Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$ . What is the sum of the digits of $N$
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair be $(x-10)$ and $(x+10)$ as well as $(y-\frac{23}{2})$ and $(y+\frac{23}{2})$ . We also know the product of both the complementary divisors give the same number so $(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})$ . 
Now we let $y=\frac{a}{2}$ . Then we substitute and get $x^2-100=\frac{(a^2-529)}{4}$ . Finally we multiply by 4 and get $4x^2-a^2=-129, a^2-4x^2=129$ . 
Then we use differences of squares and get $a$ $2x$ =129, $a$ $2x$ =1. We finish by getting $a=$ 65 and $x=32$ . So $(42)(22) = 924$ Adding the digits, we have $9+2+4 = \boxed{15}$

Answer: C 15
Problem: If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$ . What is the sum of the digits of $N$
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: $N$ can be written $N = \left( a - 10 \right) \left( a + 10 \right)$ with a positive integer $a > 10$ and $N = \left( \frac{2b + 1}{2} - \frac{23}{2} \right) \left( \frac{2b + 1}{2} + \frac{23}{2} \right)$ with a positive integer $b > 11$
The above equations can be reorganized as \[ \left( 2b + 1 + 2 a \right) \left( 2 b + 1 - 2 a \right) = 43 \cdot 3 . \]
The only solution is $2b + 1 + 2a = 129$ and $2b + 1 - 2a = 1$ .
Thus, $a = b = 32$ .
Therefore, $N = 924$ .
So the sum of the digits of $N$ is $9 + 2 + 4 = \boxed{15}$

Answer: C 15
Problem: If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$ . What is the sum of the digits of $N$
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: We can write $N$ as $a(a+20)$ or $b(b+23)$ where $a$ and $b$ are divisors of $N.$ Since $a(a+20) = b(b+23),$ we know that $a^2 + 20a - b^2 - 23b = 0$ , and we can view this as a quadratic in $a.$
Since the solution for $a$ must be an integer, the discriminant for this quadratic must be a perfect square and therefore $20^2 - 4(-b^2 - 23b) = (2c)^2 = 400 + 4b^2 + 92b$ so $b^2 + 23b -c^2 + 100 = 0.$
Since the discriminant of this quadratic in $b$ must also be a perfect square we know that $23^2 - 4(-c^2+100) = d^2$ which we can simplify as $d^2 - 4c^2 = (d-2c)(d+2c) = 129.$ Since they are both positive integers $d - 2c$ and $d + 2c$ are factors of $129 = 3 \cdot 43$ so $d - 2c = 1$ and $d + 2c = 129$ or $d - 2c = 3$ and $d - 2c = 43.$
These systems of equations give us $(c,d) = (32,65)$ and $(c,d) = (10,23)$ respectively, if we plug our values for $c$ into the equation for $b$ we get $b^2 + 23b - 924 = 0$ and $b^2 + 23b = 0$ respectively. The first equation gives us $b = 21$ or $b = -44$ and the second gives us $b = 0$ or $b = -23$ , since $b$ is positive we know that $b = 21$ and $N = (21)(21 + 23) = 924$ , therefore the sum of the digits of $N$ is $9 + 2 + 4 = \boxed{15}.$

Answer: C 15
Problem: If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$ , then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$ . Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$ . What is the sum of the digits of $N$
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Consider the numbers of the form $a(a+20)$ . Since $b(b+23)$ is always even, $a$ is even. Thus, for $a \ge 2$ , we calculate $a(a+20)$ for even values of $a$ . Then, we check if it can also be represented as a product of numbers that differ by $23$ . Checking, we see that $22 \cdot 42 = 21 \cdot 44 = 924$ works. Thus, the answer is $9 + 2 + 4 = \boxed{15}$

Answer: C 15
Problem: If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$ . For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$ , but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ is not an edge, then $d(A, B) = 2$ . Let $Q$ $R$ , and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that $d(Q, R) > d(R, S)$
$\textbf{(A) } \frac{7}{22} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{3}{8} \qquad \textbf{(D) } \frac{5}{12} \qquad \textbf{(E) } \frac{1}{2}$

Solution: Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and 
S basically the same, we can first count the probability that $d(Q,R) = d(R,S)$
$\mathfrak{Case} \ \mathfrak{1}: d(Q,R) = d(R,S) = 1$
There are 5 points $P$ such that $d(Q,P) = 1$ . There is $5 \times 4 = \boxed{20}$ ways to choose Q and S in this case.

Answer:  20
Problem: Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?
$\textbf{(A) } \frac{1}{12} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{6} \qquad\textbf{(D) } \frac{1}{8} \qquad\textbf{(E) } \frac{2}{9}$

Solution: Given that the first three glasses are full and the fourth is only $\frac{1}{3}$ full, let's represent their contents with a common denominator, which we'll set as 6. This makes the first three glasses $\dfrac{6}{6}$ full, and the fourth glass $\frac{2}{6}$ full.
To equalize the amounts, Mrs. Jones needs to pour juice from the first three glasses into the fourth. Pouring $\frac{1}{6}$ from each of the first three glasses will make them all $\dfrac{5}{6}$ full. Thus, all four glasses will have the same amount of juice. Therefore, the answer is $\boxed{16}.$

Answer: C 16
Problem: Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?
$\textbf{(A) } \frac{1}{12} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{6} \qquad\textbf{(D) } \frac{1}{8} \qquad\textbf{(E) } \frac{2}{9}$

Solution: We let $x$ denote how much juice we take from each of the first $3$ children and give to the $4$ th child.
We can write the following equation: $1-x=\dfrac13+3x$ , since each value represents how much juice each child (equally) has in the end. (Each of the first three children now have $1-x$ juice, and the fourth child has $3x$ more juice on top of their initial $\dfrac13$ .)
Solving, we see that $x=\boxed{16}.$

Answer: C 16
Problem: Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$

Solution: We can create the equation: \[0.8x \cdot 1.075 = 43\] using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get \[\frac{4}{5} \cdot x \cdot \frac{43}{40} = 43\] \[\frac{4}{5} \cdot x \cdot \frac{1}{40} = 1\] \[\frac{1}{5} \cdot x \cdot \frac{1}{10} = 1\] \[x  = \boxed{50}\]

Answer:  50
Problem: Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$

Solution: The discounted shoe is $20\%$ off the original price. So that means $1 - 0.2 = 0.8$ . There is also a $7.5\%$ sales tax charge, so $0.8 * 1.075 = 0.86$ . Now we can set up the equation $0.86x = 43$ , and solving that we get $x=\boxed{50}$ ~ kabbybear

Answer: B 50
Problem: Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$

Solution: Let the original price be $x$ dollars. 
After the discount, the price becomes $80\%x$ dollars. 
After tax, the price becomes $80\% \times (1+7.5\%) = 86\% x$ dollars.
So, $43=86\%x$ $x=\boxed{50}.$

Answer: B 50
Problem: Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$

Solution: We can assign a variable $c$ to represent the original cost of the shoes. Next, we set up the equation $80\%\cdot107.5\%\cdot c=43$ . We can solve this equation for $c$ and get $\boxed{50}$

Answer: B 50
Problem: Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$

Solution: We know the discount price will be 5/4, and 0.075 is equal to 3/40. So we look at answer choice $\textbf{(B) }$ , see that the discount price will be 40, and with sales tax applied it will be 43, so the answer choice is $\boxed{50}$

Answer: B 50
Problem: Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250$

Solution: $6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$ , so the answer is $\boxed{1,625}$

Answer: C 1,625
Problem: Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250$

Solution: $6.5$ millimeters can be represented as $65 \times 10^{-2}$ centimeters. $25$ meters is $25 \times 10^{2}$ centimeters. Multiplying out these results in $(65 \times 10^{-2}) \times (25 \times 10^{2})$ , which is $65 \times 25$ making the answer $\boxed{1,625}$

Answer: C 1,625
Problem: Maddy and Lara see a list of numbers written on a blackboard. Maddy adds $3$ to each number in the list and finds that the sum of her new numbers is $45$ . Lara multiplies each number in the list by $3$ and finds that the sum of her new numbers is also $45$ . How many numbers are written on the blackboard?
$\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution: Let there be $n$ numbers in the list of numbers, and let their sum be $S$ . Then we have the following
\[S+3n=45\]
\[3S=45\]
From the second equation, $S=15$ . So, $15+3n=45$ $\Rightarrow$ $n=\boxed{10}.$

Answer: A 10
Problem: Maddy and Lara see a list of numbers written on a blackboard. Maddy adds $3$ to each number in the list and finds that the sum of her new numbers is $45$ . Lara multiplies each number in the list by $3$ and finds that the sum of her new numbers is also $45$ . How many numbers are written on the blackboard?
$\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution: Let $x_1,x_2,x_3,...,x_n$ where $x_n$ represents the $n$ th number written on the board. Lara's multiplied each number by $3$ , so her sum will be $3x_1+3x_2+3x_3+...+3x_n$ . This is the same as $3\cdot (x_1+x_2+x_3+...+x_n)$ . We are given this quantity is equal to $45$ , so the original numbers add to $\frac{45}{3}=15$ . Maddy adds $3$ to each of the $n$ terms which yields, $x_1+3+x_2+3+x_3+3+...+x_n+3$ . This is the same as the sum of the original series plus $3 \cdot n$ . Setting this equal to $45$ $15+3n=45 \Rightarrow n =\boxed{10}.$

Answer: A 10
Problem: Let $L_{1}=1, L_{2}=3$ , and $L_{n+2}=L_{n+1}+L_{n}$ for $n\geq 1$ . How many terms in the sequence $L_{1}, L_{2}, L_{3},...,L_{2023}$ are even?
$\textbf{(A) }673\qquad\textbf{(B) }1011\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }674$

Solution: We calculate more terms: \[1,3,4,7,11,18,\ldots.\] We find a pattern: if $n$ is a multiple of $3$ , then the term is even, or else it is odd. There are $\left\lfloor \frac{2023}{3} \right\rfloor =\boxed{674}$ multiples of $3$ from $1$ to $2023$

Answer: E 674
Problem: Let $L_{1}=1, L_{2}=3$ , and $L_{n+2}=L_{n+1}+L_{n}$ for $n\geq 1$ . How many terms in the sequence $L_{1}, L_{2}, L_{3},...,L_{2023}$ are even?
$\textbf{(A) }673\qquad\textbf{(B) }1011\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }674$

Solution: Like in the other solution, we find a pattern, except in a more rigorous way.
Since we start with $1$ and $3$ , the next term is $4$
We start with odd, then odd, then (the sum of odd and odd) even, (the sum of odd and even) odd, and so on. Basically the pattern goes: odd, odd, even, odd odd, even, odd, odd even…
When we take $\frac{2023}{3}$ we get $674$ with a remainder of one. So we have $674$ full cycles, and an extra odd at the end.
Therefore, there are $\boxed{674}$ evens.

Answer: E 674
Problem: Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$ , as shown below. IMG 1031.jpeg
What is the degree measure of $\angle EAB$
$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$

Solution: First, let's call the center of both squares $I$ . Then, $\angle{AIE} = 20$ , and since $\overline{EI} = \overline{AI}$ $\angle{AEI} = \angle{EAI} = 80$ . Then, we know that $AI$ bisects angle $\angle{DAB}$ , so $\angle{BAI} = \angle{DAI} = 45$ . Subtracting $45$ from $80$ , we get $\boxed{35}$

Answer: B 35
Problem: Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$ , as shown below. IMG 1031.jpeg
What is the degree measure of $\angle EAB$
$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$

Solution: First, label the point between $A$ and $H$ point $O$ and the point between $A$ and $H$ point $P$ . We know that $\angle{AOP} = 20$ and that $\angle{A} = 90$ . Subtracting $20$ and $90$ from $180$ , we get that $\angle{APO}$ is $70$ . Subtracting $70$ from $180$ , we get that $\angle{OPB} = 110$ . From this, we derive that $\angle{APE} = 110$ . Since triangle $APE$ is an isosceles triangle, we get that $\angle{EAP} = (180 - 110)/2 = 35$ . Therefore, $\angle{EAB} = 35$ . The answer is $\boxed{35}$

Answer: B 35
Problem: Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$ , as shown below. IMG 1031.jpeg
What is the degree measure of $\angle EAB$
$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$

Solution: Call the center of both squares point $O$ , and draw circle $O$ such that it circumscribes the squares. $\angle{EOF} = 90$ and $\angle{BOF} = 20$ , so $\angle{EOB} = 70$ . Since $\angle{EAB}$ is inscribed in arc $\overset \frown {EB}$ $\angle{EAB} = 70/2 = \boxed{35}$

Answer: B 35
Problem: Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$ , as shown below. IMG 1031.jpeg
What is the degree measure of $\angle EAB$
$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$

Solution: Draw $EA$ : we want to find $\angle EAB$ . Call $P$ the point at which $AB$ and $EH$ intersect. Reflecting $\triangle APE$ over $EA$ , we have a parallelogram. Since $\angle EPB = 70^{\circ}$ , angle subtraction tells us that two of the angles of the parallelogram are $110^{\circ}$ . The other two are equal to $2\angle EAB$ (by properties of reflection).
Since angles on the transversal of a parallelogram sum to $180^{\circ}$ , we have $2\angle EAB + 110 = 180$ , yielding $\angle EAB = \boxed{35}$

Answer: B 35
Problem: Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$ , as shown below. IMG 1031.jpeg
What is the degree measure of $\angle EAB$
$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$

Solution: We call the point where $AB$ and $EH$ intersect I. We can make an educated guess that triangle AEI is isosceles so $AI=EI$ $\angle AIE = 110^{\circ}$ $\angle AIH = 20^{\circ}$ , and $\angle EIB = 70^{\circ}$ . So, we get $\angle EAI$ is $(180^{\circ} - 110^{\circ})/2  = \boxed{35}$

Answer: B 35
Problem: What is the units digit of $2022^{2023} + 2023^{2022}$
$\text{(A)}\ 7 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 3$

Solution: $2022^{2023} + 2023^{2022} \equiv 2^3 + 3^2 \equiv 17 \equiv 7$ (mod 10). $\boxed{7}$ ~andliu766

Answer: A 7
Problem: What is the units digit of $2022^{2023} + 2023^{2022}$
$\text{(A)}\ 7 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 3$

Solution: When looking at the units digit patterns of the powers of $2$ , we see that
$2^1=$ ,  units digit $2$
$2^2=$ ,  units digit $4$
$2^3=$ ,  units digit $8$
$2^4=$ ,  units digit $6$
$2^5=$ ,  units digit $2$
And the pattern repeats. This pattern will apply for the powers of $2022$ as well, since the units digit of $2022$ is $2$ . We can find the pattern for the powers of $3$ too. The pattern follows with units digits, $3$ $9$ $7$ $1$ $3$ $9$ , ...
Similarly, the units digit of $2023$ will follow the same pattern as the powers of $3$
Both of these powers cycle in groups of $4$ . When diving $2023$ by $4$ , we get $505$ remainder $3$ , meaning $505$ complete cycles; or the power being a multiple of $4$ $505$ times, and $3$ extra. So the units digit of $2022^{2023}$ is $8$ $2022$ divided by $4$ is $505$ reminder $2$ , which means $505$ complete cycles, or the power being a multiple of $4$ $505$ times, and $2$ extra. So the units digit of $2023^{2022}$ is $9$
We only need to find the units digit in the end, so we just add those $2$ already found units digits, to get a new units digit of $7$ . Therefore the answer is $\boxed{7}$

Answer: A 7
Problem: What is the units digit of $2022^{2023} + 2023^{2022}$
$\text{(A)}\ 7 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 3$

Solution: Note that the units digit will be the same regardless of the tens, hundreds, and thousands digits, so we can simplify this problem to finding the last digit of $2^{2023} + 3^{2022}$ . We can find the units digit of $2^{2023}$ , by listing the units digits of the first few powers of two, and trying to find a pattern.
$2^1=2$
$2^2=4$
$2^3=8$
$2^4=6+10$
$2^5=2+30$
$2^6=4+60$
As we can see the units digits of powers of two repeat after every four iterations. Now we know the units digit of $2^{2020}$ is $6$ and the units digit of $2^{2023} \Rightarrow 2^3\cdot 2^{2020} \Rightarrow 6\cdot 8 \Rightarrow 8$ . Similarly we can find the last digits of powers of three repeat after every four, so the units digit of $3^{2022}$ is $1\cdot 3^2 = 9$ . Adding these together, the ones digit is the same as the ones digit of $9+8$ which is $7$ $\boxed{7}$

Answer: A 7
Problem: The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$ . How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$
$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$

Solution: Let x be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$ . Thus, $x \le 1011$ . So there are $\boxed{1011}$ numbers that satisfy the equation.

Answer: B 1011
Problem: The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$ . How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$
$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$

Solution: The smallest number that can be expressed as the difference of a pair of consecutive positive squares is $3$ , which is $2^2-1^2$ . The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to $2023$ is $2023$ , which is $1012^2-1011^2$ . These numbers are in the form $(x+1)^2-x^2$ , which is just $2x+1$ . These numbers are just the odd numbers from 3 to 2023, so there are $[(2023-3)/2]+1=1011$ such numbers. The answer is $\boxed{1011}$

Answer: B 1011
Problem: You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$

Solution: First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ guesses, and one with $4$ . Since the problem is asking for the minimum number, the answer is $\boxed{4}$

Answer: C 4