AMC 10 2021 Spring 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
$$(2^2-2)-(3^2-3)+(4^2-4)?$$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~12$
$\textbf{D}$
\begin{align*} \left(2^2-2\right)-\left(3^2-3\right)+\left(4^2-4\right) &= (4-2)-(9-3)+(16-4) \\ &= 2-6+12 \\&=8 \end{align*}
Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$
$\textbf{C}$
The number of students in Portia's high school is $\dfrac{3}{1+3}=\dfrac34$ of the total number. So the answer is $\dfrac34\times2600=1950$.
The sum of two natural numbers is $17{,}402$. One of the two numbers is divisible by $10$. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426$
$\textbf{D}$
The units digit of a multiple of $10$ will always be $0$. We add a $0$ at the end of the number whenever we multiply it by $10$. Therefore, removing its units digit is equal to dividing by $10$.
Let the smaller number (the one we get after removing the units digit) be $a$. This means the bigger number would be $10a$.
We know the sum is $10a+a = 11a=17402$. Hence, we get $a=1582$. The difference is $10a-a = 9a=9\times1582=14238$.
A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$-second time interval, it travels $7$ inches more than during the previous $1$-second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$
$\textbf{D}$
The distance the cart traveled in each second is an arithmetic sequence with the first term $a_1=5$ and common difference $d=7$. The sum of the first 30 terms is $$a_1+a_2+\cdots+a_{30}=\dfrac12\times[a_1+(a_1+29d)]\times30=\dfrac12\times[5+(5+29\times7)]\times30=3195$$
The quiz scores of a class with $k > 12$ students have a mean of $8$. The mean of a collection of $12$ of these quiz scores is $14$. What is the mean of the remaining quiz scores in terms of $k$?
$\textbf{(A)} ~\dfrac{14-8}{k-12} \qquad\textbf{(B)} ~\dfrac{8k-168}{k-12} \qquad\textbf{(C)} ~\dfrac{14}{12} - \dfrac{8}{k} \qquad\textbf{(D)} ~\dfrac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\dfrac{14(k-12)}{8k}$
$\textbf{B}$
The total score of the class is $8k,$ and the total score of the $12$ quizzes is $12\times14=168.$ Therefore, for the remaining $k-12$ quizzes, the total score is $8k-168.$ Their mean score is $\dfrac{8k-168}{k-12}.$
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at $4$ miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to $2$ miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at $3$ miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?
$\textbf{(A)} ~\dfrac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\dfrac{13}{12} \qquad\textbf{(D)} ~\dfrac{24}{13} \qquad\textbf{(E)} ~2$
$\textbf{A}$
Let $2d$ miles be the distance from the trailhead to the fire tower. It takes Chantal $\dfrac{d}{4}$ hours for the first half, $\dfrac{d}{2}$ hours for the second half, and $\dfrac{d}{3}$ hours coming back to the middle point. The total time of travel is $\dfrac{d}{4}+\dfrac{d}{2}+\dfrac{d}{3}=\dfrac{13}{12}d$. It takes Jean the same amount of hours to travel a distance of $d$, so the average speed is $d\div\dfrac{13}{12}d=\dfrac{12}{13}$ miles per hour.
Tom has a collection of $13$ snakes, $4$ of which are purple and $5$ of which are happy. He observes that
$\quad\bullet$ all of his happy snakes can add,
$\quad\bullet$ none of his purple snakes can subtract, and
$\quad\bullet$ all of his snakes that can't subtract also can't add.
Which of these conclusions can be drawn about Tom's snakes?
$\textbf{(A) }$ Purple snakes can add.
$\textbf{(B) }$ Purple snakes are happy.
$\textbf{(C) }$ Snakes that can add are purple.
$\textbf{(D) }$ Happy snakes are not purple.
$\textbf{(E) }$ Happy snakes can't subtract.
$\textbf{D}$
According to statements 2 and 3, all purple snakes can neither subtract nor add. Given that all happy snakes can add, we know that happy snakes are not purple.
When a student multiplied the number $66$ by the repeating decimal,\[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\]where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$-digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$
$\textbf{E}$
It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\dfrac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\dfrac{\underline{a} \ \underline{b}}{100}.$
Let $x=\underline{a} \ \underline{b}.$ We have\[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5\] Hence, we get $\dfrac{x}{150}=0.5,$ so $x=75.$
What is the least possible value of $(xy-1)^2+(x+y)^2$ for all real numbers $x$ and $y?$
$\textbf{(A) }0 \qquad \textbf{(B) }\dfrac14 \qquad \textbf{(C) }\dfrac12 \qquad \textbf{(D) }1 \qquad \textbf{(E) }2$
$\textbf{D}$
We expand the original expression, then factor the result by grouping:\begin{align*} (xy-1)^2+(x+y)^2&=\left(x^2y^2-2xy+1\right)+\left(x^2+2xy+y^2\right) \\ &=x^2y^2+x^2+y^2+1 \\ &=x^2\left(y^2+1\right)+\left(y^2+1\right) \\ &=\left(x^2+1\right)\left(y^2+1\right) \end{align*}Clearly, both factors are positive. Hence, we have\[\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=1\]The least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$
Which of the following is equivalent to\[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\]
$\textbf{(A) }3^{127}+2^{127} \qquad \textbf{(B) }3^{127}+2^{127}+2\cdot 3^{63}+3\cdot 2^{63} \qquad \textbf{(C) }3^{128}-2^{128} \qquad \textbf{(D) }3^{128}+2^{128} \qquad \textbf{(E) }5^{127}$
$\textbf{C}$
By multiplying $3-2=1$ at the beginning of the original expression, all terms can be simplified by the formula for the difference of squares $a^2-b^2=(a+b)(a-b)$, one by one. The final result is $3^{128}-2^{128}$.
For which of the following integers $b$ is the base-$b$ number $2021_b - 221_b$ not divisible by $3$?
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$
$\textbf{E}$
$$2021_b-221_b=2\times b^3+2\times b^1+1-2\times b^2-2\times b^1-1=2b^2(b-1)$$ which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is E.
Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3 \text{ cm}$ and $6 \text{ cm}$. Into each cone is dropped a spherical marble of radius $1 \text{ cm}$, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$
$\textbf{E}$
Let the heights of the liquid in the narrow cone and the wide cone be $h_1$ and $h_2,$ respectively. The volume of the liquid in the narrow cone is $\dfrac13\pi\times3^2\times h_1=3\pi h_1$. The volume of the liquid in the wide cone is $\dfrac13\pi\times6^2\times h_2=12\pi h_2$. Since the volumes of liquid are the same, we have $$3\pi h_1=12\pi h_2$$ After dropping the marbles, let the rises of the liquid levels in the narrow cone and the wide cone be $\Delta h_1$ and $\Delta h_2$. Now the volume is $$3\pi(h_1+\Delta h_1)=12\pi(h_2+\Delta h_2)$$ So we have \begin{align*}3\pi\Delta h_1&=12\pi\Delta h_2\\ \dfrac{\Delta h_1}{\Delta h_2}&=4\end{align*}
What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$, $AC = 3$, $AD = 4$, $BC = \sqrt{13}$, $BD = 2\sqrt{5}$, and $CD = 5$ ?
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$
$\textbf{C}$
We notice that $AB^2+AC^2=BC^2$. So $\triangle ABC$ is a right triangle. Furthermore, we also notice that $AC^2+AD^2=CD^2$, $AB^2+AD^2=BD^2$. Hence, we put point $A$ at the origin of the $xyz$ coordinate system, then $B$, $C$, $D$ on the $x$, $y$, and $z$ axis as $B(2,0,0)$, $C(0,3,0)$, $D(0,0,4)$.
The volume of tetrahedron $ABCD$ is $\dfrac13\times\dfrac12(2\times3)\times4=4$.
All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$?
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$
$\textbf{A}$
The polynomial can be reorganized as $$z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16=(z-z_1)(z-z_2)(z-z_3)(z-z_4)(z-z_5)(z-z_6)$$ where $z_1,z_2.z_3,z_4,z_5$ and $z_6$ are six positive integer roots. By polynomial expansion (or Vieta's formulas), the sum of the six roots is $10$ (the opposite number of the coefficient of $z^5$ term), and the product of the six roots is $16$ (the constant term). By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$. So the polynomial is $$(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$$ Therefore, calculating just the coefficient of the $z^3$ terms, we get $B = -32 - 2\times24 - 8 = -88$.
Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$)
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$
$\textbf{C}$
Both parabolas have $y$-axis symmetry and go upwards. Parabola $y=Ax^2+B$ intersects the $y$-axis at point $(0,B)$. Parabola $y=Cx^2+D$ intersects the $y$-axis at point $(0,D)$. Since order doesn't matter, we assume the first curve is above the second on the $y$-axis. Now we have $B>D$. To make the two curves intersect, the lower curve must have a larger growth rate, which means $C>A$. Therefore, we need to choose two pairs of numbers $(A,C)$ and $(B,D)$. The number of ways to choose the two pairs is $\dbinom{6}{2}\dbinom{4}{2}=90$.
In the following list of numbers, the integer $n$ appears $n$ times in the list for $1\le n \le 200$.\[1,2,2,3,3,3,4,4,4,...,200,200,...,200\]What is the median of the numbers in this list?
$\textbf{(A) }100.5 \qquad \textbf{(B) }134 \qquad \textbf{(C) }142 \qquad \textbf{(D) }150.5\qquad \textbf{(E) }167$
$\textbf{C}$
There are $1+2+\dots+199+200=\dfrac{200\times201}{2}=20100$ numbers in total. Let the median be $k$. We want to find the median $k$ such that \[\dfrac{k(k-1)}{2}<\dfrac{20100}2\leq\dfrac{k(k+1)}{2}\] or \[k(k-1)<20100\leq k(k+1)\] Note that $\sqrt{20100}\approx100\sqrt2 \approx 141$. We try $$141\times142=20022$$ $$142\times143=20306$$ Hence, we get $$141\times142<20100\leq142\times143$$ $142$ is the $10050$th and $10051$st numbers in the list. Therefore, the median is 142.
Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
$\textbf{D}$
Since $BC=CD$ and $P$ is the midpoint of $BD$, we know that $CP\perp BD$. Given that $AB\parallel CD$, we find that $\triangle ABD\sim\triangle CDP$. So $$\dfrac{AD}{CP}=\dfrac{BD}{DP}=2$$ We also find that $\triangle AOD\sim\triangle COP$, so $$\dfrac{OD}{OP}=\dfrac{AD}{CP}=2\rightarrow OD=2OP=22$$ By the Pythagorean theorem $$CP=\sqrt{CD^2-DP^2}=\sqrt{43^2-(11+22)^2}=2\sqrt{190}$$ So $AD=2CP=4\sqrt{190}$. The answer is $m+n=4+190=194$.
Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b)=f(a)+f(b)$ for all positive rational numbers $a$ and $b$. Suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following numbers $x$ is $f(x)<0$?
$\textbf{(A) }\dfrac{17}{32} \qquad \textbf{(B) }\dfrac{11}{16} \qquad \textbf{(C) }\dfrac79 \qquad \textbf{(D) }\dfrac76\qquad \textbf{(E) }\dfrac{25}{11}$
$\textbf{E}$
We know that $f(p) = f(p \cdot 1) = f(p) + f(1)$. Hence, we get $f(1)=0.$ Also \[f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2\]\[f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3\]\[f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11\] In $\textbf{(A)}$ we have $f\left(\dfrac{17}{32}\right)=17+5f\left(\dfrac{1}{2}\right)=17-5\times2=7$.
In $\textbf{(B)}$ we have $f\left(\dfrac{11}{16}\right)=11+4f\left(\dfrac{1}{2}\right)=11-4\times2=3$.
In $\textbf{(C)}$ we have $f\left(\dfrac{7}{9}\right)=7+2f\left(\dfrac{1}{3}\right)=7-2\times3=1$.
In $\textbf{(D)}$ we have $f\left(\dfrac{7}{6}\right)=7+f\left(\dfrac{1}{2}\right)+f\left(\dfrac{1}{3}\right)=7-2-3=2$.
In $\textbf{(E)}$ we have $f\left(\dfrac{25}{11}\right)=10+f\left(\dfrac{1}{11}\right)=10-11=-1$.
Thus, the answer is $\textbf{(E)}$.
The area of the region bounded by the graph of
$$x^2+y^2 = 3|x-y| + 3|x+y|$$
is $m+n\pi$, where $m$ and $n$ are integers. What is $m + n$?
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$
$\textbf{E}$
The line $x-y=0$ and $x+y=0$ divide the $xy$ plane into 4 regions.
$\textbf{Case 1:}$ $x-y>0$, $x+y>0$
Now we have $|x-y|=x-y, |x+y|=x+y$. Substituting and simplifying, we have $x^2-6x+y^2=0$, i.e. $(x-3)^2+y^2=3^2$, which gives us a circle of radius $3$ centered at $(3,0)$.
$\textbf{Case 2:}$ $x-y<0$, $x+y>0$
Now we have $|x-y|=y-x, |x+y|=x+y$. Substituting and simplifying again, we have $x^2+y^2-6y=0$, i.e. $x^2+(y-3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,3)$.
$\textbf{Case 3:}$ $x-y>0$, $x+y<0$
Now we have $|x-y|=x-y, |x+y|=-x-y$. Doing the same process as before, we have $x^2+y^2+6y=0$, i.e. $x^2+(y+3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,-3)$.
$\textbf{Case 4:}$ $x-y<0$, $x+y<0$
Now we have $|x-y|=y-x, |x+y|=-x-y$. One last time: we have $x^2+y^2+6x=0$, i.e. $(x+3)^2+y^2=3^2$. This gives us a circle of radius $3$ centered at $(-3,0)$.
After combining all the cases and drawing them on the $xy$ plane, this is what the diagram looks like:
The bounded region is a square with side length $6$ and four semicircles with radius $3$. The area is $6^2+4\times \dfrac{9\pi}{2} = 36+18\pi$.
In how many ways can the sequence $1, 2, 3, 4, 5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~32\qquad\textbf{(E)} ~44$
$\textbf{D}$
By symmetry with respect to $3,$ note that $(x_1,x_2,x_3,x_4,x_5)$ is a valid sequence if and only if $(6-x_1,6-x_2,6-x_3,6-x_4,6-x_5)$ is a valid sequence. We enumerate the valid sequences that start with 1, 2, 3-1 or 3-2, as shown below:
There are $16$ valid sequences that start with 1, 2, 3-1 or 3-2. By symmetry, there are $16$ valid sequences that start with 5, 4, 3-5 or 3-4. So the answer is $16+16=32.$
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 $m + n + p$?
$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$
$\textbf{C}$
Let $P,Q,R,X,Y,$ and $Z$ be the intersections $AB\cap CD,CD\cap EF,EF\cap AB,BC\cap DE,DE\cap FA,$ and $FA\cap BC,$ respectively.
The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ\div6=120^\circ.$ Furthermore, we find 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 $36+16+3=55.$
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?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$
$\textbf{B}$
Suppose the roommate took sheets $a$ through $b$, which means $b-a+1$ sheets are taken. The page numbers are $2a-1, 2a,\cdots, 2b$, for a total of $2(b-a+1)$ pages. The sum of these page numbers is $$\dfrac12(2a-1+2b)\cdot2(b-a+1)=(2a-1+2b)(b-a+1)$$ The number of rest pages is $50-2(b-a+1)$, with an average of 19. So the sum of numbers on the rest pages is $$19[50-2(b-a+1)]$$ The sum of page numbers for the whole notes is $$1+2+\cdots+50=\dfrac12\cdot50\cdot51$$ Hence, we have $$(2a-1+2b)(b-a+1)+19[50-2(b-a+1)]=\dfrac12\cdot50\cdot51$$which can be simplified as $$ (2a+2b-39)(b-a+1)=5^2\cdot13$$ Trying all possible solutions and we find $a=10$, $b=22$. The answer is $22-10+1=13$.
Frieda the frog begins a sequence of hops on a $3\times3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
$\textbf{(A) }\dfrac{9}{16} \qquad \textbf{(B) }\dfrac{5}{8} \qquad \textbf{(C) }\dfrac34 \qquad \textbf{(D) }\dfrac{25}{32}\qquad \textbf{(E) }\dfrac{13}{16}$
$\textbf{D}$
We can find the probability that the frog doesn't land on the corner within 4 steps, then subtracting it by 1 to get the answer.
After the first step, the frog lands on the middle edge.
After the second step, the frog has a probability of $\dfrac12$ to land on the corner, $\dfrac14$ to land on the center, and $\dfrac14$ to land on the middle edge. Here we have 2 cases to discuss.
$\textbf{Case 1:}$ the frog lands on the center with a probability of $\dfrac14$. After the third step, the frog lands on the middle edge. For the fourth step, the frog has a probability of $\dfrac12$ to not land on the corner. So in this case, the probability that the frog doesn't land on the corner within 4 steps is $\dfrac14\cdot\dfrac12=\dfrac18$.
$\textbf{Case 2:}$ the frog lands on the middle edge with a probability of $\dfrac14$. For the third step, she has a probability of $\dfrac14$ to land on the center, which means she can't land on the corner in the fourth step, or a probability of $\dfrac14$ to land on the middle edge, which means she has a probability of $\dfrac12$ to not land on the corner in the fourth step. So in this case, the probability that the frog doesn't land on the corner within 4 steps is $\dfrac14\left(\dfrac14+\dfrac14\cdot\dfrac12\right)=\dfrac{3}{32}$.
In conclusion, the probability that the frog doesn't land on the corner within 4 steps is $\dfrac18+\dfrac{3}{32}=\dfrac{7}{32}$. Our answer is $1-\dfrac{7}{32}=\dfrac{25}{32}$.
The interior of a quadrilateral is bounded by the graphs of $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$, where $a$ is a positive real number. What is the area of this region in terms of $a$, valid for all $a > 0$?
$\textbf{(A)} ~\dfrac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\dfrac{4a}{a+1}\qquad\textbf{(C)} ~\dfrac{8a}{a+1}\qquad\textbf{(D)} ~\dfrac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\dfrac{8a}{a^2+1}$
$\textbf{D}$
The cases for $(x+ay)^2 = 4a^2$ are $x+ay = \pm2a,$ or two parallel lines. We rearrange each case and construct the table below:
The cases for $(ax-y)^2 = a^2$ are $ax-y=\pm a,$ or two parallel lines. We rearrange each case and construct the table below:
Since the slopes of intersecting lines $(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),$ and $(2)\cap(2*)$ are negative reciprocals, we get four right angles, from which the quadrilateral is a rectangle.
Recall that for constants $A,B,C_1$ and $C_2,$ the distance $d$ between parallel lines $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ is\[d=\frac{\left|C_2-C_1\right|}{\sqrt{A^2+B^2}}.\]From this formula:
The distance between lines $(1)$ and $(2)$ is $\dfrac{4a}{\sqrt{1+a^2}},$ the length of this rectangle.
The distance between lines $(1*)$ and $(2*)$ is $\dfrac{2a}{\sqrt{a^2+1}},$ the width of this rectangle.
The area we seek is\[\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\frac{8a^2}{a^2+1}\]
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?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$
$\textbf{E}$
Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors in certain configurations, 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 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 three B's and 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. Hence, there are $6\times6=36$ ways in total.