Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$
$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$

Solution: Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively.
The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ\div6=120^\circ.$ By angle chasing, we conclude that the interior angles of $\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,$ and $\triangle ZAB$ are all $60^\circ.$ Therefore, these triangles are all equilateral triangles, from which $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles.
We are given that \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} so we get $PQ=16\sqrt3$ and $YZ=36,$ respectively.
By equilateral triangles and segment addition, we find the perimeter of hexagon $ABCDEF:$ \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} Finally, the answer is $36+16+3=\boxed{55}.$

Answer: C 55
Problem: Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$
$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$

Solution: Let the length $AB=x, BC=y.$ Then, we have \begin{align*} (y+2x)^2\cdot\frac{\sqrt 3}{4}&=324\sqrt3, \\ (x+2y)^2\cdot\frac{\sqrt 3}{4}&=192\sqrt3. \end{align*} We get \begin{align*} y+2x&=36, \\ x+2y&=16\sqrt3. \end{align*} We want $3x+3y,$ and it follows that \[3x+3y=(y+2x)+(x+2y)=36+16\sqrt3.\] Finally, the answer is $36+16+3=\boxed{55}.$

Answer: C 55
Problem: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Solution: Suppose the roommate took sheets $a$ through $b$ , or equivalently, page numbers $2a-1$ through $2b$ . Because there are $(2b-2a+2)$ numbers taken, \[\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.\] The first possible solution that comes to mind is if $2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12$ , which indeed works, giving $b=22$ and $a=10$ . The answer is $22-10+1=\boxed{13}$

Answer: B 13
Problem: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Solution: Suppose the smallest page number borrowed is $k,$ and $n$ pages are borrowed. It follows that the largest page number borrowed is $k+n-1.$
We have the following preconditions:
Together, we have \begin{align*} \frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&=19 \\ 1275-\frac{(2k+n-1)n}{2}&=19(50-n) \\ 2550-(2k+n-1)n&=38(50-n) \\ 2550-(2k+n-1)n&=1900-38n \\ 650&=(2k+n-39)n. \end{align*} The factors of $650$ are \[1,2,5,10,13,25,26,50,65,130,325,650.\] Since $n$ is even, we only have a few cases to consider: \[\begin{array}{c|c|c} & & \\ [-2.25ex]  \boldsymbol{n} & \boldsymbol{2k+n-39} & \boldsymbol{k} \\ [0.5ex] \hline  & & \\ [-2ex]  2 & 325 & 181 \\    10 & 65 & 47 \\  26 & 25 & 19 \\  50 & 13 & 1 \\ 130 & 5 & -43 \\ 650 & 1 & -305 \\ \end{array}\] Since $1\leq k \leq 49,$ only $k=47,19,1$ are possible:
Therefore, the only possibility is $k=19.$ We conclude that $n=26$ pages, or $\frac n2=\boxed{13}$ sheets, are borrowed.

Answer: B 13
Problem: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Solution: Let $n$ be the number of sheets borrowed, with an average page number $k+25.5$ . The remaining $25-n$ sheets have an average page number of $19$ which is less than $25.5$ , the average page number of all $50$ pages, therefore $k>0$ . Since the borrowed sheets start with an odd page number and end with an even page number we have $k \in \mathbb N$ . We notice that $n < 25$ and $k \le (49+50)/2-25.5=24<25$
The weighted increase of average page number from $25.5$ to $k+25.5$ should be equal to the weighted decrease of average page number from $25.5$ to $19$ , where the weights are the page number in each group (borrowed vs. remained), therefore
\[2nk=2(25-n)(25.5-19)=13(25-n) \implies 13 | n \text{ or } 13 | k\]
Since $n, k < 25$ we have either $n=13$ or $k=13$ . If $n=13$ then $k=6$ . If $k=13$ then $2n=25-n$ which is impossible. Therefore the answer should be $n=\boxed{13}$

Answer: B 13
Problem: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Solution: Let $(2k-1)-2n$ be pages be borrowed, the sum of the page numbers on those pages is $(2n+2k+1)(n-k)$ while the sum of the rest pages is $1275-(2n+2k+1)(n-k)$ and we know the average of the rest is $\frac{1275-(2n+2k+1)}{50-2n+2k}$ which equals to $19$ ; multiply this out we got $950-38(n-k)=1275-(2n+2k+1)(n-k)$ and we got $(2n+2k-37)(n-k)=325$ . As $325=25\cdot13$ , we can see $n-k=13$ and that is desired $\boxed{13}$

Answer: B 13
Problem: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Solution: Let $c$ be the number of consecutive sheets Hiram’s roommate borrows, and let $b$ be the number of sheets preceding the $c$ borrowed sheets (i.e. if the friend borrows sheets $3$ $4$ , and $5$ , then $c=3$ and $b=2$ ).
The sum of the page numbers up till $b$ sheets is $1+2+3+\cdots + 2b=\frac{2b\cdot(2b+1)}{2} = b(2b+1)$ .
The last page number of the borrowed sheets would be $2(b+c)$ . Therefore, the sum of the remaining page numbers of the sheets after the $c$ borrowed sheets would be $2(b+c)+1 + 2(b+c)+2+\cdots+50$
The total number of page numbers after the borrow would be $50-2c$
Thus the average of the page numbers after the borrow would be: \[\frac{b(2b+1)+ 2(b+c)+1 + 2(b+c)+2+\cdots+50}{50-2c} =19.\] By the arithmetic series formula, this turns out to be: \[\frac{b(2b+1)+ \frac{(2(b+c)+1+50)\cdot(50-2c-2b)}{2}}{50-2c} =19\] because in the changed sum, there are $50$ numbers minus $2c$ borrowed numbers and $2b$ numbers from the first $b$ sheets.
This simplifies to \[\frac{b(2b+1)+ (2(b+c)+51)\cdot(25-c-b)}{50-2c} =19 \implies 19(50-2c)= b(2b+1)+ (2b+2c+51)\cdot(25-c-b).\] Noticing that some terms will cancel, we expand, leading to: \[950-38c=2b^2+b+50b-2bc-2b^2+50c-2c^2-2bc+1275-51c-51b\] \[\implies 950-38c=1275-4bc-c-2c^2 \implies 2c^2+4b-37c=325.\] Factoring, we get \[c(2c+4b-37)=325.\] The prime factorization of 325 is $5^2\cdot13$ . Recall that $0\leq c \leq 25$ , so $c$ could be $1$ $5$ $13$ , or $25$
We can rule out $c=25$ since Hiram would have no paper left over, so the average of the page numbers he has would be $0$ . We can now plug in the other answers for $c$ and we if we get a valid answer for $b$
Finally, just to make sure, we test $c=13 \implies 26+4b-37=25 \implies 4b=36 \implies b=9$
$b$ is an integer and $b+c=22$ , so everything checks out. The number of consecutive sheets borrowed by Hiram’s friend is $c=\boxed{13}$

Answer: B 13
Problem: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Solution: The sum of all the page numbers is \[1+2+3+\cdots+50 = 1275.\] If we add the page numbers on each sheet, we get this sequence: \[3, 7, 11, \ldots, 99.\] So we can write the sum of the numbers on the first sheet that the roommate borrowed as $4n+3$ for some nonnegative integer, $n$ . If the roommate borrowed $k$ sheets, he borrowed sheets \[4n+3, 4n+7, \ldots, 4n+4k-1.\] The sum of the numbers in this sequence is \[\frac{k(8n+4k+2)}{2} = 4nk+2k^2+k.\] Since there are $2$ pages per sheet, there are $50-2k$ pages remaining, so the average page number of the remaining sheets is \[\frac{1275 - (4nk+2k^2+k)}{50-2k}.\] Therefore, \[\frac{1275 - (4nk+2k^2+k)}{50-2k} = 19,\] which simplifies to \[2k^2-37k-325 = -4nk.\] Factoring the left-hand side, \[(2k+13)(k-25)=-4nk.\] Since the right-hand side of this equation is divisible by $k$ , the left-hand side must also be divisible by $k$
In order for $(2k+13)(k-25)$ to be divisible by $k$ , either $2k+13$ or $k-25$ must be divisible by $k$ \[\frac{2k+13}{k}=2+\frac{13}{k},\] so $2k+13$ is divisible by $k$ only if $k$ is a factor of $13$ \[\frac{k-25}{k}=1-\frac{25}{k},\] so $k-25$ is divisible by $k$ only if $k$ is a factor of $25$
None of the answer choices are factors of $25$ , but answer choice B is a factor of $13$ . Hence, the answer is $\boxed{13}$

Answer: B 13
Problem: How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$

Solution: Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid. 
WLOG assume that A is in the center. \[\begin{tabular}{ c c c }  ? & ? & ? \\   ? & A & ? \\    ? & ? & ?     \end{tabular}\] In this configuration, there are two cases, either all the A's lie on the same diagonal: \[\begin{tabular}{ c c c }  ? & ? & A \\   ? & A & ? \\    A & ? & ?     \end{tabular}\] or all the other two A's are on adjacent corners: \[\begin{tabular}{ c c c }  A & ? & A \\   ? & A & ? \\    ? & ? & ?     \end{tabular}\] In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.
In each case there is only one way to put the three B's and the three C's as shown in the diagrams. \[\begin{tabular}{ c c c }   C & B & A \\    B & A & C \\     A & C & B     \end{tabular}\] \[\begin{tabular}{ c c c }  A & B & A \\   C & A & C \\    B & C & B     \end{tabular}\] This means that there are $4+2=6$ ways to arrange A,B, and C in the grid, and there are $6$ ways to rearrange the colors. Therefore, there are $6\cdot6=36$ ways in total, which is $\boxed{36}$

Answer: E 36
Problem: How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$

Solution: We consider all possible configurations of the red chips for which rotations matter: [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, red);  fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, red); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red);  fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, red); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red);  fill((18,2)--(19,2)--(19,3)--(18,3)--cycle, red); fill((20,1)--(21,1)--(21,2)--(20,2)--cycle, red); fill((19,0)--(20,0)--(20,1)--(19,1)--cycle, red);  fill((24,1)--(25,1)--(25,2)--(24,2)--cycle, red); fill((26,1)--(27,1)--(27,2)--(26,2)--cycle, red); fill((25,2)--(26,2)--(26,3)--(25,3)--cycle, red);  draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5));  draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5));  draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5));  draw((18,0)--(21,0)--(21,3)--(18,3)--cycle, linewidth(1.5)); draw((18,1)--(21,1), linewidth(1.5)); draw((18,2)--(21,2), linewidth(1.5)); draw((19,0)--(19,3), linewidth(1.5)); draw((20,0)--(20,3), linewidth(1.5));  draw((24,0)--(27,0)--(27,3)--(24,3)--cycle, linewidth(1.5)); draw((24,1)--(27,1), linewidth(1.5)); draw((24,2)--(27,2), linewidth(1.5)); draw((25,0)--(25,3), linewidth(1.5)); draw((26,0)--(26,3), linewidth(1.5));  label("Rotational",(1.5,4.5)); label("Symmetry",(1.5,3.75)); label("$2$ Configurations",(1.5,-0.75)); label("$4$ Configurations",(7.5,-0.75)); label("$4$ Configurations",(13.5,-0.75)); label("$4$ Configurations",(19.5,-0.75)); label("$4$ Configurations",(25.5,-0.75)); [/asy] As there are $2!=2$ permutations of blue and green for each configuration, the answer is $2\cdot(2+4+4+4+4)=\boxed{36}.$

Answer: E 36
Problem: How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$

Solution: $(1) \quad$ $\begin{tabular}{ c c c }  R & G & ? \\   B & R & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & B \\   B & R & G \\    R & G & B     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & R \\   B & R & B \\    G & B & G     \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$
There are $3$ choices for $R$ $2$ choices for $G$ $R$ on the down left corner can be switched with $B$ on the upper right corner.
$(2) \quad$ $\begin{tabular}{ c c c }  R & G & ? \\   B & R & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & B \\   B & R & G \\    G & B & R     \end{tabular}$ $\quad 3 \cdot 2 = 6$
There are $3$ choices for $R$ $2$ choices for $G$
$(3) \quad$ $\begin{tabular}{ c c c }  G & R & ? \\   R & B & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & B \\   R & B & G \\    G & R & B     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & G \\   R & B & R \\    B & G & B     \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$
Note that $(3)$ is a $180$ ° rotation of $(1)$
$(4) \quad$ $\begin{tabular}{ c c c }  G & R & ? \\   R & B & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & B \\   R & B & G \\    B & G & R     \end{tabular}$ $\quad 3 \cdot 2 = 6$
Note that $(4)$ is a $90$ ° rotation of $(2)$
Therefore, the answer is $2 \cdot (12 + 6) = \boxed{36}$

Answer: E 36
Problem: How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$

Solution: Case (1) : We have a permutation of R, B, and G as all of the rows. There are $3!$ ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, $\frac{3!}{e} \approx 2$ , so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are $2$ possible permutations for the last row. Thus, there are $3!\cdot2\cdot2=24$ possibilities.
Case (2) : All of the rows have two chips that are the same color and one that is different. There are obviously $3$ possible configurations for the first row, $2$ for the second, and $2$ for the third. Thus, there are $3\cdot2\cdot2=12$ possibilities.
Therefore, our answer is $24+12=\boxed{36}.$

Answer: E 36
Problem: How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$

Solution: Since $3\pi\approx9.42$ , we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\boxed{19}$

Answer: D 19
Problem: How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$

Solution: $|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$ . Since $\pi$ is approximately $3.14$ $3\pi$ is approximately $9.42$ . We are trying to solve for $-9.42<x<9.42$ , where $x\in\mathbb{Z}$ . Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$ , for $x\in\mathbb{Z}$ . The number of integer values of $x$ is $9-(-9)+1=19$ . Therefore, the answer is $\boxed{19}$

Answer: D 19
Problem: How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$

Solution: $3\pi \approx 9.4.$ There are two cases here.
When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$
When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$ . Dividing by $-1$ and flipping the sign, we get $x>-9.4.$
From case 1 and 2, we know that $-9.4 < x < 9.4$ . Since $x$ is an integer, we must have $x$ between $-9$ and $9$ . There are a total of \[9-(-9) + 1 = \boxed{19}.\] PureSwag

Answer: D 19
Problem: How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$

Solution: Looking at the problem, we see that instead of directly saying $x$ , we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions: $9+9=18.$ But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is $9+9+1=18+1=\boxed{19}.$

Answer: D 19
Problem: How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$

Solution: There are an odd number of integer solutions $x$ to this inequality since if any non-zero integer $x$ satisfies this inequality, then so does $-x,$ and we must also account for $0,$ which gives us the desired. Then, the answer is either $\textbf{(A)}$ or $\textbf{(D)},$ and since $3 \pi > 3 \cdot 3 > 9,$ the answer is at least $9 \cdot 2 + 1 = 19,$ yielding $\boxed{19}.$

Answer: D 19
Problem: In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$

Solution: Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations \[5j=2s\] \[j+s=28,\] and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is \[\boxed{8}.\]

Answer: C 8