Answer: C

Solution(s):

We have that \[ \begin{align*}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{align*} \] and \[ \begin{align*}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997.\end{align*} \]

As such, their difference is: \[ 9700 - 9997 = - 297. \]

Thus, C is the correct answer.

Answer: C

Solution(s):

Note that \(45\) minutes is \[ \dfrac{45}{60} = \dfrac{3}{4} \] hours. The second meeting is then \[ 2 \cdot \dfrac{3}{4} = \dfrac{3}{2} \] hours. Both meetings take a total of \[ \dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4} \] hours then. The percent of Makarla's work day spent attending meetings is \[ 100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%. \]

Thus, C is the correct answer.

Answer: C

Solution(s):

To maximize the number of socks, we want to grab as many single socks as possible before getting a pair.

There are \(4\) colors, which means that we can draw one sock of each color before drawing a pair.

This means that it takes at least \(4 + 1 = 5\) socks to be draw before a pair is guaranteed.

Thus, C is the correct answer.

Answer: C

Solution(s):

We have \[ \heartsuit(1) = \dfrac{1 + 1^2}{2} = 1, \] \[ \heartsuit(2) = \dfrac{2 + 2^2}{2} = 3, \] and \[ \heartsuit(3) = \dfrac{3 + 3^2}{2} = 6. \]

Then \[ 1 + 3 + 6 = 10. \]

Thus, C is the correct answer.

Answer: B

Solution(s):

Note that \(31\) days leaves a remainder of \(3\) when divided by \(7.\)

This means that if the month starts on a Saturday, Sunday, Tuesday, or Wednesday, there will be an uneven number of Mondays and Wednesdays.

Then the month can only start on a Monday, Thursday, or Friday.

Thus, B is the correct answer.

Answer: B

Solution(s):

Consider the following diagram:

We have that \[ \angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}. \]

Since \(\triangle AOC\) is isosceles, we have that \[ \angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}. \]

Since \(\angle CAB = \angle CAO,\) we have that \(\angle CAB = 25^{\circ}.\)

Thus, B is the correct answer.

Answer: D

Solution(s):

To find the area of the triangle, we can drop the altitude to the side of length \(12.\)

Then we have a right triangle with one leg \(12 \div 2 = 6\) and hypotenuse \(10.\)

The other leg has length \[ \sqrt{10^2 - 6^2} = \sqrt{64} = 8. \]

The area of the triangle is then \[ \dfrac{8 \cdot 12}{2} = 48. \]

The length of the rectangle is then \(48 \div 4 = 12.\) Its perimeter is \[ 2(4 + 12) = 2 \cdot 16 = 32. \]

Thus, D is the correct answer.

Answer: E

Solution(s):

Note that \(x\) must divide both \(48\) and \(64,\) which means that must divide their greatest common divisor.

The greatest common divisor of \(48\) and \(64\) is \(16,\) which means \(x\) divides \(16.\)

\(16\) has \(5\) factors, namely all the powers of \(2\) up to \(16.\)

Thus, E is the correct answer.

Answer: D

Solution(s):

Ignoring the parentheses, Larry would get \[ 1 - 2 - 3 - 4 + e = e - 8. \]

Evaluating with the parentheses, one would get \[\begin{align*} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e.\end{align*}\]

Both of these values are the same, so \[ e - 8 = -2 - e \]\[ e = 3. \]

Thus, D is the correct answer.

Answer: C

Solution(s):

Note that \(40\) minutes is \(\frac{2}{3}\) hours. Let Shelby drive \(h\) hours in the sun.

Then she drives \(\frac{2}{3} - h\) hours in the rain, which means she travels a total of \[ 30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3} \] miles. We have this equals \(16,\) so \[ 10h + \dfrac{40}{3} = 16 \]\[ h = \dfrac{4}{15}. \]

Then Shelby drives \[ 60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24 \] minutes in the rain.

Thus, C is the correct answer.

Answer: A

Solution(s):

Let \(p\) be the price of the item. Then coupon A saves \(.15p.\) Coupon B saves \($ 30.\)

Coupon C will save \[ .25(p - 100) = .25p - 25. \]

We must have that \[ .15p \gt 30 \]\[ p \geq 200 \] and \[ .15p \gt .25p - 25 \]\[ 250 \geq p. \]

This shows that \(x = 200\) and \(y = 250.\) Therefore \(y - x = 50.\)

Thus, A is the correct answer.

Answer: D

Solution(s):

To minimize \(x,\) we want to have as many kids as possible maintain their answer.

We then need at least \(70 - 50 = 20\) percent of the students to change their answer.

To maximize \(x,\) we can have everybody that answered no change their answer, but some who answered yes must stay yes.

We need \(70 - 50 = 20\) percent of the people who said yes to stay yes, which means only \(50 - 20 = 30\) percent can switch.

This makes the maximum percent of students that can switch \(50 + 30 = 80\) percent. The difference is then \(80 - 20 = 60.\)

Thus, D is the correct answer.

Answer: C

Solution(s):

We first take care of the outer absolute value. We have either \[ x = 2x - |60 - 2x| \] or \[ x = -2x + |60 - 2x|. \]

These simplify to \[ x = |60 - 2x| \text{ and } 3x = |60 - 2x|. \]

We have \(2\) cases for each equation. For the first one, we have \[ x = 60 - 2x \text{ and } x = 2x - 60. \]

Solving both gives us \[ 3x = 60 \]\[ x = 20 \] and \[ -x = -60 \]\[ x = 60. \]

For the other equation, we have \[ 3x = 60 - 2x \text{ and } 3x = 2x - 60. \]

Again solving both, we ahve \[ 5x = 60 \]\[ x = 12 \] and \(x = -60.\) Note that \(x\) cannot be negative since that means the original absolute value is negative, which is not possible.

Adding up all the solutions gives us \[ 20 + 60 + 12 = 92. \]

Thus, C is the correct answer.

Answer: B

Solution(s):

Recall that the sum of the first \(n\) integers is \(\dfrac{n(n + 1)}{2}.\)

Then, we have that \[ \dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x, \] which simplifies to \[ 99 \cdot 50 = (100^2 - 1)x\]\[ = 101 \cdot 99x, \] by difference of squares. Dividing gives us \(x = \dfrac{50}{101}.\)

Thus, B is the correct answer.

Answer: C

Solution(s):

Let \(x\) be the number of questions Jesse answered correctly and \(y\) be the number he answered incorrectly.

Then \[ 4x - y = 99 \text{ and } x + y \leq 50. \]

We have that \[ y = 4x - 99 \]\[ 5x - 99 \leq 50. \]

Rearranging and simplifying tells us that \(x \leq 29.8.\) Since \(x\) is an integer, its maximum value is \(29.\)

Thus, C is the correct answer.

Answer: B

Solution(s):

Drop the altitude from \(O\) to \(\overline{AB}\) at \(X.\)

Looking at the side lengths, we see that \(\triangle OBX\) is a \(30-60-90\) triangle.

This means that the area of sector \(AOB\) is \[ \dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}. \]

We also have that the area of \(\triangle AOB\) is \[ \dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12.}. \]

The area of the sector outside the square and inside the circle is then \[ \dfrac{\pi}{18} - \dfrac{\sqrt3}{12}. \]

There are \(4\) of these regions, which gives us a total area of \[ 4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}. \]

Thus, B is the correct answer.

Answer: B

Solution(s):

Let there be \(x\) schools. Then there are \(3x\) participants in the contest.

Since everyone has a unique score, there is also a unique median. This means \(x\) is odd.

Note there are at least \(23\) teams, since otherwise a \(64^\text{th}\) placed wouldn't exist.

We also have at most \(23\) teams, since otherwise we have Andrea's place as being greater than Beth's.

This means that there are \(23\) teams.

Thus, B is the correct answer.

Answer: E

Solution(s):

Note that \[ abc + ab + a = a(bc + b + 1). \] This means that if \(a\) is divisible by \(3,\) the whole expression is as well.

Since \(2010\) is divisible by \(3,\) we have that \(a\) is divisible by \(3\) with probability \(\frac{1}{3}.\)

Now consider \(a\) not divisible by \(3.\) For the expression to be divisible by \(3,\) we must have that \(bc + b + 1\) is divisible by \(3.\)

This means that \[ bc + b = b(c + 1) \equiv 2 \pmod{3}. \]

The only possibility for this is that one of the factors is \(2\) mod \(3\) and the other is \(1\) mod \(3.\)

For each of the two cases, there is a \(\frac{1}{3}\) chance that each of the factors is the desired modulus, for a probability of \[\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.\]

There are two cases, which means that this happens with a \(\dfrac{2}{9}\) probability.

The total probability is then \[ \dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}. \]

Thus, E is the correct answer.

Answer: B

Solution(s):

Consider the following diagram:

Using the formula for the area of a circle, we have that \(BO = \sqrt{156}\) since it is a radius.

Extend \(\overline{AO}\) to intersect \(\overline{BC}\) at \(X.\) Let \(s\) be the side length of \(\triangle ABC.\)

Then we have that \[ BX = \dfrac{s}{2} \] and \[ AX = \dfrac{s\sqrt3}{2}. \]

We have that \(\triangle OXB\) is right, which means that we can apply the Pythagorean Theorem. This gives us \[ (\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2\]\[ + \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2. \]

Simplifying, we get \[ 156 = s^2 + 12s + 48 \]\[ \] \[ s^2 + 12s - 108 = 0 \]\[ \] \[ (s - 6)(s + 18) = 0. \]

Since \(s\) is positive, we must have that \(s = 6.\)

Thus, B is the correct answer.

Answer: D

Solution(s):

Consider the following diagram:

We can see that the smaller circle is inscribed within an equilateral triangle of side length \(1.\)

The inradius of this equilateral triangle is \(\dfrac{\sqrt3}{6}.\) The area of the circle is then \[ \pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}. \]

Let \(O\) be the center of the larger circle. Drop the perpendicular from \(O\) to \(\overline{GH}\) at \(J.\) Draw \(\overline{OG}.\)

We have that \(\triangle OJG\) is right. Since \(\angle HGI = 60^{\circ},\) we also have that \(\triangle OJG\) is a \(30-60-90\) triangle.

Let \(OJ = r.\) Then \(OG = 2r.\) We also have that \(OG\) is the sum of the height of the hexagon, equilateral triangle, and radius of the circle.

Then \[ OG = \dfrac{\sqrt3}{2} + \sqrt3 + r. \]

Substituting in \(OG,\) we get \[ 2r = \dfrac{\sqrt3}{2} + \sqrt3 + r. \] Simplifying gives us \[ r = \dfrac{3\sqrt3}{2}. \]

The area of the larger circle is then \[ \pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi. \]

The desired ratio is then \[ \dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81. \]

Thus, D is the correct answer.

Answer: E

Solution(s):

Note that we can express any \(4\) digit number as \(abcd.\) This can be expressed in long form as \[ 10^3a + 10^2b + 10c + d. \] Since in a palindrome, we have that \(a = d\) and \(b = c.\) We can simplify this to get \[ 1001a + 110b. \] Note that \(1001\) is divisible by \(7.\) This means that \(110b\) must also be divisible by \(7.\)

The only way for this to happen is if \(b\) is \(0\) or \(7\) since \(110\) is not divisible by \(7.\)

There are \(9\) options for \(a\) and \(2\) options for \(b,\) for a total of \(9 \cdot 2 = 18\) palindromes.

The total number of palindromes is \(9 \cdot 10\) since there are \(9\) options for the thousands digit and \(10\) options for the hundreds digit.

The desired probability is then \[ \dfrac{18}{9 \cdot 10} = \dfrac{1}{5}. \]

Thus, E is the correct answer.

Answer: C

Solution(s):

We can count this with complementary counting. The total number of ways to distribute the candies with no restrictions is \[ 3^7 = 2187. \]

To find the number of invalid arrangements, we have to count the number of ways where either the red or blue bag is empty.

For the case where the red bag is empty, each candy has \(2\) options for the bag that goes into. There are then \[ 2^7 = 128 \] arrangements for this case. Similarly, there are \(128\) arrangements for the case where the blue bag is empty.

There is an overlap of one case where both bags are empty. The final answer is then \[ 2187 - (128 + 128 - 1) = 1932. \]

Thus, C is the correct answer.

Answer: D

Solution(s):

Note that \(1\) and \(9\) must be in the top left and bottom right corners respectively. We must also have that \(2\) and \(8\) are next to these squares.

We can then case on the center square. Note that the only possible values are \(4,\)\(5,\) or \(6.\)

Case 1: the center is \(4\)

The \(3\) is necessarily next to the \(1,\) since there is no other option that is less than \(4.\)

Any number can be in the square next to the \(8,\) but the other two squares are then fixed. There are \[ 2 \cdot 2 \cdot 3 = 12 \] cases (two places for the \(2,\) two places for the \(8,\) and three choices for the square adjacent to \(8\)).

Case 2: the center is \(5\)

We can case on the position of the \(3.\) If the \(3\) is in the top right square, the \(4\) is necessarily next to the \(1.\)

If the \(8\) is above the \(9,\) then the other two squares are fixed. If it is to the left of the \(9,\) the other two squares can be filled arbitrarily.

Now consider when the \(3\) is below the \(1.\) There are two spots for the \(8,\) and the square next to the \(8\) can be any number.

The other two squares are then fixed. This means that this case has a total of \[ 2(1 + 2 + 2 \cdot 3) = 18. \] We multiply by two since the \(2\) can be either to the right of or below the \(1.\)

Case 3: the center is \(6\)

This is similar to case \(1\) since the \(7\) is fixed instead of the \(3.\)

The total number of arrangements is then \[ 12 + 18 + 12 = 42. \]

Thus, D is the correct answer.

Answer: E

Solution(s):

Let \(a\) be the number of points scored in the first quarter. Let \(r\) be the common ratio for the Raiders and \(d\) the difference for the Wildcats.

First, we can establish that \(r\) is an integer. Assume for the sake of contradiction that it is not and \(r = \frac{m}{n}\) where \(\gcd(m, n) = 1.\)

For \(ar, ar^2,\) and \(ar^3\) to all be integers, we must have that \(n, n^2,\) and \(n^3\) divide \(a.\) We can then let \(a = n^3k\) for some integer \(k.\)

Then we have that \[ k\left(n^3 + n^2m + nm^2 + m^3\right) \lt 100 \]

Minimizing \(n\) and \(k,\) we can assume that \(n = 2\) and \(k = 1.\) This gives us \(m \lt 4,\) which means that \(r = \dfrac{3}{2}.\)

Testing out this value shows that it does not work. Since this value does not work, no other non-integer rational ratio will work. This means \(r\) is integral.

Then the sum of the quarter scores for the Wildcats is \[ a + (a + d) + (a + 2d) + (a + 3d) \]\[ = 4a + 6d. \]

We must have that \[ a(1 + r + r^2 + r^3) = \] \[ 4a + 6d + 1. \]

Trying out values, let \(r = 2.\) Then we have \[ 15a = 4a + 6d + 1, \] which simplifies to \[ 11a = 6d + 1. \]

Looking for the smallest multiple of \(11\) that leaves a remainder of \(1\) when divided by \(6,\) we get \(55.\)

With this value, we get \(a = 5\) and \(d = 9.\) The sum of the scores of the first two quarters for both teams is \[ 5 + 10 + 5 + 14 = 34. \]

Thus, E is the correct answer.

Answer: B

Solution(s):

We can define a new function \[ Q(x) = P(x) - a \] so that \(Q\) has roots at \(1, 3, 5,\) and \(7.\) Then we can factor \(Q\) to get \[ Q(x) = (x - 1)(x - 3)(x - 5)\]\[(x - 7)(x). \]

We can plug in the values of \(2, 4, 6,\) and \(8\) to get \[ R(2) = (2 - 1)(2 - 3)(2 - 5) \] \[ (2 - 7)Q(2) = -15Q(2) = -2a\] And \[ R(4) = (4 - 1)(4 - 3)(4 - 5)\]\[(4 - 7)Q(4) = 9Q(4) = -2a \] And \[ R(6) = (6 - 1)(6 - 3)(6 - 5)\]\[(6 - 7)Q(6) = -15Q(6) = -2a \] And \[ R(8) = (8 - 1)(8 - 3)(8 - 5) \] \[ (8 - 7)Q(8) = 105Q(8) = -2a \]

To minimize \(a,\) we can set \(a = \text{lcm}(15, 9, 15, 105).\) Then \(a = 315.\)

Thus, B is the correct answer.