Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$ . What is the value of
\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]
$\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$

Solution: We can use Newton Sums to solve this problem.
We start by noticing that we can rewrite the equation as $\frac{x^3}{y^2} + \frac{y^3}{x^2} + x + y.$ Then, we know that $x + y = 4,$ so we have $\frac{x^3}{y^2} + \frac{y^3}{x^2} + 4.$ We can use the equation $x \cdot y = -2$ to write $x = \frac{-2}{y}$ and $y = \frac{-2}{x}.$ Next, we can plug in these values of $x$ and $y$ to get $\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5}{4} + \frac{y^5}{4},$ which is the same as \[\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4}.\] Then, we use Newton sums where $S_n$ is the elementary symmetric sum of the sequence and $P_n$ is the power sum ( $x^n + y^n$ ). Using this, we can make the following Newton sums: \[P_1 = S_1\] \[P_2 = P_1 S_1 - 2S_2\] \[P_3 = P_2 S_1 - P_1 S_2\] \[P_4 = P_3 S_1 - P_2 S_2\] \[P_5 = P_4 S_1 - P_3 S_2.\] We also know that $S_1$ is 4 because $x + y$ is four, and we know that $S_2$ is $-2$ because $x \cdot y$ is $-2$ as well.
Then, we can plug in values! We have \[P_1 = S_1 = 4\] \[P_2 = P_1 S_1 - 2S_2 = 16 - (-4) = 20\] \[P_3 = P_2 S_1 - P_1 S_2 = 80 - (-8) = 88\] \[P_4 = P_3 S_1 - P_2 S_2 = 88 \cdot 4 - (-40) = 392\] \[P_5 = P_4 S_1 - P_3 S_2 = 392 \cdot 4 - (-2) \cdot 88 = 1744.\] We earlier noted that $\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4},$ so we have that this equals $\frac{1744}{4},$ or $436.$ Then, plugging this back into the original equation, this is $436 + 4$ or $440,$ so our answer is $\boxed{440}.$

Answer: D 440
Problem: Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$ . What is the value of
\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]
$\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$

Solution: As in the first solution, we get the expression to be $\frac{x^3+y^3}{x^2} + \frac{x^3+y^3}{y^2}.$
Then, since the numerators are the same, we can put the two fractions as a common denominator and multiply the numerator by $x^2y^2.$ This gets us $\frac{(x^2 + y^2)(x^3 + y^3)}{x^2y^2}.$
Now, since we know $x+y=4$ and $xy=-2,$ instead of solving for $x$ and $y,$ we will try to manipulate the above expression them into a manner that we can substitute the sum and product that we know. Also, another form of $x^3+y^3$ is $(x+y)(x^2-xy+y^2).$
Thus, we can convert the current expression to $\frac{(x^2 + y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}.$
Doing some algebraic multiplications, we get $\frac{((x+y)^2 - 2xy)(x+y)((x+y)^2 - 3xy)}{(xy)^2}.$
Since we know $x+y=4$ and $xy=-2,$ we have $\frac{(16-(-4))(4)(16-(-6))}{4} = \frac{20 \cdot 4 \cdot 22}{4} = 20 \cdot 22 = 440.$
Therefore the answer is $\boxed{440}.$

Answer: D 440
Problem: Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$ . What is the value of
\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]
$\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$

Solution: We give all of the terms in this expression a common denominator. $x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = \frac{x^3y^2 + x^5 + y^5 + x^2y^3}{x^2y^2}$ . We can find $x^2y^2 = (xy)^2 = (-2)^2 = 4$ and we can find $x^3y^2 + x^2y^3 = x^2y^2(x + y) = 16$ . Our expression $\frac{xy^2 + x^5 + y^5 + x^2y}{x^2y^2}$ is now $\frac{16 + x^5 + y^5}{4}$ . Now all we need to find is $x^5 + y^5$ . Using the binomial theorem, $(x + y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5 = 4^5 = 1024$ . The extra terms that we don't need is $5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 = 5xy(x^3 + y^3) + 10x^2y^2(x + y) = -10(x^3 + y^3) + 160$ . What's $x^3 + y^3$ ? We use the same method. Using the binomial theorem, $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 + 3xy(x + y) = x^3 + y^3 - 24 = 4^3 = 64$ . Now we know that $x^3 + y^3 = 64 + 24 = 88$ , and plugging that into $-10(x^3 + y^3) + 160$ gives $-10(88) + 160 = -880 + 160 = -720$ . Now we see that those extra terms have a sum of $-720$ . Thus $(x + y)^5 = x^5 + y^5 - 720 = 1024$ so $x^5 + y^5 = 1024 + 720 = 1744$ . Remember our goal: we want to find $\frac{-8 + x^5 + y^5}{4}$ . Using $x^5 + y^5 = 1744$ $\frac{16 + x^5 + y^5}{4} = \frac{16 + 1744}{4} = \frac{1760}{4} = \boxed{440}$

Answer: D 440
Problem: Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$ . What is the value of
\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]
$\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$

Solution: Since we know $x+y=4$ , we can simplify to $4 + \frac{x^3}{y^2} + \frac{y^3}{x^2}$ . Then, when you add the fractions, you get $\frac{x^5 + y^5}{x^2y^2} = \frac{x^5 + y^5}{4}$ . To find $x^5 + y^5$ , we expand $(x+y)^5 = 4^5$ and then simplify. $x^5+y^5+5x^4y+5xy^4+10x^3y^2+10x^2y^3 = 1024$ . We can use the fact that $xy = -2$ to simplify to $x^5 + y^5-10x^3 - 10y^3 + 40x + 40y = 1024$ . We can factor to get $x^5 + y^5 -10(x^3 + y^3) +40(x+y)$ . To find $x^3 + y^3$ , we simplify $(x+y)^3 = 4^3$ $x^3+y^3+3x^2y+3y^2x = 64$ . We can use the same process as we did before and get $x^3 + y^3 -6(x+y) = 64$ (using the fact that $xy = -2$ ). Therefore, $x^3 + y^3 = 88$ . Now, plugging it back in to our other expansion, we get $x^5 + y^5 + 160 - 880 = 1024$ $x^5 +y^5 = 1744$ . Now to get the final answer, all we need to do is to plug it back into the original equation and get $\frac{1744}{4} + 4 = \boxed{440}$

Answer: D 440
Problem: A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23$

Solution: The prime factorization of $12!$ is $2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$ . 
This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12!.$ In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that the divisor can't have any factors of $7$ and $11$ in the prime factorization because there is only one of each in $12!.$ Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$ , you can have $0$ $2$ $4$ $6$ $8$ , or $10$ $2$ s, etc.)
The probability that the divisor chosen is a perfect square is \[\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{23}\]

Answer: E 23
Problem: A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

Solution: [asy] size(10cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.3989, 90, 180)); filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.3989, 180, 270)); filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1), 0.3989, 270, 360)); filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray); [/asy]
The diagram represents each unit square of the given $2020 \times 2020$ square.
We consider an individual one-by-one block.
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$ , the area covered by the circles should be $0.5$ . Because of this, and the fact that there are four circles, we write
\[4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}\]
Solving for $d$ , we obtain $d = \frac{1}{\sqrt{2\pi}}$ , where with $\pi \approx 3$ , we get $d \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}$ , and from here, we see that $d \approx 0.4 \implies \boxed{0.4}.$

Answer: B 0.4
Problem: A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

Solution: As in the previous solution, we obtain the equation $4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}$ , which simplifies to $\pi d^2 = \frac{1}{2} = 0.5$ . Since $\pi$ is slightly more than $3$ $d^2$ is slightly less than $\frac{0.5}{3} = 0.1\bar{6}$ . We notice that $0.1\bar{6}$ is slightly more than $0.4^2 = 0.16$ , so $d$ is roughly $\boxed{0.4}.$ emerald_block

Answer: B 0.4
Problem: A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

Solution: As above, we find that we need to estimate $d = \frac{1}{\sqrt{2\pi}}$
Note that we can approximate $2\pi \approx 6.28318 \approx 6.25$ and so $\frac{1}{\sqrt{2\pi}}$ $\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4$
And so our answer is $\boxed{0.4}$

Answer: B 0.4
Problem: A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

Solution: We only need to figure out the probability for a unit square, as it will scale up to the $2020\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\frac{1}{2}$ , we can find which answer will come the closest to covering $\frac{1}{2}$ of the area.
Since the closest is $0.4$ which turns out to be $(0.4)^2\times \pi = 0.16 \times \pi$ which is about $0.502$ , we find that the answer rounded to the nearest tenth is $0.4$ or $\boxed{0.4}$

Answer: B 0.4
Problem: A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

Solution: As per the above diagram, realize that $\pi d^2 = \frac{1}{2}$ , so $d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}$
$\sqrt{2} \approx 1.4 = \frac{7}{5}$
$\sqrt{\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$ , so we can say $\sqrt{\pi} \approx 1.75 = \frac{7}{4}$
So $d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}$ . This is slightly above $\boxed{0.4}$

Answer: B 0.4
Problem: A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

Solution: As above, we have the equation $\pi d^2 = \frac{1}{2}$ , and we want to find the most accurate value of $d$ . We resort to the answer choices and can plug those values of $d$ in and see which value of $d$ will lead to the most accurate value of $\pi$
Starting off in the middle, we try option C with $d=0.5$ . Plugging this in, we get $\pi \left(\frac{1}{2}\right)^2 = \frac{1}{2},$ and after simplifying we get $\pi = \frac{1}{2} \cdot 4 = 2.$ That's not very good. We know $\pi \approx 3.14.$
Let's see if we can do better. Trying option A with $d = 0.3,$ we get $\pi = \frac{1}{2} \cdot \frac{100}{9} = \frac{50}{9} = 5 \frac{5}{9}.$
Hm, let's try option B with $d = 0.4.$ We get $\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}$ . This is very close to $\pi$ and is the best estimate for $\pi$ of the 5 options.
Therefore, the answer is $\boxed{0.4}.$ ~ epiconan

Answer: B 0.4
Problem: Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$
$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Solution: We perform casework on $P(n)\leq0:$
Together, the answer is $100+5000=\boxed{5100}.$

Answer: E 5100
Problem: Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$
$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Solution: Notice that $P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2, 98^2$ and $97^2, \cdots$ $2^2$ and $1^2$ (inclusive), because there are an odd number of negatives, which means that the number of values equals \[((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \cdots + ((2+1)(2-1)+1).\] This reduces to \[200 + 196 + 192 + \cdots + 4 = 4(1+2+\cdots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{5100}.\] ~Zeric

Answer: E 5100
Problem: Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$
$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Solution: We know that $P(x)$ is a $100$ -degree function with a positive leading coefficient. That is, $P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}$
Since the degree of $P(x)$ is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach $\infty$ as $x$ goes in either direction, from which \[\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty.\] So the first time $P(x)$ is going to be negative is when it intersects the $x$ -axis at an $x$ -intercept and it's going to dip below. This happens at $1^2$ , which is the smallest intercept.
However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at $2^2$ . And when it hits $3^2$ , it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until $100^2$
To get the amount of integers below and/or on the $x$ -axis, we simply need to count the integers. For example, the amount of integers in between the $[1^2,2^2]$ interval we got earlier, we subtract and add one. $(2^2-1^2+1)=4$ integers, so there are four integers in this interval that produce a negative result.
Doing this with all of the other intervals, we have \[(2^2-1^2+1)+(4^2-3^2+1)+\cdots+(100^2-99^2+1)=\boxed{5100}\] from Solution 2's result.

Answer: E 5100
Problem: Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$
$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Solution: We know $P(x) \leq 0$ when an odd number of its factors are positive and negative. For example, to make the first factor positive, $x \in [1^2, 2^2]$ . then there will be a even number of positive factors. We would do $2^2 - 1^2 + 1 (inclusive)$ to find all integers that work. In short we can generalize too: \begin{align*} x^2 - (x-1)^2 + 1 &= 2x \\ x^2 - (x^2 - 2x + 1) + 1 &= 2x \\ x^2 - x^2 + 2x - 1 + 1 &= 2x. \\ \end{align*} But remember this only works when $x \in \{2, 4, 6, 8 \cdots 98, 100\}$ because only then will there be a odd amount of positive and negative factors. So we can set $x = 2k$ , for $k \in \{1, 2, 3, 4, \cdots 49, 50\}$ Now we only have to solve: \[\sum_{k=1}^{k=50}2(2k) = 2\sum_{k = 1}^{k = 50}2k = 4\sum_{k = 1}^{k = 50}k = 4 \cdot \dfrac{(50)(51)}{2} = 2 \cdot (50)(51) = \boxed{5100}.\] ~Wiselion

Answer: E 5100
Problem: Let $(a,b,c,d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set $\{0,1,2,3\}.$ For how many such quadruples is it true that $a\cdot d-b\cdot c$ is odd? (For example, $(0,3,1,1)$ is one such quadruple, because $0\cdot 1-3\cdot 1 = -3$ is odd.)
$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

Solution: In order for $a\cdot d-b\cdot c$ to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are $2(2 + 4) = 12$ ways to pick numbers to obtain an even product. There are $2 \cdot 2 = 4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a\cdot d-b\cdot c$ odd is $2 \cdot (12 \cdot 4) = \boxed{96}$

Answer: C 96
Problem: Let $(a,b,c,d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set $\{0,1,2,3\}.$ For how many such quadruples is it true that $a\cdot d-b\cdot c$ is odd? (For example, $(0,3,1,1)$ is one such quadruple, because $0\cdot 1-3\cdot 1 = -3$ is odd.)
$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

Solution: Consider parity. We need exactly one term to be odd, one term to be even. Because of symmetry, we can set $ad$ to be odd and $bc$ to be even, then multiply by $2.$ If $ad$ is odd, both $a$ and $d$ must be odd, therefore there are $2\cdot2=4$ possibilities for $ad.$ Consider $bc.$ Let us say that $b$ is even. Then there are $2\cdot4=8$ possibilities for $bc.$ However, $b$ can be odd, in which case we have $2\cdot2=4$ more possibilities for $bc.$ Thus there are $12$ ways for us to choose $bc$ and $4$ ways for us to choose $ad.$ Therefore, also considering symmetry, we have $2 \cdot 4 \cdot 12= \boxed{96}$ total values of $ad-bc.$

Answer: C 96
Problem: Let $(a,b,c,d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set $\{0,1,2,3\}.$ For how many such quadruples is it true that $a\cdot d-b\cdot c$ is odd? (For example, $(0,3,1,1)$ is one such quadruple, because $0\cdot 1-3\cdot 1 = -3$ is odd.)
$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

Solution: There are 4 ways to choose any number independently and 2 ways to choose any odd number independently.
To get an even products, we count: $\text{P(any number)} \cdot \text{P(any number)}-\text{P(odd)}\cdot\text{P(odd)}$ , which is $4 \cdot 4 - 2 \cdot 2=12$ .
The number of ways to get an odd product can be counted like so: $\text{P(odd)}\cdot\text{P(odd)}$ , which is $2 \cdot 2$ , or $4$ .
So, for one product to be odd the other to be even: $2 \cdot 4 \cdot 12=\boxed{96}$ (order matters). 
~ Anonymous and Arctic_Bunny

Answer: C 96
Problem: Let $(a,b,c,d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set $\{0,1,2,3\}.$ For how many such quadruples is it true that $a\cdot d-b\cdot c$ is odd? (For example, $(0,3,1,1)$ is one such quadruple, because $0\cdot 1-3\cdot 1 = -3$ is odd.)
$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

Solution: For parity reasons, if $ad - bc$ is to be odd, we must have $ad$ odd and $bc$ even or $ad$ even and $bc$ odd. By symmetry, these cases are identical, so we consider the first one and multiply by two at the end. For $ad$ to be odd, we must have both $a$ and $d$ odd, and there are $2 \cdot 2$ ways to do so. To count the cases where $bc$ is odd, we use PIE. there are $2 \cdot 4 = 8$ ways for $b$ to be odd and $4 \cdot 2 = 8$ ways for $c$ to be odd, and there are $2 \cdot 2 = 4$ ways for both to be odd. Thus, there are $8 + 8 - 4 = 12$ ways for $bc$ to be even. Multiplying out, there are $2 \cdot 4 \cdot 12$ ways to have $ad - bc$ odd for a total of $\boxed{96}$

Answer: C 96
Problem: As shown in the figure below, a regular dodecahedron (the polyhedron consisting of $12$ congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring? [asy] import graph; unitsize(5cm); pair A = (0.082, 0.378); pair B = (0.091, 0.649); pair C = (0.249, 0.899); pair D = (0.479, 0.939); pair E = (0.758, 0.893); pair F = (0.862, 0.658); pair G = (0.924, 0.403); pair H = (0.747, 0.194); pair I = (0.526, 0.075); pair J = (0.251, 0.170); pair K = (0.568, 0.234); pair L = (0.262, 0.449); pair M = (0.373, 0.813); pair N = (0.731, 0.813); pair O = (0.851, 0.461); path[] f; f[0] = A--B--C--M--L--cycle; f[1] = C--D--E--N--M--cycle; f[2] = E--F--G--O--N--cycle; f[3] = G--H--I--K--O--cycle; f[4] = I--J--A--L--K--cycle; f[5] = K--L--M--N--O--cycle; draw(f[0]); axialshade(f[1], white, M, gray(0.5), (C+2*D)/3); draw(f[1]); filldraw(f[2], gray); filldraw(f[3], gray); axialshade(f[4], white, L, gray(0.7), J); draw(f[4]); draw(f[5]); [/asy] $\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810$

Solution: Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.
We have $5$ choices for which face we visit first on the top ring. From there, we have $9$ choices for how far around the top ring we go before moving down: $1,2,3,$ or $4$ faces around clockwise, $1,2,3,$ or $4$ faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.
We then have $2$ choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly $2$ lower-ring faces) and then once again $9$ choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.
Multiplying together all the numbers of choices we have, we get $5 \cdot 9 \cdot 2 \cdot 9 = \boxed{810}$

Answer: E 810