Find the total amount of juice, then divide it equally among the four glasses

Each full glass ends at the common level, so it must pour out $1$ minus that level

There’s $3 + \frac{1}{3} = \frac{10}{3}$ glasses of juice in all. Split four ways, each glass ends up with $\frac{10}{3} \div 4 = \frac{5}{6}.$ So a full glass has to pour out $1 - \frac{5}{6} = \frac{1}{6}.$ Thus, C is the correct answer.

A $20\%$ discount multiplies the price by $0.8;$ a $7.5\%$ tax multiplies by $1.075$

Set the final cost $0.8 \times 1.075 \times P$ equal to $43$ and solve for $P$

Let $P$ be the original price. After the discount and tax, Carlos pays $0.8P \times 1.075 = 0.86P.$ He can afford it when $0.86P \le 43,$ which means $P \le 50.$ So the priciest shoes he can swing start at $\$50.$ Therefore, the answer is B.

The hypotenuse of a right triangle is a diameter of its circumscribed circle

The radii are $\frac{5}{2}$ and $\frac{13}{2};$ area scales as the square of the radius

The hypotenuse of a right triangle is a diameter of the circle around it. So circle $A$ has diameter $5$ and circle $B$ has diameter $13.$ Areas scale as the square of that, giving $\left(\frac{5}{13}\right)^2 = \frac{25}{169}.$ Thus, D is the correct answer.

Convert everything to centimeters: $6.5$ mm $= 0.65$ cm and $25$ m $= 2500$ cm

Area is width $\times$ length once both are in the same units

Put everything in centimeters first. The strip is $6.5$ mm $= 0.65$ cm wide and $25$ m $= 2500$ cm long. Its area is $0.65 \times 2500 = 1625$ square centimeters. Therefore, the answer is C.

Let the count be $n$ and the original sum be $S;$ adding $3$ to each number adds $3n$

Lara’s condition gives $3S = 45,$ so $S = 15$

Let $S$ be the original sum and $n$ the number of entries. Maddy adds $3$ to each, so her total is $S + 3n = 45.$ Lara triples each, so hers is $3S = 45,$ giving $S = 15.$ Then $15 + 3n = 45,$ so $n = 10.$ Thus, A is the correct answer.

Track only the parity of the terms: odd, odd, even, and then it repeats

Even terms occur exactly at positions that are multiples of $3$

Track the parities: $1, 3, 4, 7, 11, 18, \ldots$ run odd, odd, even, then repeat with period $3.$ So $L_n$ is even exactly when $n$ is divisible by $3.$ Among $1 \le n \le 2023,$ that’s $\lfloor \frac{2023}{3} \rfloor = 674$ multiples of $3.$ Therefore, the answer is E.

Let $O$ be the center; then $OA = OE$ and $\angle AOE = 20^\circ$

Triangle $OAE$ is isosceles, and the diagonal $AO$ makes a $45^\circ$ angle with $AB$

Let $O$ be the shared center. The rotation carries $A$ to $E,$ so $OA = OE$ and $\angle AOE = 20^\circ.$ That makes triangle $OAE$ isosceles, with base angles $\angle OAE = \frac{180^\circ - 20^\circ}{2} = 80^\circ.$ The diagonal $AC$ splits the right angle at $A,$ so $\angle OAB = 45^\circ.$ Subtracting, $\angle EAB = 80^\circ - 45^\circ = 35^\circ.$ Thus, B is the correct answer.

Units digits of powers of $2$ cycle $2, 4, 8, 6;$ powers of $3$ cycle $3, 9, 7, 1$

Reduce each exponent modulo $4,$ then add the two units digits

Only the units digits matter. Powers of $2$ cycle $2, 4, 8, 6$ with period $4,$ and $2023 \equiv 3 \pmod 4,$ so $2022^{2023}$ ends in $8.$ Powers of $3$ cycle $3, 9, 7, 1,$ and $2022 \equiv 2 \pmod 4,$ so $2023^{2022}$ ends in $9.$ Add them: $8 + 9 = 17,$ so the units digit is $7.$ Therefore, the answer is A.

Consecutive squares $k^2$ and $(k+1)^2$ differ by $2k+1$

Solve $2k + 1 \le 2023$ for positive integers $k$

Consecutive squares $k^2$ and $(k+1)^2$ differ by $(k+1)^2 - k^2 = 2k + 1.$ We need $2k + 1 \le 2023,$ which gives $k \le 1011.$ So $k$ runs $1, 2, \ldots, 1011,$ for $1011$ pairs. Thus, B is the correct answer.

If none of your guesses is covered, the two covered squares lie among the squares you did not guess

The unguessed squares must have no two adjacent; the most non-adjacent squares in a $3\times 3$ grid is $5$

Suppose every guess misses. Then the domino lies entirely on unguessed squares, so those squares include two adjacent cells. To force a hit, we need the unguessed squares to have no two adjacent. Color the grid like a checkerboard. The biggest set of pairwise non-adjacent squares has $5$ cells, one color’s worth: the four corners and the center. So we must guess at least $9 - 5 = 4$ squares. And $4$ is enough: guess the four edge midpoints, since every domino covers one square of each color, hence one edge midpoint. Therefore, the answer is C.

Divide $20a + 50b + 100c = 800$ by $10;$ parity then forces the number of $\$50$ bills to be even

Writing $b = 2t,$ count positive integer pairs $(t, c)$ with $t + c \le 7$

Let $a, b, c \ge 1$ count the $\$20, \$50, \$100$ bills. Then $20a + 50b + 100c = 800,$ which divides down to $2a + 5b + 10c = 80.$ Both $2a$ and $10c$ are even, so $5b$ is too, forcing $b = 2t.$ Now $a = 40 - 5t - 5c \ge 1$ means $t + c \le 7.$ With $t, c \ge 1,$ the pairs number $1 + 2 + \cdots + 6 = 21.$ Thus, B is the correct answer.

For $x \gt 10$ every factor is positive; the sign of $P$ can only change at a root

Crossing the root $x = i$ flips the sign exactly when the exponent $i$ is odd

For $x \gt 10,$ every factor $(x - i)^i$ is positive, so $P(x) \gt 0.$ Now move left. Crossing $x = i$ flips the sign only when $i$ is odd, that is at $i = 9, 7, 5, 3, 1.$ So the eleven intervals, right to left, carry signs $+, +, -, -, +, +, -, -, +, +, -.$ Six are positive. Therefore, the answer is C.

Substitute $u = |x|$ and $v = |y|;$ then $|u - 1| + |v - 1| \le 1$ is a diamond

Reflecting the first-quadrant piece across both axes copies it $4$ times

Substitute $u = |x|,$ $v = |y|.$ Then $|u - 1| + |v - 1| \le 1$ is a diamond centered at $(1, 1)$ with diagonals of length $2,$ so it has area $2,$ and it sits entirely in $u, v \ge 0.$ The map $(x, y) \mapsto (|x|, |y|)$ is four-to-one over $u, v \gt 0,$ so the full region has area $4 \times 2 = 8.$ Thus, B is the correct answer.

If $m = 0$ then $n = 0;$ otherwise assume $|m| \le |n|$

Then $m^2 n^2 = m^2 + mn + n^2 \le 3n^2$ forces $m^2 \le 3$