PROBLEM 1 :
What is the unit digit of:
$$
222,222-22,222-2,222-222-22-2 ?
$$
(A) 0
(B) 2
(C) 4
(D) 8
(E) 10
SOLUTION :
$$
\begin{gathered}
222,222-22,222=200,000 \
200,000-2,222=197778 \
197778-222=197556 \
197556-22=197534 \
197534-2=197532
\end{gathered}
$$
So our answer is (B) 2.
PROBLEM 2 :
What is the value of this expression in decimal form?
$$
\frac{44}{11}+\frac{110}{44}+\frac{44}{1100}
$$
(A) 6.4
(B) 6.504
(C) 6.54
(D) 6.9
(E) 6.94
SOLUTION :
We see that $\frac{44}{11}$ is $4 ; \frac{110}{44}$ simplifies to $\frac{5}{2}$, which is 2.5 ; and $\frac{44}{1100}$ simplifies to $\frac{1}{25}$, which is 0.04 ; $4+2.5+0.04$ reveals
$$
\frac{44}{11}+\frac{110}{44}+\frac{44}{1100}
$$
is (C) $6.54 .
PROBLEM 3 :
Four squares of side length $4,7,9$, and 10 are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units?
(A) 42
(B) 45
(C) 49
(D) 50
(E) 52
SOLUTION :
We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is
$$
10^2-9^2+7^2-4^2=100-81+49-16=19+33=\text { (E) } 52
$$
PROBLEM 4 :
When Yunji added all the integers from 1 to 9 , she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
SOLUTION :
Adding numbers 1 through 9 gives us 45 . This was her expected sum, but what she got was a perfect square. Since she got that perfect square sum by forgetting a number, that sum is less than 45 . The square number right under 45 is 36 . So $45-36=9$. So the solution is (E) 9
PROBLEM 5 :
Aaliyah rolls two standard 6 -sided dice. She notices that the product of the two numbers rolled is a multiple of 6 . Which of the following integers cannot be the sum of the two numbers?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
SOLUTION :
First, figure out all pairs of numbers whose product is 6 . Then, using the process of elimination, we can find the following:
$(\mathrm{A})$ is possible: $2 \times 3$
$(\mathrm{C})$ is possible: $1 \times 6$
$(\mathrm{D})$ is possible: $2 \times 6$
The only integer that cannot be the sum is (B) 6 .
We can prove that $B$ is correct because the only ordered pairs that add to 6 is $(1,5),(5,1),(2,4),(4,2)$, and , $(3,3)$, and these multiply to 5,8 , and 9.
PROBLEM 6 :
Sergai skated around an ice rink, gliding along different paths. The gray lines in the figures below show four of the paths labeled $P, Q, R$, and $S$. What is the sorted order of the four paths from shortest to longest?
(A) $P, Q, R, S$
(B) $P, R, S, Q$
(C) $Q, S, P, R$
(D) $R, P, S, Q$
(E) $R, S, P, Q$
SOLUTION :
Obviously Path Q is the longest path, followed by Path S.
So, it is down to Paths P and R.
Notice that curved lines are always longer than the straight ones that meet their endpoints, therefore Path P is longer than Path R.
Thus, the order from shortest to longest is (D) R, P, S, Q.
PROBLEM 7 :
A $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2,1 \times 4$, and $1 \times 1$, shown below. What is the minimum possible number of $1 \times 1$ tiles used?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
SOLUTION :
Let $x$ be the number of 1 by 1 tiles. There are 21 squares and each 2 by 2 or 1 by 4 tile takes up 4 squares, so $x \equiv 1(\bmod 4)$, so it is either 1 or 5 . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are 12 red squares and 9 blue squares, but each 2 by 2 and 1 by 4 shape takes up an equal number of blue and red squares, so there must be 3 more 1 by 1 tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is □ (E) which can easily be confirmed to work.
PROBLEM 8 :
On Monday, Taye has $\$ 2$. Every day, he either gains $\$ 3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, 3 days later?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
SOLUTION :
How many dollar values could be on the first day? Only 2 dollars. The second day, you can either add 3 dollars, or double, so you can have 5 dollars, or 4 . For each of these values, you have 2 values for each. For 5 dollars, you have 10 dollars or 8 , and for 4 dollars, you have 8 dollars or 7 dollars. Now, you have 2 values for each of these. For 10 dollars, you have 13 dollars or 20 , for 8 dollars, you have 16 dollars or 11 , for 8 dollars, you have 16 dollars or 11, and for 7 dollars, you have 14 dollars or 10 .
On the final day, there are $11,11,16$, and 16 repeating, leaving you with $8-2=$ □ (D) 6 different values.
PROBLEM 9 :
All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles, and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
(A) 24
(B) 25
(C) 26
(D) 27
(E) 28
SOLUTION :
Since Maria has half as many red marbles as green, we can call the number of red marbles $x$, and the number of green marbles $2 x$. Since she has half as many green marbles as blue, we can call the number of blue marbles $4 x$. Adding them up, we have: $7 x$ marbles. The number of marbles therefore must be a multiple of 7 , as $x$ represents an integer, so the only possible answer is □ (E) 28 .
PROBLEM 10 :
In January 1980 the Mauna Loa Observatory recorded carbon dioxide (CO2) levels of 338 ppm (parts per million). Over the years the average $C O 2$ reading has increased by about 1.515 ppm each year. What is the expected $C O 2$ level in ppm in January 2030 ? Round your answer to the nearest integer.
(A) 399
(B) 414
(C) 420
(D) 444
(E) 459
SOLUTION :
This is a time period of $2030-1980=50$ years, so we can expect the ppm to increase by $50 * 1.515=75.75 \approx 76 \mathrm{ppm}$. Since we started with 338 ppm , we have $76+338= (\text{B}) {4 1 4}$.
PROBLEM 11 :
The coordinates of $\triangle A B C$ are $A(5,7), B(11,7)$, and $C(3, y)$, with $y>7$. The area of $\triangle A B C$ is 12 . What is the value of $y$ ?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
SOLUTION :
By the Shoelace Theorem, $\triangle A B C$ has area
$$
\frac{1}{2}|(y \cdot 11+7 \cdot 5+7 \cdot 3)-(3 \cdot 7+11 \cdot 7+5 \cdot y)|=\frac{1}{2}|(11 y+56)-(98+5 y)|=\frac{1}{2}|6 y-42| .
$$
From the problem, this is equal to 12 . We now solve for y .
$$
\begin{aligned}
& \frac{1}{2}|6 y-42|=12 \
& |6 y-42|=24 \
& 6 y-42=24 \text { OR } 6 y-42=-24 \
& 6 y=66 \text { OR } 6 y=18 \
& y=11 \text { OR } y=3
\end{aligned}
$$
However, since, as stated in the problem, $y>7$, our only valid solution is (D) 11 .
PROBLEM 12 :
Rohan keeps 90 guppies in 4 fish tanks.
How many guppies are in the 4th tank?
(A) 20
(B) 21
(C) 23
(D) 24
(E) 26
SOLUTION :
Let $x$ denote the number of guppies in the first tank.
Then, we have the following for the number of guppies in the rest of the tanks:
The number of guppies in all of the tanks combined is 90 , so we can write the equation
$$
(x)+(x+1)+(x+1+2)+(x+1+2+3)=90 .
$$
Simplifying the equation gives
$$
4 x+10=90
$$
Solving the resulting equation gives $x=20$, so the number of guppies in the fourth tank is $20+1+2+3=(\mathbf{E}) 26$.
PROBLEM 13 :
Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz Bunny start on the ground, make a sequence of 6 hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.)
(A) 4
(B) 5
(C) 6
(D) 8
(E) 12
SOLUTION :
Any combination can be written as some re-arrangement of $U U U D D D$. Clearly we must end going down, and start going up, so we need the number of ways to insert $2 U$ 's and $2 D$ 's into $U \ldots D$. There are $\binom{4}{2}=6$ ways, but we have to remove the case UDDUUD , giving us (B) 5 .
We know there are no more cases since there will be at least one $U$ before we have a $D$ (from the first $U$ ), at least two $U$ 'S before two $D$ 's (since we removed the one case), and at least three $U$ 's before three $D$ 's, as we end with the third $D$.
PROBLEM 14 :
The one-way routes connecting towns $A, M, C, X, Y$, and $Z$ are shown in the figure below(not drawn to scale). The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
(A) 28
(B) 29
(C) 30
(D) 31
(E) 32
SOLUTION :
We can simply see that path $A \rightarrow X \rightarrow M \rightarrow Y \rightarrow C \rightarrow Z$ will give us the smallest value. Adding, $5+2+6+5+10=$ □ 28 This is nice as it's also the smallest value, solidifying our answer.
You can also simply brute-force it or sort of think ahead - for example, getting from A to M can be done 2 ways; $A \rightarrow X \rightarrow M(5+2)$ or $A \rightarrow M(8)$, so you should take the shorter route ( $5+2$ ). Another example is M to C , two ways - one is $6+5$ and the other is 14 . Take the shorter route. After this, you need to consider a few more times - consider if $5+10(Y \rightarrow C \rightarrow Z)$ is greater than $17(Y \rightarrow Z)$, which it is not, and consider if $25(M \rightarrow Z)$ is greater than $14+10(M \rightarrow C \rightarrow Z)$ or $6+5+10(M \rightarrow Y \rightarrow C \rightarrow Z)$ which it is not. TLDR: $5+2+6+5+10=$ □ 28 [Note: This is probably just the thinking behind the solution.] {Double-note: As MaxyMoosy said, since this answer is the smallest one, it has to be the right answer.}
PROBLEM 15 :
Let the letters $F, L, Y, B, U, G$ represent distinct digits. Suppose $\underline{F} \underline{L} \underline{Y} \underline{F} \underline{L} \underline{Y}$ is the greatest number that satisfies the equation
$$
8 \cdot \underline{F} \underline{L} \underline{Y} \underline{F} \underline{L} \underline{Y}=\underline{B} \underline{U} \underline{G} \underline{B} \underline{U} \underline{G} .
$$
What is the value of $\underline{F} \underline{L} \underline{Y}+\underline{B} \underline{U} \underline{G}$ ?
(A) 1089
(B) 1098
(C) 1107
(D) 1116
(E) 1125
SOLUTION :
Notice that $\underline{F} \underline{L} \underline{Y} \underline{F} \underline{L} \underline{Y}=1000(\underline{F} \underline{L} \underline{Y})+\underline{F} \underline{L} \underline{Y}$.
Likewise, $\underline{B} \underline{U} \underline{G} \underline{B} \underline{U} \underline{G}=1000(\underline{B} \underline{U} \underline{G})+\underline{B} \underline{U} \underline{G}$.
Therefore, we have the following equation:
$$
8 \times 1001(\underline{F} \underline{L} \underline{Y})=1001(\underline{B} \underline{U} \underline{G})
$$
Simplifying the equation gives
$$
8(\underline{F} \underline{L} \underline{Y})=(\underline{B} \underline{U} \underline{G})
$$
We can now use our equation to test each answer choice.
We have that $123123 \times 8=984984$, so we can find the sum:
$$
\underline{F} \underline{L} \underline{Y}+\underline{B} \underline{U} \underline{G}=123+984=1107 .
$$
So, the correct answer is (C) 1107 .
PROBLEM 16 :
Minh enters the numbers 1 through 81 into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by 3 ?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
SOLUTION :
Note you can swap/rotate any configuration of rows, such that all the rows and columns that have a product of 3 are in the top left. Hence the points are bounded by a $a \times b$ rectangle. This has $a b$ area and $a+b$ rows and columns divisible by 3 . We want $a b \geq 27$ and $a+b$ minimized.
If $a b=27$, we achieve minimum with $a+b=9+3=12$.
If $a b=28$,our best is $a+b=7+4=11$. Note if $a+b=10, a b=25$. Because $25<27$, there is no smaller answer, and we get (D) 11 .
PROBLEM 17 :
A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a $3 \times 3$ grid attacks all 8 other squares, as shown below. Suppose a white king and a black king are placed on different squares of a 3 $x 3$ grid so that they do not attack each other (in other words, not right next to each other). In how many ways can this be done?
(A) 20
(B) 24
(C) 27
(D) 28
(E) 32
SOLUTION :
If you place a king in any of the 4 corners, the other king will have 5 spots to go and there are 4 corners, so $5 \times 4=20$. If you place a king in any of the 4 edges, the other king will have 3 spots to go and there are 4 edges so $3 \times 4=12$. That gives us $20+12=32$ spots for the other king to go into in total. So □ (E) 32 is the answer.
PROBLEM 18 :
Three concentric circles centered at $O$ have radii of 1, 2, and 3 . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angles $B O C$, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle B O C$ in degrees?
SOLUTION :
Let $x=\angle B O C$.
We see that the shaded region is the inner ring plus a sector $x^{\circ}$ of the outer ring. Using the formula for the area of a circle ( $A=\pi r^2$ ), we find that the area of $x$ is $(4 \pi-\pi)+\frac{x}{360}(9 \pi-4 \pi)$. This simplifies to $3 \pi+\frac{x}{360}(5 \pi)$.
The unshaded portion is comprised of the smallest circle plus the sector $(360-x)^{\circ}$ of the outer ring, which evaluates to $\pi+\frac{360-x}{360}(5 \pi)$.
We are told these are equal. Therefore, $3 \pi+\frac{x}{360}(5 \pi)=\pi+\frac{360-x}{360}(5 \pi)$. Solving for $x$ reveals $x=(\mathbf{A}) 108$.
PROBLEM 19 :
Jordan owns 15 pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction?
(A) 0
(B) $\frac{1}{5}$
(C) $\frac{4}{15}$
(D) $\frac{1}{3}$
(E) $\frac{2}{5}$
SOLUTION :
We first start by finding the number of red and white sneakers. $\frac{3}{5} \times 15=9$ red sneakers, so 6 are white. Then $\frac{2}{3} \times 15=10$ are high top sneakers, so 5 are low top sneakers. Now think about 15 slots, and the first 10 are labeled high-top sneakers. If we insert the last 5 sneakers as red sneakers, there are 4 leftover red sneakers. Putting those 4 sneakers as high top sneakers, we have our answer as C , or $\frac{4}{15}$.
PROBLEM 20 :
Any three vertices of the cube $P Q R S T U V W$, shown in the figure below, can be connected to form a triangle. (For example, vertices $P, Q$, and $R$ can be connected to form isosceles $\triangle P Q R$.) How many of these triangles are equilateral and contain $P$ as a vertex?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 6
SOLUTION :
The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have 3 possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are 3 possible triangles. So the answer is (D).
PROBLEM 21 :
A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was $3: 1$. Then 3 green frogs moved to the sunny side and 5 yellow frogs moved to the shady side. Now the ratio is $4: 1$. What is the difference between the number of green frogs and the number of yellow frogs now?
(A) 10
(B) 12
(C) 16
(D) 20
(E) 24
SOLUTION :
Let the initial number of green frogs be $g$ and the initial number of yellow frogs be $y$. Since the ratio of the number of green frogs to yellow frogs is initially $3: 1, g=3 y$. Now, 3 green frogs move to the sunny side and 5 yellow frogs move to the shade side, thus the new number of green frogs is $g+2$ and the new number of yellow frogs is $y-2$. We are given that $\frac{g+2}{y-2}=\frac{4}{1}$, so $g+2=4 y-8$, since $g=3 y$, we have $3 y+2=4 y-8$, so $y=10$ and $g=30$. Thus the answer is $(g+2)-(y-2)=32-8=(E) 24$.
PROBLEM 22 :
A roll of tape is 4 inches in diameter and is wrapped around a ring that is 2 inches in diameter. A cross section of the tape is shown in the figure below. The tape is 0.015 inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest 100 inches.
(A) 300
(B) 600
(C) 1200
(D) 1500
(E) 1800
SOLUTION :
The roll of tape is $1 / 0.015 \approx 66$ layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by 66 . Since the diameter of the small circle is 2 inches and the diameter of the large one is 4 inches, the "middle value" (or mean) is 3 . Therefore, the average circumference is $3 \pi$. Multiplying $3 \pi \cdot 66$ gives approximately $(B) 600$.
PROBLEM 23 :
Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the 4 cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$. How many cells will he color this time?
(A) 6000
(B) 6500
(C) 7000
(D) 7500
(E) 8000
SOLUTION :
Let $f(x, y)$ be the number of cells the line segment from $(0,0)$ to $(x, y)$ passes through. The problem is then equivalent to finding
$$
f(5000-2000,8000-3000)=f(3000,5000) .
$$
Sometimes the segment passes through lattice points in between the endpoints, which happens $\operatorname{gcd}(3000,5000)-1=999$ times. This partitions the segment into 1000 congruent pieces that each pass through $f(3,5)$ cells, which means the answer is
$$
1000 f(3,5) .
$$
Note that a new square is entered when the lines pass through one of the lines in the coordinate grid, which for $f(3,5)$ happens $3-1+5-1=6$ times. Because 3 and 5 are relatively prime, no lattice point except for the endpoints intersects the line segment from $(0,0)$ to $(3,5)$. This means that including the first cell closest to $(0,0)$, The segment passes through $f(3,5)=6+1=7$ cells. Thus, the answer is (C)7000.
PROBLEM 24 :
Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is 8 feet high while the other peak is 12 feet high. Each peak forms a $90^{\circ}$ angle, and the straight sides form a $45^{\circ}$ angle with the ground. The artwork has an area of 183 square feet. The sides of the mountain meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h$ ?
(A) 4
(B) 5
(C) $4 \sqrt{2}$
(D) 6
(E) $5 \sqrt{2}$
SOLUTION :
Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown.
The side length of the largest right triangle is $12 \sqrt{2}$, which means its area is 144 . Similarly, the area of the second largest right triangle is 64 (the side length is $8 \sqrt{2}$ ), and the area of the overlap is $h^2$ (the side length is $h \sqrt{2}$ ). Because the right triangles have a side ratio of $1: 1: \sqrt{2}$. Thus,
$$
144+64-h^2=183,
$$
which means that the answer is (B) 5 .
PROBLEM 25 :
A small airplane has 4 rows of seats with 3 seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?
(A) $\frac{8}{15}$
(B) $\frac{32}{55}$
(C) $\frac{20}{33}$
(D) $\frac{34}{55}$
(E) $\frac{8}{11}$
SOLUTION :
Suppose the passengers are indistinguishable. There are $\binom{12}{8}=495$ total ways to distribute the passengers. We proceed with complementary counting, and instead, will count the number of passenger arrangements such that the couple cannot sit anywhere. Consider the partitions of 8 among the rows of 3 seats, to make our lives easier, assuming they are non-increasing. We have $(3,3,2,0),(3,3,1,1),(3,2,2,1),(2,2,2,2)$.
For the first partition, clearly, the couple will always be able to sit in the row with 0 occupied seats, so we have 0 cases here.
For the second partition, there are $\frac{4!}{2!2!}=6$ ways to permute the partition. Now the rows with exactly 1 passenger must be in the middle, so this case generates 6 cases.
For the third partition, there are $\frac{4!}{2!}=12$ ways to permute the partition. For rows with 2 passengers, there are $\binom{3}{2}=3$ ways to arrange them in the row so that the couple cannot sit there. The row with 1 passenger must be in the middle. We obtain $12 \cdot 3^2=108$ cases.
For the fourth partition, there is 1 way to permute the partition. As said before, rows with 2 passengers can be arranged in 3 ways, so we obtain $3^4=81$ cases.
Collectively, we obtain a total of $6+108+81=195$ cases. The final probability is $1-\frac{195}{495}=$ (C) $\frac{20}{33}$.
