Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A permutation $(a_1,a_2,a_3,a_4,a_5)$ of $(1,2,3,4,5)$ is heavy-tailed if $a_1 + a_2 < a_4 + a_5$ .  What is the number of heavy-tailed permutations?
$\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52$

Solution: We use case work on the value of $a_3$
Case 1: $a_3 = 1$ . Since $a_1 + a_2 < a_4 + a_5$ $(a_1, a_2)$ can only be a permutation of $(2, 3)$ or $(2, 4)$ . The values of $a_1$ and $a_2$ , as well as the values of $a_4$ and $a_5$ , are interchangeable, so this case produces a total of $2(2 \cdot 2) = 8$ solutions.
Case 2: $a_3 = 2$ . Similarly, we have $(a_1, a_2)$ is a permutation of $(1, 3)$ $(1, 4)$ , or $(1, 5)$ , which gives a total of $3(2 \cdot 2) = 12$ solutions.
Case 3: $a_3 = 3$ $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 4)$ , which gives a total of $2(2 \cdot 2) = 8$ solutions.
Case 4: $a_3 = 4$ $(a_1, a_2)$ is a permutation of $(1, 2)$ $(1, 3)$ , or $(2, 3)$ , which gives a total of $3(2 \cdot 2) = 12$ solutions.
Case 5: $a_3 = 5$ $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 3)$ , which gives a total of $2(2 \cdot 2) = 8$ solutions.
Therefore, our answer is $8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{48}$

Answer: D 48
Problem: A class collects $50$ dollars to buy flowers for a classmate who is in the hospital. Roses cost $3$ dollars each, and carnations cost $2$ dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50$ dollars?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

Solution: The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \boxed{9}$

Answer: C 9
Problem: Postman Pete has a pedometer to count his steps. The pedometer records up to $99999$ steps, then flips over to $00000$ on the next step. Pete plans to determine his mileage for a year. On January $1$ Pete sets the pedometer to $00000$ . During the year, the pedometer flips from $99999$ to $00000$ forty-four times. On December $31$ the pedometer reads $50000$ . Pete takes $1800$ steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
$\textbf{(A)}\ 2500 \qquad \textbf{(B)}\ 3000 \qquad \textbf{(C)}\ 3500 \qquad \textbf{(D)}\ 4000 \qquad \textbf{(E)}\ 4500$

Solution: Every time the pedometer flips, Pete has walked $100,000$ steps. Therefore, Pete has walked a total of $100,000 \cdot 44 + 50,000 = 4,450,000$ steps, which is $4,450,000/1,800 = 2472.2$ miles, which is the closest to the answer choice $\boxed{2500}$

Answer: A 2500
Problem: For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008$ th term of the sequence?
$\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064$

Solution: Letting the sum of the sequence equal $a_1+a_2+\cdots+a_n$ yields the following two equations:
$\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008$ and
$\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007$
Therefore:
$a_1+a_2+\cdots+a_{2008}=2008^2$ and $a_1+a_2+\cdots+a_{2007}=2007^2$
Hence, by substitution, $a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{4015}$

Answer: B 4015
Problem: A pyramid has a square base $ABCD$ and vertex $E$ .  The area of square $ABCD$ is $196$ , and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91$ , respectively.  What is the volume of the pyramid?
$\textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784$

Solution: Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$ . The side length of the base is $14$ . The heights of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$ , respectively. Consider a side view of the pyramid from $\triangle BCE$ .  We have a systems of equations through the Pythagorean Theorem:
$13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$
Setting them equal to each other and simplifying gives $-27+28a=225 \implies a=9$
Therefore, $h=\sqrt{15^2-9^2}=12$ , and the volume of the pyramid is $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784}$

Answer: E 784
Problem: The sum of the base- $10$ logarithms of the divisors of $10^n$ is $792$ . What is $n$
$\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15$

Solution: Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$ . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$ . For any factor $2^a \times 5^b$ , there will be another factor $2^{n-a} \times 5^{n-b}$ . Note this is not true if $10^n$ is a perfect square. When these are multiplied, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$ $\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792.
There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $(n+1)^2$ total factors. $\frac{(n+1)^2}{2}*n = 792$ . We then plug in answer choices and arrive at the answer $\boxed{11}$

Answer:  11
Problem: Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$

Solution: Let $a_n=|A_{n-1}A_n|$ . We need to rewrite the recursion into something manageable. The two strange conditions, $B$ 's lie on the graph of $y=\sqrt{x}$ and $A_{n-1}B_nA_n$ is an equilateral triangle, can be compacted as follows: \[\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1\] which uses $y^2=x$ , where $x$ is the height of the equilateral triangle and therefore $\frac{\sqrt{3}}{2}$ times its base.
The relation above holds for $n=k$ and for $n=k-1$ $(k>1)$ , so \[\left(a_k\frac{\sqrt{3}}{2}\right)^2-\left(a_{k-1}\frac{\sqrt{3}}{2}\right)^2=\] \[=\left(\frac{a_k}{2}+a_{k-1}+a_{k-2}+\cdots+a_1\right)-\left(\frac{a_{k-1}}{2}+a_{k-2}+a_{k-3}+\cdots+a_1\right)\] Or, \[a_k-a_{k-1}=\frac23\] This implies that each segment of a successive triangle is $\frac23$ more than the last triangle. To find $a_{1}$ , we merely have to plug in $k=1$ into the aforementioned recursion and we have $a_{1} - a_{0} = \frac23$ . Knowing that $a_{0}$ is $0$ , we can deduce that $a_{1} = 2/3$ .Thus, $a_n=\frac{2n}{3}$ , so $A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{2}{3} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{3}$ . We want to find $n$ so that $n^2<300<(n+1)^2$ $n=\boxed{17}$ is our answer.

Answer:  17
Problem: Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$

Solution: Consider two adjacent equilateral triangles obeying the problem statement. For each, drop an altitude to the $x$ axis and denote the resulting heights $h_n$ and $h_{n+1}$ . From 30-60-90 rules, the distance between the points where these altitudes meet the x-axis is \[\frac{h_{n+1}}{\sqrt{3}}+\frac{h_n}{\sqrt{3}} = \frac{h_{n+1}+h_n}{\sqrt{3}}\]
But the square root curve means that this distance is also expressible as $h_{n+1}^2-h_n^2$ (the $x$ coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by $h_{n+1}+h_n$ leaves $h_{n+1}-h_n=\frac{1}{\sqrt{3}}$ . So the difference in height of successive triangles is $\frac{1}{\sqrt{3}}$ , meaning their bases are wider by $2/3$ units each time. From here, one can proceed as in Solution 1 to arrive at $n=\boxed{17}$

Answer:  17
Problem: Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$

Solution: Note that $A_{1}$ is of the form $(2x,0)$ for some $x$ , and thus $B_{1}$ is of the form $(x, x \sqrt{3}).$ Then, we are told that $B_{1}$ lies on the graph of $y = \sqrt{x}$ , so \[(x \sqrt{3})^{2} = x.\] Solving for x, we get that $x = \frac{1}{3},$ and so $A_{1} = (2/3,0)$
Now, similarly to before, let $|A_{1}A_{2}|=2y.$ Then, $B_{2} = (y+\frac{2}{3},y \sqrt{3})$ , and so \[(y \sqrt{3})^{2} = (y+\frac{2}{3}).\] Solving using the quadratic formula gives \[y = \frac{-(-1) \pm \sqrt{1^{2}-4(3)(-\frac{2}{3})}}{6} = \frac{1 \pm \sqrt{1^{2}+8}}{6} = \frac{2}{3}.\] Then, $2y = \frac{4}{3},$ so $A_{2} = (2,0).$
In general, if $A_{n} = (a_{n},0)$ for all integers $n$ , and $|A_{n}A_{n+1}| = 2x$ for some real number $x$ , we have the following equation for $x$ \[(x \sqrt{3})^{2} = x+a_{n},\] which give us when plugged into the quadratic formula gives \[x = \frac{1 \pm \sqrt{1+12a_{n}}}{6},\] but since $x$ must be positive, we have that \[x = \frac{1 + \sqrt{1+12a_{n}}}{6},\] and so \[a_{n+1} = a_{n}+2x = a_{n}+\frac{1 + \sqrt{1+12a_{n}}}{3}.\] Computing a few terms of $a_{n}$ using this method gives $a_{3} = 4,$ $a_{4} = \frac{20}{3},$ and $a_{5} = 10$
Notice how all our $a_{n}$ terms so far are rational, even though there is an abundance of radicals in the recurrence. This motivates us to look at our discriminants in the quadratic formula that is solved for $x$
The discriminant of \[(x \sqrt{3})^{2} = x+a_{2}\] is \[1+12 \cdot \frac{2}{3} = 9.\] Similarly, the discriminant of \[(x \sqrt{3})^{2} = x+a_{3}\] is $1+12 \cdot a_{3} = 25,$ and $1+12 \cdot a_{4} = 49$
Note how our results keep coming out as the squares of the odd integers. Moreover, it seems that \[1+12a_{n} = (2n+1)^{2}.\] We will prove this with induction. The base case, $n=2$ , we have already verified.
Now, for the Inductive step, assume that \[1+12a_{n-1} = (2n-1)^{2}.\] for some integer $n-1$ . We will prove that this is true for $n$ as well.
Plugging this into our recurrence formula gives us \[a_{n} = a_{n}+2x = a_{n-1}+\frac{1 + \sqrt{1+12a_{n-1}}}{3}\] \[a_{n} = \frac{(2n-1)^{2}-1}{12} + \frac{1 + (2n-1)}{3}\] \[a_{n} = \frac{4n^{2}-4n+1-1}{12} + \frac{8n}{12}\] \[a_{n} = \frac{4n^{2}+4n+1-1}{12}\] \[a_{n} = \frac{(2n+1)^{2}-1}{12}.\] Therefore, we have proved our claim. Now, we have that $|A_{0}A_{n}|=a_{n},$ so we just need the least integer $n$ so that \[a_{n} > 100,\] or \[(2n+1)^{2} > 1201.\] Then, we see that $35^{2} = 1225$ is the smallest odd square larger than $1201.$ Therefore, we have $2n+1=35$ , so $n = \boxed{17}.$

Answer:  17
Problem: Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$

Solution: We can iteratively calculate out the first few $A_i$ and $B_i$ . We know that $A_0 = (0,0)$ and the line through $A_0$ and $B_1$ needs to make a $60^{\circ}$ angle with the x-axis (because the triangle is equilateral). The equation of a line that makes an angle $\theta$ with the x-axis and passes through the origin has equation $\tan(\theta)x$ , so the line through $A_0$ and $B_1$ has equation $\sqrt{3}x$ . We set this equal to $\sqrt{x}$ (to find where the two curves intersect) and, when solving, find that $x = \frac{1}{3}$ . Therefore, $B_1 = (\frac{1}{3}, \frac{\sqrt{3}}{3})$ . By equilateral triangle properties, we must then have that $A_1 = (\frac{2}{3}, 0)$ . To find $B_2$ , we find the equation of the line through $A_1$ that makes a $60^{\circ}$ angle with the $x$ -axis. This is the same line as the one through the origin (which we already found) shifted $\frac{2}{3}$ to the right, so it has equation $y = \sqrt{3}(x - \frac{2}{3})$ . Setting this equal to $\sqrt{x}$ and solving for $x$ , we get $x = \frac{4}{3}$ , so $B_2 = (\frac{4}{3}, \frac{2\sqrt{3}}{3})$ . By equilateral triangle properties, we have that $A_2 = (2, 0)$ . Repeating this process, we find that $B_3 = (3, \sqrt{3})$ and $A_3 = (4,0)$ . At this point, we notice that the $y$ -coordinate of each $B_i$ is $\frac{\sqrt{3}}{3}$ more than that of $B_{i-1}$ , so $B_4 = (\frac{16}{3}, \frac{4\sqrt{3}}{3})$ and $A_4 = (\frac{20}{3}, 0)$ . It may be helpful to continue calculating out the $A_i$ and $B_i$ if the pattern in the $x$ -coordinates of the $A_i$ isn't visible yet. At this point, it can be noticed that the differences between the x coordinates of consecutive $A_i$ form an arithmetic sequence ( $x_{A_1} - x_{A_0} = \frac{2}{3}$ $x_{A_2} - x_{A_1} = \frac{4}{3}$ , etc.). Additionally, the numerator of each of the fractions is a consecutive even number, and the sum of the first $n$ even numbers is $n(n+1)$ . Thus, we must find the least integral solution to $\frac{n(n+1)}{3} \geq 100$ . This least solution is $\boxed{17}$

Answer: C 17
Problem: Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?
$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

Solution: \[16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14\] \[\frac{14}2=7\Rightarrow\boxed{7}\]

Answer: A 7
Problem: A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$

Solution: The point $(a,b)$ is the foot of the perpendicular from $(12,10)$ to the line $y=-5x+18$ . The perpendicular has slope $\frac{1}{5}$ , so its equation is $y=10+\frac{1}{5}(x-12)=\frac{1}{5}x+\frac{38}{5}$ . The $x$ -coordinate at the foot of the perpendicular satisfies the equation $\frac{1}{5}x+\frac{38}{5}=-5x+18$ , so $x=2$ and $y=-5\cdot2+18=8$ . Thus $(a,b) = (2,8)$ , and $a+b = \boxed{10}$

Answer:  10
Problem: A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$

Solution: If the mouse is at $(x, y) = (x, 18 - 5x)$ , then the square of the distance from the mouse to the cheese is $(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).$ The value of this expression is smallest when $x = 2$ , so the mouse is closest to the cheese at the point $(2, 8)$ , and $a+b=2+8 = \boxed{10}$

Answer:  10
Problem: A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$

Solution: We are trying to find the point where distance between the mouse and $(12, 10)$ is minimized. This point is where the line that passes through $(12, 10)$ and is perpendicular to $y=-5x+18$ intersects $y=-5x+18$ . By basic knowledge of perpendicular lines, this line is $y=\frac{x}{5}+\frac{38}{5}$ . This line intersects $y=-5x+18$ at $(2,8)$ . So $a+b=\boxed{10}$ . - MegaLucario1001

Answer:  10
Problem: A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$

Solution: If the mouse is at $(x, y) = (x, 18 - 5x)$ , then the square of the distance from the mouse to the cheese is $(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4)$ .
The value of this expression is smallest when $x = 2$ , so the mouse is closest to the cheese at the point $(2, 8)$ , and $a+b=2+8 = \boxed{10}$ .
-Paixiao

Answer:  10
Problem: Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$

Solution: Let $S_{n}$ denote the number of spacy subsets of $\{ 1, 2, ... n \}$ .  We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$
The spacy subsets of $S_{n + 1}$ can be divided into two groups:
Hence,
From this recursion , we find that
And so the answer is $\boxed{129}$

Answer: E 129
Problem: Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$

Solution: Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.
From the set $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents ${1,4,7,10}$
For subsets of size $k$ there must be $2(k - 1)$ dividers between the balls, leaving $12 - k - 2(k - 1) = 12 - 3k + 2$ dividers to be be placed in $k + 1$ spots between the balls. The number of way this can be done is $\binom{(12 - 3k + 2) + (k + 1) - 1}k = \binom{12 - 2k + 2}k$
Therefore, the number of spacy subsets is $\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = \boxed{129}$

Answer:  129
Problem: Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$

Solution: A shifting argument is also possible, and is similar in spirit to Solution 2. Clearly we can have at most $4$ elements. Given any arrangment, we subract $2i-2$ from the $i-th$ element in our subset, when the elements are arranged in increasing order. This creates a bijection with the number of size $k$ subsets of the set of the first $14-2k$ positive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive numbers. Therefore, we can easily plug in the possible integers 0, 1, 2, 3, 4, 5 for $k$ ${14 \choose 0} + {12 \choose 1} + {10 \choose 2} + {8 \choose 3} + {6 \choose 4} = \boxed{129}$

Answer:  129
Problem: Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$

Solution: Let us consider each size of subset individually. Since each integer in the subset must be at least $3$ away from any other integer in the subset, the largest spacy subset contains $4$ elements.
First, it is clear that there is $1$ spacy set with $0$ elements in it, the empty set. Next, there are $12$ spacy subsets with $1$ element in them, one for each integer $1$ through $12$
Now, let us consider the spacy subsets with $2$ elements in them. If the smaller integer is $1$ , the larger integer is any of the $9$ integers from $4$ to $12$ . If the smaller integer is $2$ , the larger integer is any of the $8$ integers from $5$ to $12$ . This continues, up to a smaller integer of $9$ and $1$ choice for the larger integer, $12$ . This means that there are $9 + 8 + \cdots + 1 = 45$ spacy subsets with $2$ elements.
For spacy subsets with $3$ elements, we first consider the middle integer. The smallest such integer is $4$ , and it allows for $1$ possible value for the smaller integer ( $1$ ) and possible $6$ values for the larger integer ( $7$ through $12$ ), for a total of $1 \cdot 6 = 6$ possible subsets. The next middle integer, $5$ , allows for $2$ smaller integers and $5$ larger integers, and this pattern continues up until the middle integer of $9$ , which has $6$ values for the smaller integer and $1$ value for the larger integer. This means that there are $1 \cdot 6 + 2 \cdot 5 + \cdots + 6 \cdot 1 = 56$ spacy subsets with $3$ elements.
Lastly, there are $3$ main categories for spacy subsets with $4$ elements, defined by the difference between their smallest and largest values. The difference ranges from $9$ to $11$ . If it is $9$ , there is only $1$ set of places to put the two middle values ( $n + 3$ and $n + 6$ , where $n$ is the smallest value). Since there are $3$ possible sets of smallest and largest values, there are $1 \cdot 3 = 3$ sets in this category. If the difference is $10$ , there are now $3$ sets of places to put the two middle values ( $n + 3$ and $n + 6$ or $7$ , and $n + 4$ and $n + 7$ ). There are $2$ possible sets of smallest and largest values, so there are $3 \cdot 2 = 6$ sets in this category. Finally, if the difference is $11$ , there are $6$ possible sets of places to put the two middle values ( $n + 3$ and $n + 6$ $7$ , or $8$ $n + 4$ and $n + 7$ or $8$ , and $n + 5$ and $n + 8$ ) and one possible set of smallest and largest values, meaning that there are $6 \cdot 1 = 6$ sets in this category. Adding them up, there are $3 + 6 + 6 = 15$ spacy subsets with $4$ elements.
Adding these all up, we have a total of $1 + 12 + 45 + 56 + 15 = \boxed{129}$ spacy subsets. ~ emerald_block

Answer: E 129