### 2022 AMC 10B

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Used with permission of the Mathematical Association of America.

1:15:00

1.

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

$$-2$$

$$-1$$

$$0$$

$$1$$

$$2$$

###### Solution(s):

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

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.) $$3\sqrt 5$$

$$10$$

$$6\sqrt 5$$

$$20$$

$$25$$

###### Solution(s):

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 $$5\cdot 4 = 20.$$

3.

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

$$150$$

$$250$$

$$350$$

$$450$$

$$550$$

###### Solution(s):

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 $$90\cdot 5 = 450.$$

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?

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

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

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

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

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

###### Solution(s):

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.

5.

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

$$\sqrt3$$

$$2$$

$$\sqrt{15}$$

$$4$$

$$\sqrt{105}$$

###### Solution(s):

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*}

6.

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

$$0$$

$$1$$

$$2$$

$$3$$

$$4$$

###### Solution(s):

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.

7.

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

$$\ 6$$

$$\ 8$$

$$\ 9$$

$$\ 14$$

$$\ 16$$

###### Solution(s):

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 $$7?$$

$$\ 40$$

$$\ 42$$

$$\ 43$$

$$\ 49$$

$$\ 50$$

###### Solution(s):

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.

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 $$a+b?$$

$$\ 2020$$

$$\ 2021$$

$$\ 2022$$

$$\ 2023$$

$$\ 2024$$

###### Solution(s):

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 $$2023.$$

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?

$$\ 5$$

$$\ 7$$

$$\ 9$$

$$\ 11$$

$$\ 13$$

###### Solution(s):

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.$$

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.

###### Solution(s):

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.

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 $$\dfrac{1}{2}?$$

$$2$$

$$3$$

$$4$$

$$5$$

$$6$$

###### Solution(s):

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 $$4.$$

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?

$$\ 8$$

$$\ 10$$

$$\ 11$$

$$\ 13$$

$$\ 16$$

###### Solution(s):

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 $$16.$$

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?

$$\ 12$$

$$\ 13$$

$$\ 14$$

$$\ 15$$

$$\ 16$$

###### Solution(s):

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 $$13.$$

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 $$S_{20}?$$

$$340$$

$$360$$

$$380$$

$$400$$

$$420$$

###### Solution(s):

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 $S_{20} = -1(20)+20^2+20 = 400.$

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? $$15\dfrac{1}{8}$$

$$15\dfrac{3}{8}$$

$$15\dfrac{1}{2}$$

$$15\dfrac{5}{8}$$

$$15\dfrac{7}{8}$$

###### Solution(s):

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 $$15 \dfrac 58.$$

17.

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

$$2^{606}-1$$

$$2^{606}+1$$

$$2^{607}-1$$

$$2^{607}+1$$

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

###### Solution(s):

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 $$10.$$

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.)

$$\ 302$$

$$\ 338$$

$$\ 340$$

$$\ 343$$

$$\ 344$$

###### Solution(s):

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.

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.) $$\ 14$$

$$\ 18$$

$$\ 22$$

$$\ 26$$

$$\ 30$$

###### Solution(s):

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.

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 $$\angle BFC?$$

$$\ 110$$

$$\ 111$$

$$\ 112$$

$$\ 113$$

$$\ 114$$

###### Solution(s): 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, $\angle BFC = 180^\circ - 67^\circ = 113^\circ.$

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?

$$\ 10$$

$$\ 13$$

$$\ 19$$

$$\ 20$$

$$\ 23$$

###### Solution(s):

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 $$23.$$

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 $$S?$$

$$\ 48 \pi$$

$$\ 68 \pi$$

$$\ 96 \pi$$

$$\ 102 \pi$$

$$\ 136 \pi$$

###### Solution(s): 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 $$1?$$

$$\dfrac 13$$

$$\dfrac 12$$

$$\dfrac 23$$

$$\dfrac 34$$

$$\dfrac 56$$

###### Solution(s):

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 $$\frac 14 + \frac 5{12} = \frac 23.$$

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 $f(f(800))-f(f(400))?$

$$25$$

$$50$$

$$100$$

$$150$$

$$200$$

###### Solution(s):

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.

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}?$

$$6$$

$$7$$

$$12$$

$$14$$

$$15$$

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.$