Answer: D

Solution(s):

Converting them all to decimals and adding, we get the average to be \[\begin{align*} \dfrac{6 + .5 + 1 + 2.5 + 10}{5} &= \dfrac{20}{5} \\&= 4. \end{align*}\]

Thus, D is the correct answer.

Answer: B

Solution(s):

WLOG, let the side lengths of the squares be \(1.\)

This means that the length of the rectangle is \(4.\) We also have that the width must be \(4 - 1 = 3.\)

The desired ratio is then \(\dfrac{4}{3}.\)

Thus, B is the correct answer.

Answer: D

Solution(s):

Let \(x\) the number of marbles that Eric ends up with. Then Tyrone ends up with \(2x.\)

The total number of marbles is \(97 + 11 = 108,\) so \[ 3x = 108 \]\[x = 36. \]

Then, Tyrone ends up with \(36 \cdot 2 = 72\) marbles. This means he has to give away \(97 - 72 = 25\) marbles.

Thus, D is the correct answer.

Answer: B

Solution(s):

Note that \(7 \cdot 56 = 392\) and \(8 \cdot 56 = 448,\) which means that the minimum number of discs needed is \(8.\)

Then the minutes of reading that each disc contains is \[ 412 \div 8 = 51.5. \]

Thus, B is the correct answer.

Answer: E

Solution(s):

Recall that the formula for the circumference of a circle is \(2 \pi r.\) We then have that \[ 24\pi = 2\pi r \]\[ r = 12. \]

The area of a circle is \(\pi r^2,\) so we have that \[ k \pi = \pi 12^2 \]\[ k = 144. \]

Thus, E is the correct answer.

Answer: C

Solution(s):

Evaluating the inner expression, we get \[ \spadesuit (2, 2) = 2 - \dfrac{1}{2} = \dfrac{3}{2}. \] Then we have \[ \spadesuit \left(2, \dfrac{3}{2}\right) = 2 - \dfrac{1}{\frac{3}{2}} = \dfrac{4}{3}. \]

Thus, C is the correct answer.

Answer: C

Solution(s):

From the diagram, we see that the distance traveled is the hypotenuse of a right triangle.

One of the legs is just \(1\) from running due north. The other leg is \[ \sqrt{1^2 + 1^2} = \sqrt2. \]

The final distance is then \[ \sqrt{\sqrt2^2 + 1^2} = \sqrt3. \]

Thus, C is the correct answer.

Answer: D

Solution(s):

Since \(2 \cdot .5 = 1,\) we have that Tony makes a dollar per day per full year of his age.

If he is \(12\) at the end of the period, then Tony can make a maximum of \[ 12 \cdot 50 = 600 \] dollars in the period. If he was \(13\) at the end, then he could have made \[ 13 \cdot 50 = 650. \] By this, we can see that Tony is \(13\) the end of the period, since otherwise he would make too much or too little money.

Thus, D is the correct answer.

Answer: E

Solution(s):

Note that \(x\) is at most \(999.\) This means that \(x + 32\) has a maximum of \(1031.\)

Similarly, we have that the minimum value of \(x + 32\) is \(1000.\)

The only palindrome in this range is \(1001,\) so this is what \(x + 32\) equals.

Then \[ x + 32 = 1001 \]\[ x = 969. \]

The sum of the digits is then \[ 9 + 6 + 9 = 24. \]

Thus, E is the correct answer.

Answer: E

Solution(s):

Note that on a normal year, we have that \[ 365 = 52 \cdot 7 + 1, \] which means that for a specific day, it moves to the day after the next year.

On a leap year, the day of the week moves forward two since there is an extra day.

Then in \(2009,\) this day falls on a Wednesday. In \(2010,\) it falls on a Thursday.

Similarly, in \(2011,\) it falls on a Friday. In \(2012,\) however, since it is a leap year, it falls on a Sunday.

Now, for the next three years, the day moves forward one. Then in \(2016,\) it moves forward two, landing on a Friday.

Finally, in \(2017,\) the day of the week is a Saturday.

Thus, E is the correct answer.

Answer: D

Solution(s):

Splitting the inequality into two of them and solving gives us \[ a \le 2x + 3 \]\[ x \ge \dfrac{a - 3}{2} \] and \[ 2x + 3 \le b \]\[ x \le \dfrac{b - 3}{2}. \]

The range of the solutions is then \[ \dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10, \] which then simplifying gives us \[ (b - 3) - (a - 3) = 20\]\[ \]\[ b - a = 20. \]

Thus, D is the correct answer.

Answer: C

Solution(s):

The miniature tower holds \[ \dfrac{100,000}{.1} = 1,000,000 \] times less water than the actual tower. Since this is the ratio for volumes, the ratio of heights is \[ (1,000,000)^{1 / 3} = 100. \] This means that the height of the miniature tower is \[ \dfrac{40}{100} = .4. \]

Thus, C is the correct answer.

Answer: A

Solution(s):

Before the stop, Angelina drove for \(80t\) km using the distance formula.

The stop takes \(\frac{1}{3}\) of an hour, which means that Angelina travels for \[ 2 - \dfrac{1}{3} = \dfrac{8}{3} \] hours after the stop. Then after the stop, Angelina drives for \[ 100\left(\dfrac83 - t\right) \] km. Since the total distance driven is \(250\) km, which makes the final equation \[ \dfrac{8}{3}t + 100\left(\dfrac83 - t\right) = 250. \]

Thus, A is the correct answer.

Answer: C

Solution(s):

Let \(\angle BAE = \angle ACD = x.\) Note that \(\angle CFE = 60^{\circ}\) since \(\triangle CFE\) is equilateral.

We then have that \[ \angle AFC = 180^{\circ} - \angle CFE = 120^{\circ}. \]

Then: \[\begin{align*} \angle FAC &= 180^{\circ} - 120^{\circ} - x\\ &=60^{\circ} - x \\ &= \angle EAC.\end{align*} \]

We then get that \[ \begin{align*} \angle BAC &= \angle BAE + \angle EAC\\ &= x + 60^{\circ} - x \\&= 60^{\circ}. \end{align*}\]

Since \(AB = 2 \cdot AC\) and \(\angle BAC = 60^{\circ},\) we have that \(\triangle ABC\) is a \(30-60-90\) triangle.

Thus, C is the correct answer.

Answer: D

Solution(s):

If Brian is a frog, then he must be lying, which means that Mike must be a frog.

If Brian is a toad, then he must be telling the truth, which also means that Mike is a frog.

Therefore, Mike is a frog, which means that Mike is lying. This means that there is at most one toad.

Then, at least one of LeRoy and Chris is a frog. This means the other is telling the truth, which makes them a toad.

This means there is one toad, which makes there be \(3\) frogs.

Thus, D is the correct answer.

Answer: B

Solution(s):

Using the Angle Bisector Theorem, we have that \[ \dfrac{AB}{3} = \dfrac{BC}{8} \]\[ AB = \dfrac{3}{8} BC. \]

For \(AB\) and \(BC\) to be integers, we must have that \(BC\) is a multiple of \(8.\)

To minimize the perimeter, we can set \(BC = 8\) and \(AB = 3.\) This, however, makes the triangle degenerate.

\(BC\) must then be \(16\) and \(AB = 6.\) Since \(DC = 11,\) the perimeter is \[ 16 + 6 + 11 = 33. \]

Thus, B is the correct answer.

Answer: A

Solution(s):

Note that all the cut out solids intersect in the middle of the cube.

This region of intersection is a cube with side length \(2.\) Then the area of the cutout region is \[ \begin{align*}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20.\end{align*} \]

We have to subtract out the center region twice since it is included in all \(3\) regions.

The remaining volume is then \[ 3^3 - 20 = 27 - 20 = 7. \]

Thus, A is the correct answer.

Answer: B

Solution(s):

There are two cases: Bernardo picks a \(9\) or he doesn't.

Case 1: Bernardo picks a \(9\)

Since a number is fixed, there are \(\binom{8}{2} = 28\) ways to choose the other two numbers.

There are a total of \(\binom{9}{3} = 84\) ways to pick all three numbers. The probability is then \[ \dfrac{28}{84} = \dfrac{1}{3}. \]

Note that if Bernardo picks a \(9,\) he automatically has a greater number than Silvia.

This means that Bernardo always wins in this case.

Case 2: Bernardo doesn't pick a \(9\)

There is a \(1 - \frac{1}{3} = \frac{2}{3}\) chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning.

We still need to find the probability that the numbers are the same. There is a \[ \dfrac{1}{\binom{8}{3}} = \dfrac{1}{56} \] chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then \[ \dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}. \]

The total probability of Bernardo getting a higher number is then \[ \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}. \]

Thus, B is the correct answer.

Answer: E

Solution(s):

Note that \(\triangle ACE\) is equilateral. Using the law of cosines, we get that \[ AC^2 = r^2 + 1^1 - 2r \cos{\dfrac{2\pi}{3.}} \]

The area of \(\triangle ACE\) is then \[ \dfrac{\sqrt3}{4} (r^2 + r + 1). \]

Recall that the formula for the area of a triangle can be given by \[ \dfrac{1}{2} AB \sin{\theta}. \]

Using this formula on \(\triangle ABC,\) we get its area to be \[ \dfrac{1}{2} \cdot 1 \cdot r \cdot \sin{120^{\circ}} = \dfrac{r \sqrt3}{4}. \]

The area of all three triangles is then \[ 3 \cdot \dfrac{r\sqrt3}{4} = \dfrac{3r\sqrt3}{4}. \]

The area of the entire hexagon is then \[ \dfrac{\sqrt3}{4}(r^2 + 4r + 1). \]

The problem conditions tells us that \[ \dfrac{\sqrt3}{4}(r^2 + r + 1) = \] \[ \dfrac{7}{10} \cdot \dfrac{\sqrt3}{4} (r^2 + 4r + 1). \]

Simplifying this gives us \[ r^2 - 6r + 1 = 0, \] which we can then use Vieta's formulas on to get the sum of values is \(6.\) Intutively, this means that if our solutions to this equation are \(r_0\) and \(r_1,\) we have that: \[\begin{align*}0&=r^2-6r+1\\&=(r-r_0)(r-r_1)\\ &=r^2-(r_0+r_1)+r_0r_1\end{align*}\] Which implies that the sum of possible solutions is equal to \(r_0+r_1 = 6.\)

Thus, E is the correct answer.

Answer: D

Solution(s):

Note that all the paths the fly can take have lengths of \(1, \sqrt2,\) or \(\sqrt3.\)

We want to maximize the number of longer length paths. We cannot travel any interior diagonal twice, since that would make the fly visit the same vertex twice.

It also possible to visit all the vertices by traveling along diagonals, so we will never have to travel a path of length \(1.\)

This means that we can mazimize the distance by traveling along \(4\) interior diagonals and \(4\) diagonals on the faces.

This path is possible by traveling along a face and then an interior diagonal, repeating this in a way that avoids visiting the same vertex twice.

The path has length \[ 4\sqrt2 + 4\sqrt3. \]

Thus, D is the correct answer.

Answer: A

Solution(s):

If \(r, \)\(s, \)\(t\) are the roots of the polynomial, we know that: \[\begin{align*}0&=x^3-ax^2+bx-2010\\ &=(x-r)(x-s)(x-t)\\ &= x^3-(r+s+t)x^2\\ &+(rs+st+rt)x-(rst) \end{align*}\] As such, we know that \(rst=2010,\) or in English, the product of the three roots is \(2010.\)

As \[ 2010 = 2 \cdot 3 \cdot 5 \cdot 67, \] we have that one of the three roots must have two of these prime numbers as factors.

Again using the first fact, we have that \(r+s+t=a\) is the sum of all the roots. To minimize this, we should have \(2\) and \(3\) multiplied together.

Then, we can let the roots be \(6, 5,\) and \(67,\) which makes \[ a = 6 + 5 + 67 = 78. \]

Thus, A is the correct answer.

Answer: A

Solution(s):

We need \(3\) chords to form the side lengths of the triangles. Each of these chords requires \(2\) points on the circle.

This means that we need to choose \(3 \cdot 2 = 6\) points from the \(8.\) Also note that any \(6\) points determine a triangle.

This is because we don't want to create chords that don't intersect in the circle, which leaves only one way to form the triangle.

The number of ways to choose the six points is then \[ \binom{8}{6} = \binom{8}{2} = 28. \]

Thus, A is the correct answer.

Answer: A

Solution(s):

Since there are \(k + 1\) marbles in the \(k\) th box, there is a \(\dfrac{k}{k + 1}\) chance Isabella draws a white marble from it.

The probability of drawing a red marble is then \(\dfrac{1}{k + 1}.\) To stop after drawing the \(n\) th marble, the first \(n - 1\) marbles must have been white.

This happens with a probability of \[ \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}. \]

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to \(\dfrac{1}{n(n + 1)}.\)

We have to find the smallest \(n\) such that \[ \dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010} \]\[\]\[ n(n + 1) \gt 2010. \]

Guessing and checking gives us that the smallest \(n\) that works is \(45.\)

Thus, A is the correct answer.

Answer: A

Solution(s):

We first find the number of zeros in \(90!.\) The number of zeros is determined by the number of factors of \(10.\)

The number of factors of \(10\) is given by the number of factors of \(2\) and \(5.\)

There are clearly more factors of \(5,\) which means that \(90!\) has \[ \left\lfloor\dfrac{90}{5}\right\rfloor + \left\lfloor\dfrac{90}{25}\right\rfloor = 18 + 3 = 21 \] factors of \(10.\)

Now, we need to find the two rightmost digits of \(N = \dfrac{90!}{10^{21}}.\) We can find the value mod \(100\) by finding the values mod \(4\) and mod \(25.\)

Since there are more factors of \(2\) than \(5\) in \(90!,\) the value of \(N\) mod \(4\) is just \(0.\)

Now, we just need to consider \(M = \dfrac{90!}{5^{21}}\) mod \(25.\) \(M\) is the product of all the non-multiples of \(5\) less than \(90.\)

Consider \[ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdots 24 \pmod{25}. \]

We can rewrite it is \[ (1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdots 12)^2 \] using the fact that \(-x \equiv 25 - x\) mod \(25\) and that there are even number of numbers in the square.

Then multiplying out some terms, we have that \[ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 11 \cdot 12 \equiv\] \[ -1 \cdot 7 \cdot 9 \cdot 11 \equiv 7 \pmod{25} \] using the fact that \[ 3 \cdot 8 = 2 \cdot 12 = 4 \cdot 6 \equiv -1. \]

Then \[ 7^2 = 49 \equiv -1 \pmod{25}. \]

This set of numbers appears \(3\) times in \(M,\) with the final being a partial expression of \[ 1 \cdot 2 \cdot 3 \cdots 14 \equiv -1 \pmod{25}, \] using similar tricks to above.

Then \[ M \equiv (-1)^4 \equiv 1 \pmod{25}. \] We also have that \[ N = \dfrac{M}{2^{21}} \equiv \dfrac{1}{2^{21}} \pmod{25}. \]

Then using Euler's theorem, we get that \[ \dfrac{1}{2^{21}} \equiv 12^{21} \equiv 12^{\phi(25)} \cdot 12 \equiv\]\[ 12 \pmod{25}. \]

Since \(12\) is divisible by \(4,\) we have that \(N \equiv 12\) mod \(100.\)

Thus, A is the correct answer.

Answer: B

Solution(s):

We can just work backwards starting with \(0.\) From this, we can add on \(1^1\) to get \(1.\)

We can again add on \(1^1\) to get \(2.\) Again, adding on \(1^1\) gives us \(3.\)

If we add on \(1^1\) now, we get \(4,\) but then \(1^1\) is not the greatest square less than \(4.\)

Then adding on \(2^2\) gives us \(7.\) We repeat this process for \(8\) steps to get \(7223.\)

\[\begin{array}{ccccc} {}&{}&{}&{}&7223\\ 7223&-&84^2&=&167\\ 167&-&12^2&=&23\\ 23&-&4^2&=&7\\ 7&-&2^2&=&3\\ 3&-&1^2&=&2\\2&-&1^2&=&1\\1&-&1^2&=&0\end{array}\]

Thus, B is the correct answer.