## AMC 10 2019 Test A

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have
**75 minutes**working time to complete the test.

What is the value of$$2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?$$

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

$\textbf{C}$

\begin{align*}

2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9=2^{\left(0^1\right)}+\left(1^1\right)^9=2^0+1^9=1+1=2

\end{align*}

What is the hundreds digit of $(20!-15!)?$

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

$\textbf{A}$

Since $20!-15!=15!\times(20\times19\times18\times17\times16-1)$, we know that $15!$ is a factor of $20!-15!$. The prime factorization of $15!$ gives three factor $5$ and more than three factor $2$. So $20!-15!$ is a multiple of $5^3\times2^3=1000$. The hundreds digit must be 0.

Ana and Bonita are born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$

$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$

$\textbf{D}$

Let $a$ be the age of Ana this year, and $b$ be the age of Bonita this year. Then we have \begin{align*}

a-1&=5(b-1)\\

a&=b^2

\end{align*}Hence, we get $a=16$, $b=4$. So $n=a-b=12$.

A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn$?$

$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$

$\textbf{B}$

In the worst case we draw $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $14+14+14+13+11+9=75$ balls, without drawing $15$ balls of any single color. Drawing one more ball guarantees that we will get $15$ balls of a single color — either red, green, or yellow. Thus, the answer is $75 + 1 =76$.

What is the greatest number of consecutive integers whose sum is $45?$

$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

$\textbf{D}$

To get the longest sequence of consecutive integers, we hope the positive integers and negative integers cancel each other, which is $-44, -43, \cdots, 43, 44$. Since the sum is 45, we need to add the number 45 at the end of the sequence. So the number of elements in the sequence $-44, -43, \cdots, 43, 44, 45$ is 90.

For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?

$\bullet\quad$a square$\newline$

$\bullet\quad$a rectangle that is not a square$\newline$

$\bullet\quad$a rhombus that is not a square$\newline$

$\bullet\quad$a parallelogram that is not a rectangle or a rhombus$\newline$

$\bullet\quad$an isosceles trapezoid that is not a parallelogram

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

$\textbf{C}$

If there is a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral, then the four vertices are concyclic, and the point we are looking for is the center of the circle. The four vertices of a square, a rectangle, or an isosceles trapezoid are concyclic. So the answer is 3.

Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$. What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$

$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

$\textbf{C}$

The line of slope $\dfrac12$ passing through $(2,2)$ is $y=\dfrac12x+1$. The line of slope $2$ passing through $(2,2)$ is $y=2x-2$. The intersection points of $y=\dfrac12x+1$, $y=2x-2$ and $y=-x+10$ are $(2,2)$, $(6,4)$ and $(4,6)$. According to the Shoelace formula, the area of a polygon with vertices $(x_1, y_1),(x_2, y_2),\ldots,(x_n, y_n)$ is \[A = \dfrac{1}{2} \left|(x_1y_2 + x_2y_3 + \cdots + x_ny_1) - (y_1x_2 + y_2x_3 + \cdots + y_nx_1) \right|\] Hence, the area of the triangle with vertices $(2,2)$, $(6,4)$ and $(4,6)$ is $$\dfrac12\left|(2\times4+6\times6+4\times2)-(2\times6+4\times4+6\times2)\right|=6$$

The figure below shows line $\ell$ with a regular, infinite, recurring pattern of squares and line segments.

How many of the following four kinds of rigid motion transformations of the plane in which this figure is drawn, other than the identity transformation, will transform this figure into itself?

$\quad\bullet\quad$some rotation around a point of line $\ell$

$\quad\bullet\quad$some translation in the direction parallel to line $\ell$

$\quad\bullet\quad$the reflection across line $\ell$

$\quad\bullet\quad$some reflection across a line perpendicular to line $\ell$

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

$\textbf{C}$

Statement $1$ is true. A $180^{\circ}$ rotation about the point half way between an up-facing square and a down-facing square will yield the same figure.

Statement $2$ is also true. Let the unit length be the distance between the centers of two adjacent up-facing squares. A translation to the left or right by a integer multiple of the unit length will place the image onto itself.

Statement $3$ is false. A reflection across line $\ell$ will change the direction of all segments extending from the squares.

Finally, statement $4$ is false because it will also cause the segments extending from the squares to switch direction.

Thus, only $2$ statements are true.

What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?

$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

$\textbf{B}$

The sum of the first $n$ positive integers is $\dfrac{(n)(n+1)}{2}$, and we want it not to be a divisor of $n!$. If $n+1$ is not a prime, all of its factors would be canceled by the factors in $n!$. If $n+1$ is a prime, the statement is true. According to the choices, 997 is the largest prime. So the answer is 996.

A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?

$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$

$\textbf{C}$

Every time the bug crosses a horizontal or vertical line, it moves into a new tile. There is a special case when the bug moves into a new tile from a vertex of a square, but it doesn't happen because the greatest common divisor of 10 and 17 is 1. The bug crosses $9+16=25$ lines in total. Take the first tile into account, the answer is $25+1=26$.

How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)?

$\textbf{(A) } 32 \qquad\textbf{(B) } 36 \qquad\textbf{(C) } 37 \qquad\textbf{(D) } 39 \qquad\textbf{(E) } 41$

$\textbf{C}$

Prime factorization of $201^9$ gives $3^9\times67^9$. A perfect square must have even powers of its prime factors, so our possible choices for the exponents to get perfect square are $0, 2, 4, 6, 8$ for both $3$ and $67$. This yields $5\times5 = 25$ perfect squares.

Perfect cubes must have multiples of $3$ for each of their prime factors' exponents, so we have either $0, 3, 6, 9$ for both $3$ and $67$, which yields $4\times4 = 16$ perfect cubes.

In total we have $25+16 = 41$ perfect squares and perfect cubes. Subtracting the overcounted powers of $6$ ($3^0\times67^0$, $3^0\times67^6$, $3^6\times67^0$, and $3^6\times67^6$), we get $41-4 =37$.

Melanie computes the mean $\mu$, the median $M$, and the modes of the $365$ values that are the dates in the months of $2019$. Thus her data consists of $12$ $1\text{s}$, $12$ $2\text{s}$, $\cdots$, $12$ $28\text{s}$, $11$ $29\text{s}$, $11$ $30\text{s}$, and $7$ $31\text{s}$. Let $d$ be the median of the modes. Which of the following statements is true?

$\textbf{(A) } \mu < d < M \qquad\textbf{(B) } M < d < \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d < M < \mu \qquad\textbf{(E) } d < \mu < M$

$\textbf{E}$

In the new sequence of $12$ $1\textrm{s}$, $12$ $2\textrm{s}$, $\cdots$, $12$ $28\textrm{s}$, we have $d=\mu=M$. However, when we extend the sequence to $12$ $1\textrm{s}$, $12$ $2\textrm{s}$, . . . , $12$ $28\textrm{s}$, $11$ $29\textrm{s}$, $11$ $30\textrm{s}$, $7$ $31\textrm{s}$, $d$ keeps unchanged while $\mu$ and $M$ become greater. So $d$ is the smallest.

In a sequence of 365 numbers, $M$ is the $183\textrm{rd}$ one. Since $183=12\times15+3$, we get $M=16$. If we extend the sequence to $12$ $1\textrm{s}$, $12$ $2\textrm{s}$, $\cdots$, $12$ $28\textrm{s}$, $12$ $29\textrm{s}$, $12$ $30\textrm{s}$, $12$ $31\textrm{s}$, $\mu$ will be larger. The new mean value $\mu=16$. So in the original sequence $\mu<M$.

Therefore, the answer is $d<\mu<M$.

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$. Construct the circle with diameter $\overline{BC}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$. What is the degree measure of $\angle BFC ?$

$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

$\textbf{D}$

Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. So $E$ is the midpoint of $AB$, $\angle ACE=20^\circ$. By $\triangle BEF\sim\triangle CDF$, we see $\angle ABD=\angle ACE=20^\circ$. Hence, $\angle BFC=\angle BEC+\angle ABD=90^\circ+20^\circ=110^\circ$.

For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$?

$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$

$\textbf{D}$

The maximum number of possible intersections is $ \dbinom42 = 6$, since each pair of lines can intersect at most once. It is possible to obtain $0$, $1$, $3$, $4$, $5$, and $6$ points of intersection, as demonstrated in the following figures:

It is impossible to have 2 intersections because if line 1 intersects only line 2, and line 3 intersects only line 4, we have $1\parallel3\parallel4$ and $3\parallel1\parallel2$. So we get $1\parallel2\parallel3\parallel4$, which means there is no intersection at all.

Therefore, the answer is $0+1+3+4+5+6= 19$.

A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \dfrac{3}{7}$, and\[a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}\]for all $n \geq 3$. Then $a_{2019}$ can be written as $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

$\textbf{E}$

We have $$\dfrac{1}{a_n} = \dfrac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\dfrac{2}{a_{n-1}}-\dfrac{1}{a_{n-2}}$$ in other words, $$\dfrac{1}{a_n}-\dfrac{1}{a_{n-1}} = \dfrac{1}{a_{n-1}}-\dfrac{1}{a_{n-2}}$$ So $\{\dfrac{1}{a_n}\}$ is an arithmetic sequence with step size $\dfrac{1}{a_2}-\dfrac1{a_1}=\dfrac{7}{3}-1=\dfrac{4}{3}$, which means $$\dfrac{1}{a_{2019}} = \dfrac1{a_1}+2018\times\dfrac43=1+2018 \times \dfrac{4}{3} = \dfrac{8075}{3}$$ Since the numerator and the denominator are relatively prime, the answer is $3+8075=8078$.

The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$

$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$

$\textbf{A}$

In equilateral triangles $OAB$ and $ABC$, we have $OC=2\sqrt3$. So the radius of the larger circle is $2\sqrt3+1$. The area of the shaded region is $\pi(2\sqrt3+1)^2-13\pi\cdot1^2=4\sqrt3\pi$.

A child builds towers using identically shaped cubes of different colors. How many different towers with a height $8$ cubes can the child build with $2$ red cubes, $3$ blue cubes, and $4$ green cubes? (One cube will be left out.)

$\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320$

$\textbf{D}$

Arranging eight cubes is the same as arranging the nine cubes first, and then removing the last cube. In other words, there is a one-to-one correspondence between every arrangement of nine cubes, and every actual valid arrangement. Thus, we initially get $9!$. However, we have overcounted, because the red cubes can be permuted to have the same overall arrangement, and the same applies with the blue and green cubes. Thus, we have to divide by the $2!$ ways to arrange the red cubes, the $3!$ ways to arrange the blue cubes, and the $4!$ ways to arrange the green cubes. Thus we have $$\dfrac {9!} {2! \cdot 3! \cdot 4!} = 1,260$$ different possible towers.

For some positive integer $k$, the repeating base-$k$ representation of the (base-ten) fraction $\dfrac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$. What is $k$?

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

$\textbf{D}$

We can expand the fraction $0.\overline{23}_k$ as follows:\begin{align*}

0.\overline{23}_k &= 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots\\

&=2( k^{-1} + k^{-3} + k^{-5} +\cdots) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\\

&=2\cdot\dfrac{k}{k^2-1}+3\cdot\dfrac{1}{k^2-1}\\

&=\dfrac{2k+3}{k^2-1}\\

&=\dfrac{7}{51}

\end{align*}Hence, we get $k =16$.

What is the least possible value of

$$(x+1)(x+2)(x+3)(x+4)+2019$$

where $x$ is a real number?

$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

$\textbf{B}$

Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$, which can be simplified to $(x^2+5x+5)^2-1+2019$. Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$, the answer is $2018$.

The numbers $1,2,\dots,9$ are randomly placed into the $9$ squares of a $3 \times 3$ grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?

$\textbf{(A) }\dfrac{1}{21}\qquad\textbf{(B) }\dfrac{1}{14}\qquad\textbf{(C) }\dfrac{5}{63}\qquad\textbf{(D) }\dfrac{2}{21}\qquad\textbf{(E) } \dfrac17$

$\textbf{B}$

There are 4 even numbers and 5 odd numbers in total. To get an odd sum in a row/column, there must be 2 even numbers or 0 even number in the same row/column. Where there is an even number, there must be an even number in the same row, an even number in the same column, then the location of the 4th even number is confirmed. Connecting the locations of 4 even numbers with lines and we get a rectangle. We have 9 possible rectangles in the $3\times3$ grid. Then we have 4! ways to permute the even numbers in the rectangle, and 5! way to permute the rest 5 odd numbers. The answer is $\dfrac{9\cdot4!\cdot5!}{9!}=\dfrac{1}{14}$.

A sphere with center $O$ has radius 6. A triangle with sides of length $15$, $15$, and $24$ is situated in space so that each of its sides are tangent to the sphere. What is the distance between $O$ and the plane determined by the triangle?

$\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) }4 \qquad \textbf{(C) } 3\sqrt{2} \qquad \textbf{(D) } 2\sqrt{5} \qquad \textbf{(E) } 5$

$\textbf{D}$

The 3D diagram is shown below:

Here we focus on the plane of the triangle:

The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use $\text{area} = \dfrac12\cdot\text{inradius} \cdot \text{perimeter}$. The area of the triangle can be found by drawing an altitude from the vertex between sides with length $15$ to the midpoint of the side with length $24$. The Pythagorean triple $9$ - $12$ - $15$ allows us easily to determine that the base is $24$ and the height is $9$. So the area of the triangle is $\dfrac12\cdot24\cdot9=108$. The perimeter of the triangle is $15 + 15 + 24= 54$. After plugging into the equation, we thus get $108 = \dfrac12\cdot\text{inradius} \cdot 54$, so the inradius is $4$.

Now, let the distance between $O$ and the plane of the triangle be $x$. Choose a point on the incircle and denote it by $A$. The distance $OA$ is $6$, because it is just the radius of the sphere. The distance from point $A$ to the center of the incircle is $4$, because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find $x = \sqrt{6^2-4^2}=\sqrt{20} =2 \sqrt {5}$.

Real numbers between $0$ and $1$, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is $0$ if the second flip is heads and $1$ if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \dfrac{1}{2}$?

$\textbf{(A) } \dfrac{1}{3} \qquad \textbf{(B) } \dfrac{7}{16} \qquad \textbf{(C) } \dfrac{1}{2} \qquad \textbf{(D) } \dfrac{9}{16} \qquad \textbf{(E) } \dfrac{2}{3}$

$\textbf{B}$

There are 4 cases depending on what the first coin flip is when determining $x$ and what the first coin flip is when determining $y$.

$\textbf{Case 1:}$ $x$ is either $0$ or $1$, and $y$ is either $0$ or $1$.

$\textbf{Case 2:}$ $x$ is either $0$ or $1$, and $y$ is chosen from the interval $[0,1]$.

$\textbf{Case 3:}$ $x$ is is chosen from the interval $[0,1]$, and $y$ is either $0$ or $1$.

$\textbf{Case 4:}$ $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$.

Each case has a $\dfrac{1}{4}$ chance of occurring (as it requires two coin flips).

For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\dfrac{1}{2}$.

For Case 2, if $x$ is 0, we need $y$ to be in the interval $\left(\dfrac{1}{2}, 1\right]$. If $x$ is 1, we need $y$ to be in the interval $\left[0, \dfrac{1}{2}\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\dfrac{1}{2}$.

By symmetry, Case 3 has the same success rate as Case 2.

For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $|x-y| > \dfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:

The shaded area is $\dfrac{1}{4}$, which means the probability for success for case 4 is $\dfrac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$).

Adding up the success rates from each case, we get the probability $\left(\dfrac{1}{4}\right) \cdot \left(\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{4}\right) = \dfrac{7}{16}$.

Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$, then Todd must say the next two numbers ($2$ and $3$), then Tucker must say the next three numbers ($4$, $5$, $6$), then Tadd must say the next four numbers ($7$, $8$, $9$, $10$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$th number said by Tadd?

$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$

$\textbf{C}$

Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).

We create a table to keep track of what numbers each child says for each round.

Tadd says $1$ number in round 1, $4$ numbers in round 2, $7$ numbers in round 3, and in general $3n - 2$ numbers in round $n$. At the end of round $n$, the number of numbers Tadd has said so far is $1 + 4 + 7 + \dots + (3n - 2) = \dfrac{n(3n-1)}{2}$, by the sum of arithmetic series formula.

We find that $\dfrac{37(110)}{2}=2035$, so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37. At the end of Tadd's turn in round 37, the children have, in total, completed $36+36+37=109$ turns. In general, at the end of turn $n$, $\dfrac{n(n+1)}{2}$ numbers has been said. So at the end of turn 109 (or the end of Tadd's turn in round 37), Tadd says the number $\dfrac{109(110)}{2}=5995$. Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was $5995-16=5979$.

Thus, the answer is $5979$.

Let $p$, $q$, and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67$. It is given that there exist real numbers $A$, $B$, and $C$ such that $$\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \dfrac{C}{s-r}$$ for all $s\not\in\{p,q,r\}$. What is $\dfrac1A+\dfrac1B+\dfrac1C$?

$\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247$

$\textbf{B}$

Multiplying both sides by $(s-p)(s-q)(s-r)$ yields\[1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)\]As this is a polynomial identity, and it is true for infinitely many $s$, it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial $0$). This means we can plug in $s = p$ to find that $\dfrac1A = (p-q)(p-r)$ (even though we know it is not allowed as $s\not\in\{p,q,r\}$). Similarly, we can find $\dfrac1B = (q-p)(q-r)$ and $\dfrac1C = (r-p)(r-q)$. Summing them up, we get that\[\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - pq - qr - pr\]By Vieta's Formulas, we know that \begin{align*}

p+q+r&=22\\

pq+pr+qr&=80\\

pqr&=67

\end{align*} So$$p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr) = 324$$ $$pq + qr + pr = 80$$ Hence, the answer is $324 -80 = 244$.

For how many integers $n$ between $1$ and $50$, inclusive, is

$$\frac{(n^2-1)!}{(n!)^{n}}$$

an integer? (Recall that $0!=1$.)

$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

$\textbf{D}$

We extend the original expression as $$\dfrac{1\cdot2\cdots(n^2-1)}{1^n\cdot2^n\cdots n^n}$$ For any $k^n$ in the denominator, if $k<n$, we can find $k, 2k, 3k,\cdots, nk$ in the numerator. So the $k^n$ in the denominator will be canceled when $k<n$. As for the $n^n$ in the denominator, we can find $n, 2n, \cdots, (n-1)n$ in the numerator but not $n^2$. This explains why the original expression is not an integer for some $n$. If $n$ is a prime, the number of prime factor $n$ in the denominator is $n$, and the number of prime factor $n$ in the numerator is $n-1$. So the original is not an integer when $n$ is a prime.

Now think about the case when $n$ is not a prime. If we split $n^2$ objects into $n$ groups of size $n$, we have $\dfrac{(n^2)!}{(n!)^n}$ ways. So $\dfrac{(n^2)!}{(n!)^n}$ must be an integer. But we couldn't decide whether $\dfrac{(n^2-1)!}{(n!)^n}=\dfrac{(n^2)!}{(n!)^n}\cdot\dfrac{1}{n^2}$ is an integer or not. If we split $n^2$ objects into $n$ $\textit{unordered}$ groups of size $n$, we have $\dfrac{(n^2)!}{(n!)^n}\cdot\dfrac{1}{n!}=\dfrac{(n^2)!}{(n!)^{n+1}}$ ways. This means $\dfrac{(n^2)!}{(n!)^{n+1}}$ must be an integer. Now the original expression is $$\dfrac{(n^2-1)!}{(n!)^n}=\dfrac{(n^2)!}{(n!)^{n+1}}\cdot\dfrac{n!}{n^2}=\dfrac{(n^2)!}{(n!)^{n+1}}\cdot\dfrac{(n-1)!}{n}$$ If $\dfrac{(n-1)!}{n}$ is an integer, the original expression must be an integer. We only care about the cases when $n$ is not a prime. Since the numerator is much greater than the denominator, when $n$ is not a prime, which means $n$ can be prime factorized into smaller numbers, $\dfrac{(n-1)!}{n}$ should be an integer when $n$ is not extremely small. However, extreme case happened when $n=4$. $\dfrac{(n-1)!}{n}$ is a integer except $n=4$.

Therefore, the original expression $\dfrac{(n^2-1)!}{(n!)^n}$ is not an integer when $n$ is a prime or $n=4$. There are 15 primes between 1 and 50. So the answer is $50-15-1=34$.