400))-f(f(300)))| \leq$ $|(f(f(800)) - f(f(900)))| +$ $|(f(f(400))-f(f(300)))|$$\leq 50.$

Now, we must conclude this value is attainable. We can make $f(x)$ a piecewise function such that $f(x) = 600$ if $x > 900$ or $x< 300,$ $f(x) = -\dfrac 12 (x-300)+600$ if $300 \leq x \leq 400,$ $f(x) = \dfrac 12 (x-600)+600$ if $400 < x < 800,$ and $f(x) = -\dfrac 12 (x-900)+600$ if $800 \leq x \leq 900.$ This would make $f(f(400)) = f(550) = 575$ and $f(f(800)) = f(650) = 625.$ This yields a difference of $50,$ so our result holds.

Thus, the answer is **B**.

25.

Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1.$ For each positive integer $n,$ define $S_n = \sum_{k=0}^{n-1} x_k 2^k$ Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1.$ What is the value of the sum $x_{2019} + 2x_{2020} +$$4x_{2021} + 8x_{2022}?$

$6$

$7$

$12$

$14$

$15$

###### Answer: A

###### Solution:

Note first that $\dfrac{ S_{2023} - S_{2019}}{2^{2019}} = x_{2019} +$$2x_{2020} + 4x_{2021} + 8x_{2022}.$ Therefore, we should attempt to find $S_{2019}, S_{2023}.$ Also, note that $0 \leq S_n \leq \sum_{k=0}^{n-1} 2^k < 2^n.$

Now, since $7S_n \equiv 1 \pmod{2^n},$ we know $7S_n = m2^n +1 \implies$$S_n = \dfrac {m2^n+1}{7}$ for some integer $m.$ Also, since $0 \leq S_n < 2^n,$ we know $0 \leq \dfrac {m2^n+1}7 < 2^n.$ This means $0 \leq m < 7.$ Now, we find $m$ such that $m2^n + 1$ is divisible by $7.$ This makes $m2^n + 1 \equiv 0 \mod 7.$

If $n=2019,$ then $0 \equiv m2^{2019} + 1 \equiv$ $m(2^3)^{667}+1 \equiv$ $m8^{667} + 1\equiv m+1,$ so $m=6.$ This makes $S_{2019} = \dfrac {6(2^{2019})+1}{7}.$

If $n=2023,$ then $0 \equiv m2^{2019} + 1$ $\equiv m(2^3)^{668}\cdot 2+1$ $\equiv m8^{668}\cdot 2 + 1$$\equiv 2m+1,$ so $m=3.$ This makes $S_{2023} = \dfrac {3(2^{2023})+1}{7}.$

Our answer is $\dfrac{ S_{2023} - S_{2019}}{2^{2019}} =$ $\dfrac1{2^{2019}} \left(\dfrac {3(2^{2023})+1}{7} -\right.$$\left. \dfrac {6(2^{2019})+1}{7}\right) =$ $\dfrac 1{2^{2019}} \left(\dfrac {48(2^{2023})}{7} \right.$$\left. - \dfrac {6(2^{2019})}{7}\right) = \dfrac {42}7 = 6.$

Thus, the correct answer is **A**.