Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$ , and in general,
\[a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}\]
Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$ .  What is the sum of all integers $k$ $1\le k \le 2011$ , such that $a_k=b_k?$
$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012$

Solution: First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$ (decreasing exponential function), and $g(x) = x^k$ for $k > 0$ (increasing power function for positive $x$ ). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant.
We will now examine the first few terms.
Comparing $a_1$ and $a_2$ $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$
Therefore, $0 < a_1 < a_2 < 1$
Comparing $a_2$ and $a_3$ $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$
Comparing $a_1$ and $a_3$ $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$
Therefore, $0 < a_1 < a_3 < a_2 < 1$
Comparing $a_3$ and $a_4$ $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$
Comparing $a_2$ and $a_4$ $0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$
Therefore, $0 < a_1 < a_3 < a_4 < a_2 < 1$
Continuing in this manner, it is easy to see a pattern(see Note 1).
Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$ . Solving gives $\boxed{1341}$

Answer: C 1341
Problem: Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$ , and in general,
\[a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}\]
Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$ .  What is the sum of all integers $k$ $1\le k \le 2011$ , such that $a_k=b_k?$
$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012$

Solution: Start by looking at the first few terms and comparing them to each other. We can see that $a_1 < a_2$ , and that $a_1 < a_3 < a_2$ , and that $a_3 < a_4 < a_2$ , and that $a_3 < a_5 < a_4$ ...
From this, we find the pattern that $a_{k-1} < a_{k+1} < a_k$
Examining this relationship, we see that every new number $a_k$ will be between the previous two terms, $a_{k-1}$ and $a_{k-2}$ . Therefore, we can see that $a_1$ is the smallest number, $a_2$ is the largest number, and that all odd terms are less than even terms. Furthermore, we can see that for every odd k, $a_k < a_{k+2}$ , and for every even k, $a_k > a_{k+2}$
This means that rearranging the terms in descending order will first have all the even terms from $a_2$ to $a_{2012}$ , in that order, and then all odd terms from $a_{2011}$ to $a_1$ , in that order (so $\{b_k\}_{k=1}^{2011} = {a_2, a_4, a_6, ... a_{2008}, a_{2010}, a_{2011}, a_{2009}, ... a_5, a_3, a_1}$ ).
We can clearly see that there will be no solution k where k is even, as the $k$ th term in $\{a_k\}_{k=1}^{2011}$ will appear in the same position in its sequence as the $2k$ th term does in $\{a_k\}_{k=1}^{2011}$ , where k is even. Therefore, we only have to look at the odd terms of $a_k$ , which occur in the latter part of $b_k$
Looking at the back of both sequences, we see that term k in $\{a_k\}_{k=1}^{2011}$ progresses backwards in the equation $2012 - k$ , and that term k in $\{a_k\}_{k=1}^{2011}$ progresses backwards in the equation $2k - 1$ . Setting these two expressions equal to each other, we get $671$
However, remember that we started counting from the back of both sequences. So, plugging $671$ back into either side of the equation from earlier, we get our answer of $\boxed{1341}$

Answer: C 1341
Problem: Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$

Solution: Multiplying $18$ and $2$ by $4$ we get $72$ students and $8$ rabbits. We then subtract: $72 - 8 = \boxed{64}.$

Answer: C 64
Problem: A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?
[asy] draw((0,0)--(0,10)--(20,10)--(20,0)--cycle);  draw(circle((10,5),5)); [/asy]
$\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200$

Solution: If the radius is $5$ , then the width is $10$ , hence the length is $20$ $10\times20= \boxed{200}.$

Answer: E 200
Problem: For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54$

Solution: If $x$ is the number of holes that the chipmunk dug, then $3x=4(x-4)$ , so $3x=4x-16$ , and $x=16$ . The number of acorns hidden by the chipmunk is equal to $3x = \boxed{48}$

Answer: D 48
Problem: For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54$

Solution: Trying answer choices, we see that $\boxed{48}$ works. ~Extremelysupercooldude

Answer: D 48
Problem: Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$

Solution: The ratio $\frac{400 \text{ euros}}{500 \text{ dollars}}$ can be simplified using conversion factors: \[\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04\] which means the money is greater by $\boxed{4}$ percent.

Answer: B 4
Problem: Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$

Solution: If we divide each of Etienne's and Diana's values by $100$ , the problem stays the same. Then, Etienne has $1.3$ times the amount of money Diana has, so Etienne has $5.2$ dollars. Since $\dfrac{5.2}{5} = 1.04$ , Etienne has $\boxed{4}$ percent more money than Diana. ~Extremelysupercooldude

Answer: B 4
Problem: Two integers have a sum of $26$ . when two more integers are added to the first two, the sum is $41$ . Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$ . What is the minimum number of even integers among the $6$ integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Since, $x + y = 26$ $x$ can equal $15$ , and $y$ can equal $11$ , so no even integers are required to make 26. To get to $41$ , we have to add $41 - 26 = 15$ . If $a+b=15$ , at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from $26$ to $41$ . Finally, we have the last transition is $57-41=16$ . If $m+n=16$ $m$ and $n$ can both be odd because two odd numbers sum to an even number, meaning only $1$ even integer is required. The answer is $\boxed{1}$ . ~Extremelysupercooldude (Latex, grammar, and solution edits)

Answer: A 1
Problem: Two integers have a sum of $26$ . when two more integers are added to the first two, the sum is $41$ . Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$ . What is the minimum number of even integers among the $6$ integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Just worded and formatted a little differently than above.
The first two integers sum up to $26$ . Since $26$ is even, in order to minimize the number of even integers, we make both of the first two odd.
The second two integers sum up to $41-26=15$ . Since $15$ is odd, we must have at least one even integer in these next two.
Finally, $57-41=16$ , and once again, $16$ is an even number so both of these integers can be odd.
Therefore, we have a total of one even integer and our answer is $\boxed{1}$

Answer: A 1
Problem: Small lights are hung on a string $6$ inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of $2$ red lights followed by $3$ green lights. How many feet separate the 3rd red light and the 21st red light?
Note: $1$ foot is equal to $12$ inches.
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5$

Solution: We know the repeating section is made of $2$ red lights and $3$ green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the 11th section. There would then be a total of $44$ lights in between the 3rd and 21st red light, translating to $45$ $6$ -inch gaps. Since the question asks for the answer in feet, the answer is $\frac{45*6}{12} \rightarrow \boxed{22.5}$

Answer: E 22.5
Problem: A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$

Solution: We can count the number of possible foods for each day and then multiply to enumerate the number of combinations.
On Friday, we have one possibility: cake.
On Saturday, we have three possibilities: pie, ice cream, or pudding. This is the end of the week.
On Thursday, we have three possibilities: pie, ice cream, or pudding. We can't have cake because we have to have cake the following day, which is the Friday with the birthday party.
On Wednesday, we have three possibilities: cake, plus the two things that were not eaten on Thursday.
Similarly, on Tuesday, we have three possibilities: the three things that were not eaten on Wednesday.
Likewise on Monday: three possibilities, the three things that were not eaten on Tuesday.
On Sunday, it is tempting to think there are four possibilities, but remember that cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday.
So the number of menus is $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 1 \cdot 3 = 729.$ The answer is $\boxed{729}$

Answer: A 729
Problem: A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$

Solution: We can perform casework as an understandable means of getting the answer. We can organize our counting based on the food that was served on Wednesday, because whether cake is or is not served on Wednesday seems to significantly affect the number of ways the chef can make said foods for that week.
Case 1: Cake is served on Wednesday. Here, we have three choices for food on Thursday and Saturday since cake must be served on Friday, and none of these choices are cake, which was served Wednesday. Likewise, we have three choices (pie, ice cream, and pudding) for the food served on Tuesday and thus three choices for those served on Monday and Sunday, with these three choices being whatever was not served on Tuesday and Monday, respectively. Hence, for this case, there are $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 243$ possibilities.
Case 2: Cake is not served on Wednesday. Obviously, this means that pie, ice cream, and pudding are our only choices for Wednesday's food. Since cake must be served on Friday, only ice cream, pudding, and cake can be served on Thursday. However, since one of those foods was chosen for Wednesday, we only have two possibilities for Thursday's food. Like our first case, we have three possibilities for the food served on Tuesday, Monday, and Sunday: whatever was not served on Wednesday, Tuesday, and Monday, respectively. $3 \cdot 3 \cdot 3 \cdot 3 \cdot 2 \cdot 3 = 486$ possibilities thus exist for this case.
Adding the number of possibilities together yields that $243 + 486 = 729$ is the total number of menus, making our answer $\boxed{729}$

Answer: A 729
Problem: A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$

Solution: Note that the choice of a food item on a given day is symmetric, i.e. the number of ways to create the meal plan with a cake on Friday is the same as the number of ways to create the meal plan with pudding on Friday, and the same reasoning holds for the other desserts. Since every meal plan is counted by the summation of the $4$ aforementioned plans (note that Friday's dessert has to be one of the $4$ given desserts) and that these cases are mutually exclusive (i.e. you cannot make both a cake and pudding on Friday), each case results in a quarter of the total meal plans (Since there are $4$ desserts, we multiply by $\tfrac{1}{4}$ ). The total number of plans with no restrictions can be counted with constructive counting, as follows:
We note that there are $4$ ways to choose the first dessert. Then, each dessert thereafter must be distinct from the prior one. Since there are $4$ options, and $1$ of them has been taken by the prior, each following dessert can be chosen in $(4 - 1) = 3$ ways. Thus, since there are $6$ desserts other than the first, the total number (without restrictions) is $4 \cdot 3^6$
Thus, by our symmetry argument derived prior, we know that the number of desired meal plans is $\frac{\text{\# plans w/o restrictions}}{4} = \frac{4 \cdot 3^6}{4} = 3^6 = 729$ , choice $\boxed{729}$

Answer: A 729
Problem: It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: She walks at a rate of $x$ units per second to travel a distance $y$ . As $vt=d$ , we find $60x=y$ and $24*(x+k)=y$ , where $k$ is the speed of the escalator. Setting the two equations equal to each other, $60x=24x+24k$ , which means that $k=1.5x$ . Now we divide $60$ by $1.5$ because you add the speed of the escalator but remove the walking, leaving the final answer that it takes to ride the escalator alone as $\boxed{40}$

Answer: B 40
Problem: It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: We write two equations using distance = rate * time. Let $r$ be the rate she is walking, and $e$ be the speed the escalator moves. WLOG, let the distance of the escalator be $120$ , as the distance is constant. Thus, our $2$ equations are $120 = 60r$ and $120 = 24(r+e)$ . Solving for $e$ , we get $e = 3$ . Thus, it will take Clea $\dfrac{120}{3} = \boxed{40}$ seconds.

Answer: B 40
Problem: It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: Clea covers $\dfrac{1}{60}$ of the escalator every second. Say the escalator covers $\dfrac{1}{r}$ of the escalator every second. Since Clea and the escalator cover the entire escalator in $24$ seconds, we can use distance $=$ rate $\cdot$ time to get $24\left(\dfrac{1}{60} + \dfrac{1}{r}\right) = 1$ . Solving gives us $r = 40$ , so if Clea were to just stand on the escalator, it would take her $\boxed{40}$ seconds to get down. ~Extremelysupercooldude

Answer: B 40
Problem: In the equation below, $A$ and $B$ are consecutive positive integers, and $A$ $B$ , and $A+B$ represent number bases: \[132_A+43_B=69_{A+B}.\] What is $A+B$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17$

Solution: Change the equation to base 10: \[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\] \[A^2 - 3A - 2B - 4=0\]
Either $B = A + 1$ or $B = A - 1$ , so either $A^2 - 5A - 6, B = A + 1$ or $A^2 - 5A - 2, B = A - 1$ . The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1) = 0, B = A + 1$ . Since A must be positive, $A = 6, B = 7$ and $A+B = \boxed{13}$

Answer: C 13
Problem: In the equation below, $A$ and $B$ are consecutive positive integers, and $A$ $B$ , and $A+B$ represent number bases: \[132_A+43_B=69_{A+B}.\] What is $A+B$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17$

Solution: We can eliminate answer choice $\textbf{(A)}$ because you can't have a $9$ in base $9$ . Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most $8$ cases. Eventually, after testing a few cases, you will find that $A=6$ and $B=7$ . The solution is $\boxed{13}$

Answer: C 13
Problem: How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$

Solution: There are $\binom{20}{2}$ selections; however, we count these twice, therefore
$2\cdot\binom{20}{2} = \boxed{380}$ . The wording of the question implies D, not E.

Answer: D 380
Problem: How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$

Solution: Consider the 20 term sequence of $0$ 's and $1$ 's.  Keeping all other terms 1, a sequence of $k>0$ consecutive 0's can be placed in $21-k$ locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are $20+19+\cdots+1=\binom{21}{2}$ strings with consecutive zeros. The same argument shows there are $\binom{21}{2}$ strings with consecutive 1's. This yields $2\binom{21}{2}$ strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases $01111...$ $00111...$ $000111...$ , ..., $000...0001$ (of which there are 19) as well as the cases $10000...$ $11000...$ $111000...$ , ..., $111...110$ (of which there are 19 as well). This yields $2\binom{21}{2}-2\cdot19=\boxed{382}$

Answer: E 382
Problem: How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$

Solution: First, we think of ways to make all the $1$ 's consecutive. If there are no consecutive $1$ 's, there are $\binom{20}{0}$ ways to order them. If there is one consecutive $1$ , there are $\binom{20}{1}$ ways to order them. If there are two consecutive $1$ 's, then there are $\binom{19}{1}$ ways to order them (We treat the two $1$ 's like a block, and then order that block with 18 other $0$ 's). Continuing in this fashion, there are $\binom{20}{0} + \binom{20}{1} + \binom{19}{1} + \cdots + \binom{1}{1} = 1 + 20 + 19 + \cdots + 2 + 1 = 210 + 1 = 211$ ways to order consecutive $1$ 's. From symmetry, there are also $211$ ways to order the $0$ 's. Now, from PIE, we subtract out the cases where both the $1$ 's and the $0$ 's are consecutive. We do this because when counting the ways to order the $1$ 's, we counted all of these cases once. Then, we did so again when ordering the $0$ 's. So, to only have all of these cases once, we must subtract them. If $1$ is the leftmost digit, then there are $20$ cases where all the $1$ 's and $0$ 's are consecutive (we basically are choosing how many $1$ 's are consecutive, and there are $20$ possibilities. All other digits become $0$ , which are automatically consecutive since the $1$ 's are consecutive. There are also $20$ cases when $0$ is the left-most digit. Thus, there are a total of $211 + 211 - 20 - 20 = \boxed{382}$ as acceptable answers.

Answer: E 382
Problem: Let $(a_1,a_2, \dots ,a_{10})$ be a list of the first 10 positive integers such that for each $2 \le i \le 10$ either $a_i+1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880$

Solution: This problem is worded awkwardly. More simply, it asks: “How many ways can you order numbers 1-10 so that each number is one above or below some previous term?”
Then, the method becomes clear. For some initial number, WLOG examine the numbers greater than it. They always must appear in ascending order later in the list, though not necessarily as adjacent terms. Then, for some initial number, the number of possible lists is just the number of combination where this number of terms can be placed in 9 slots. For 9, that’s 1 number in 9 potential slots. For 8, that’s 2 numbers in 9 potential slots.
\[\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} =512 \implies \boxed{512}\]

Answer: B 512