Navvye/Olympiad-Problems-CSV huggingface dataset:
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: We build a modified version of Markov's chain to solve this problem. First, start off by representing each element of the chosen subset with a $1$ , and every other element $0$ . The possible states are as follows; if the restriction was broken previously or not. For the sake of not over counting, we let the former be its own state (call it "C"), and the latter will be split up into more states. The previous two digits can either be $00$ $10$ , or $01$ . If a previous digit doesn't exist, it is $0$ for obvious reasons. We know that we start off at state " $00$ ". We have two options; include the first digit ( $1$ ) or not ( $0$ ). Thus, state $00$ transitions to state $01$ in $1$ case and state $00$ in $1$ case. State $01$ transitions to $10$ in $1$ case and the "C" state in $1$ case. State $10$ transitions to $00$ in $1$ case and state "C" in $1$ case. Since it doesn't matter where or how many times the restriction is broken for the subset to fail, state "C" can transition to itself in $2$ cases.
Notice how this corresponds to a transition matrix of $\begin{bmatrix}  	1 & 1 & 0 & 0\\ 	0 & 0 & 1 & 1\\ 	1 & 0 & 0 & 1 \\         0 & 0 & 0 & 2 \\ \end{bmatrix}$ . Since we start off on state $00$ , our starting state matrix is $\begin{bmatrix}  	1 & 0 & 0 & 0\\ \end{bmatrix}$ . We wish to compute the value $a+b+c = 4096-d$ when given $f^{12}(\begin{bmatrix}  	1 & 0 & 0 & 0\\ \end{bmatrix})=\begin{bmatrix}  	a & b & c & d\\ \end{bmatrix}$ , where $f(x)$ is $x\times\begin{bmatrix}  	1 & 1 & 0 & 0\\ 	0 & 0 & 1 & 1\\ 	1 & 0 & 0 & 1 \\         0 & 0 & 0 & 2 \\ \end{bmatrix}$ Through repeated applications of matrix multiplication, we obtain the output matrix of $\begin{bmatrix}  	60 & 41 & 28 & 3967\\ \end{bmatrix}$ . Our answer is thus $60+41+28=\boxed{129}$ . 
Remark: Markov's Chain is more commonly used to calculate probabilities, but it also can be very useful for counting problems with a uniform requirement.
-SigmaPiE

Answer:  129
Problem: Let $a$ $b$ , and $c$ be digits with $a\ne 0$ . The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21$

Solution: Let $k$ be the lesser of the two integers. Then the squares of the integers are $k^2$ and $k^2+2k+1$ , and the distance between them is $2k+1$ . Let this be equivalent to $3d$ , so that the one-third of the distance between the squares is equivalent to $d$ . The numbers $abc$ and $acb$ are one-third and two-thirds of the way between $k^2$ and $(k+1)^2$ . Therefore, the distance between these two numbers is also one-third the distance between the squares, or $d$ . Setting these equal to each other, we have
$\frac{2k+1}{3} = 9(c-b)$
$\Rightarrow 2k+1 = 27(c-b)$
Notice that since $c$ and $b$ are digits, their difference is at most $9$ and at least $0$ . Also notice that since $acb$ is greater than $abc$ $c > b$ . Representing this as an inequality, we have
$27 \le 27(c-b) \le 243$
Substituting $2k+1$ , we have
$27 \le 2k+1 \le 243$
$\Rightarrow 13 \le k \le 121$
However, we know that $abc$ is a $3$ -digit number, and since $k^2$ is less than $abc$ $k^2$ must be at most $961$ , or $31^2$ . Therefore $k \le 31$ . Plugging this back into our inequality, we have
$13 \le k \le 31$
$\Rightarrow 27 \le 2k+1 \le63$
$\Rightarrow 27 \le 27(c-b) \le 63$
$\Rightarrow 1 \le (c-b) \le \frac{7}{3}$
But (c-b) must be an integer, so now we have
$1 \le (c-b) \le 2$
$\Rightarrow 27 \le 27(c-b) \le 54$
$\Rightarrow 27 \le 2k+1 \le 54$
$\Rightarrow 13 \le k \le\frac{53}{2}$
$k$ is also an integer, so now we have
$\Rightarrow 13 \le k \le 26$
$\Rightarrow 27 \le 2k+1 \le 53$
$\Rightarrow 27 \le 27(c-b) \le 53$
$\Rightarrow 1 \le (c-b) \le \frac{53}{27}$
Once again, $(c-b)$ must be an integer, so we have
$1 \le (c-b) \le 1$
$\Rightarrow (c-b) = 1$
$\Rightarrow 27(c-b) = 27$
$\Rightarrow 2k+1 = 27$
$\Rightarrow k = 13$
The two squares are $13^2$ and $14^2$ , or $169$ and $196$ . A third of the distance between them is $9$ , and $169 +9 = 178$ $1 + 7 + 8 = 16 \Rightarrow \boxed{16}$

Answer: C 16
Problem: The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution: Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$ . Because $72= 4\cdot 18$ , we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$ , which gives $c=3d$ (consider a homothety , or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$ , it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$ ). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$ , so substituting $c = 3d$
Thus $3|d$ , and we verify that $d = 3$ $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$ . Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{16}$

Answer: D 16
Problem: The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution: The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$ . Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\frac{d-c}{b-a}$ $\frac{(d-c)^2}{b-a}=18$ , and the area contained by the latter is $\frac{(c+d)^2}{b-a}=72$ . Thus, $d=3c$ and $b-a$ must be even if the former quantity is to equal $18$ $c^2=18(b-a)$ so $c$ is a multiple of $3$ . Putting this all together, the minimal solution for $(a,b,c,d)=(3,1,3,9)$ , so the sum is $\boxed{16}$

Answer: D 16
Problem: The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution: Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$ ) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$ . Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is $\frac{bh}{2}$ , but since a parallelogram is two such triangles, the area becomes $bh$
Let $b_1$ and $h_1$ denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and $b_2$ and $h_2$ denote those of the larger parallelogram. Notice that $b_1$ is simple the distance from $(0, d)$ to $(0, c)$ , or $(c-d)$ . Also notice that $h_1$ is the distance from the $x$ -axis to the intersection of lines $ax+c$ and $bx+d$ . This is equivalent to the value of the $x$ -coordinate of intersection, so we solve for $x$
$ax+c=bx+d$
$\Rightarrow bx-ax = c-d$
$\Rightarrow (b-a)x = c-d$
$\Rightarrow x = \frac{c-d}{b-a}$
The area of the smaller parallelogram is $b_1*h_1$ , or
$(c-d) * \frac{c-d}{b-a}$
$\Rightarrow \frac{(c-d)^2}{b-a}$
$b_2$ is the distance from $(0, -d)$ to $(0, c)$ , or $(c+d)$ $h_2$ is the $x$ -coordinate of the intersection of the lines $ax+c$ and $bx-d$ . Again, we solve for x:
$ax+c=bx-d$
$\Rightarrow bx-ax = c+d$
$\Rightarrow (b-a)x = c+d$
$\Rightarrow x = \frac{c+d}{b-a}$
The area of the larger parallelogram is $b-1*h_1$ , or
$(c+d) * \frac{c+d}{b-a}$
$\Rightarrow \frac{(c+d)^2}{b-a}$
The areas of the parallelograms are given to us: $18$ and $72$ . Therefore we can set up a ratio:
$\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}$
$\Rightarrow 18(c+d)^2 = 72(c-d)^2$
$\Rightarrow (c+d)^2 = 4(c-d)^2$
$\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2$
$\Rightarrow 3c^2 - 10cd +3d^2 = 0$
$\Rightarrow (3c-d)(c-3d)=0$
$\Rightarrow c=3d, c=\frac{d}{3}$
We established earlier that $c>d$ , so $c=3d$ . Plugging this into the intial equations yields
$\frac{16d^2}{b-a} = 72$
and
$\frac{4d^2}{b-a} = 18$
Solving for $d$ , we get
$d = 3\sqrt{\frac{b-a}{2}}$
We want the sum of $a$ $b$ $c$ , and $d$ . We can now rewrite this
$a + b + 12\sqrt{\frac{b-a}{2}}$
We are told that $a$ $b$ $c$ , and $d$ are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this
$\frac{b-a}{2} = k^2$
where $k$ is some positive integer.
Rearranging, we get
$b = a + 2k^2$
Now we can rewrite the sum as
$a + a + 2k^2 +12k$
Since both $a$ and $k$ must be at least $1$ , the minimum value is
$1 + 1 + 2(1)^2 + 12(1) = 16 \Rightarrow \boxed{16}$

Answer: D 16
Problem: How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?
$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution: Let $a$ and $b$ be the two legs of the triangle, and $c$ be the hypotenuse.
By using $Area = \frac{r}{2} (a+b+c)$ , where $r$ is the in-radius, we get:
\[3(a+b+c) = \frac{r}{2} (a+b+c)\] \[r=6\]
In right triangle, $r = \frac{a+b-c}{2}$ \[a+b-c = 12\] \[c = a + b - 12\]
By the triangle's area we get:
\[\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}\] \[ab = 6(a+b+c)\]
By substituting $c$ in:
\[ab = 6(a+b+a + b - 12)\] \[ab - 12a - 12b + 72 = 0\] \[(a - 12)(b - 12) = 72\]
As $72 = 2^3 \cdot 3^2$ , there are $\frac{(3+1)(2+1)}{2} = 6$ solutions, $\boxed{6}$

Answer: A 6
Problem: How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?
$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution: Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is $\boxed{6}$

Answer: A 6
Problem: problem_id
c84575ab947ea92b2fa92f55386966ac    Also refer to the 2007 AMC 10B #25 (same problem)
c84575ab947ea92b2fa92f55386966ac    How many pairs of positive integers $(a,b)$ ar...
Name: Text, dtype: object

Solution: Rewrite the equation\[ $\frac{a}{b}+\frac{14b}{9a}=k$ \]in two different forms. First, multiply both sides by $b$ and subtract $a$ to obtain\[ $\frac{14b^2}{9a}=bk-a.$ \]Because $a$ $b$ , and $k$ are integers, $14b^2$ must be a multiple of $a$ , and because $a$ and $b$ have no common factors greater than 1, it follows that 14 is divisible by $a$ . Next, multiply both sides of the original equation by $9a$ and subtract $14b$ to obtain\[ $\frac{9a^2}{b}=9ak-14b.$ \]This shows that $9a^2$ is a multiple of $b$ , so 9 must be divisible by $b$ . Thus if $(a,b)$ is a solution, then $b=1$ $3$ , or $9$ , and $a=1$ , 2, 7, or 14. This gives a total of twelve possible solutions $(a,b)$ , each of which can be checked quickly. $\[
\frac{a}{b}+\frac{14b}{9a}\]$ (Error compiling LaTeX. Unknown error_msg) will only be an integer when $(a,b) \in \{(1,3),(2,3), (7,3), (14,3)\},$ for a total of $\boxed{4}$ pairs.

Answer:  4
Problem: problem_id
ff8b260c6e48fc087b54f3971592eb5d    Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d    Let us simplify this problem.  Dividing by $30...
Name: Text, dtype: object

Solution: The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD ( greatest common divisor ) of $a$ and $b$ . Therefore, the answer is $gcd(300,210)=\boxed{30}.$

Answer: C 30
Problem: problem_id
ff8b260c6e48fc087b54f3971592eb5d    Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d    Let us simplify this problem.  Dividing by $30...
Name: Text, dtype: object

Solution: Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly.  Since our supposition can be achieved, the answer is $\boxed{30}$

Answer: C 30
Problem: The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and
$f(x)+f\left(\frac{1}{x}\right)=x$
What is the largest set of real numbers that can be in the domain of $f$
$\mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}$
$\mathrm{(C) \ } \{x|x>0\}$ $\mathrm{(D) \ } \{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\}$
$\mathrm{(E) \ }  \{-1,1\}$

Solution: We know that $f(x) + f \left(\frac{1}{x}\right) = x.$ Plugging in $x = \frac{1}{x}$ we get \[f \left(\frac{1}{x}\right) + f \left(\frac{1}{\frac{1}{x}}\right) = \frac{1}{x}\] \[f \left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.\]
Also notice \[f \left(\frac{1}{x}\right) + f(x) = x\] by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set $\frac{1}{x} = x$ which gives us $x = \pm 1$ which is answer option $\boxed{1,1}.$

Answer: E 1,1
Problem: Let $S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}$ and $S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}$
What is the ratio of the area of $S_2$ to the area of $S_1$
$\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ }  102$

Solution: Looking at the constraints of $S_1$
$x+y > 0$
$\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10} 10 +\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10}(10x+10y)$
$1+x^2+y^2 \le 10x+10y$
$x^2-10x+y^2-10y \le -1$
$x^2-10x+25+y^2-10y+25 \le 49$
$(x-5)^2 + (y-5)^2 \le (7)^2$
$S_1$ is a circle with a radius of $7$ . So, the area of $S_1$ is $49\pi$
Looking at the constraints of $S_2$
$x+y > 0$
$\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)$
$\log_{10}(2+x^2+y^2)\le \log_{10} 100 +\log_{10}(x+y)$
$\log_{10}(2+x^2+y^2)\le \log_{10}(100x+100y)$
$2+x^2+y^2 \le 100x+100y$
$x^2-100x+y^2-100y \le -2$
$x^2-100x+2500+y^2-100y+2500 \le 4998$
$(x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2$
$S_2$ is a circle with a radius of $7\sqrt{102}$ . So, the area of $S_2$ is $4998\pi$
So the desired ratio is $\frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{102}$

Answer: E 102
Problem: The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

Solution: By the Multinomial Theorem , the summands can be written as
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}\]
and
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},\]
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:
\[{2006+2\choose 2} = 2015028\]
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:
if the exponent of $y$ is $1$ , the exponent of $z$ can be all even integers up to $2004$ , so there are $1003$ terms.
if the exponent of $y$ is $3$ , the exponent of $z$ can go up to $2002$ , so there are $1002$ terms.
$\vdots$
if the exponent of $y$ is $2005$ , then $z$ can only be 0, so there is $1$ term.
If we add them up, we get $\frac{1003\cdot1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003\cdot1004$ negative terms.
By subtracting this number from 2015028, we obtain $\boxed{1,008,016}$ or $1,008,016$ as our answer.

Answer: D 1,008,016
Problem: The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

Solution: Define $P$ such that $P=y+z$ . Then the expression in the problem becomes: $(x+P)^{2006}+(x-P)^{2006}$
Expanding this using binomial theorem gives $(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)$ , where $n=2006$ (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).
Simplifying gives: $2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)$ . Note that two terms that come out of different powers of $P$ cannot combine and simplify, as their exponent of $x$ will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, $P^k=(y+z)^k$ will have $k+1$ terms, so the answer is $1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{1,008,016}$

Answer: D 1,008,016
Problem: The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

Solution: Using stars and bars we know that $(x+y+z)^{2006}$ has ${2006+2\choose 2}$ or $2015028$ different terms. Now we need to find out how many of these terms are canceled out by $(x-y-z)^{2006}$ . We know that for any term(let's say $x^{a}(-y)^{b}(-z)^{c}$ ) where $a+b+c=2006$ of the expansion of $(x-y-z)^{2006}$ is only going to cancel out with the corresponding term $x^{a}y^{b}z^{c}$ if only $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. Now let's do some casework to see how many terms fit this criteria:
Case 1: $a$ is even
Now we know that $a$ is even and $a+b+c=2006$ . Thus $b+c$ is also even or both $b$ and $c$ are odd or both $b$ and $c$ are even. This case clearly fails the above criteria and there are 0 possible solutions.
Case 2: $a$ is odd
Now we know that $a$ is odd and $a+b+c=2006$ . Thus $b+c$ is odd and $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. All terms that have $a$ being odd work.
We now need to figure out how many of the terms have $a$ as a odd number. We know that $a$ can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus $\frac{1003}{2007}$ of the $2015028$ terms will cancel out and $\frac{1004}{2007}$ of the terms will work. Thus there are $(\frac{1004}{2007})(2015028)$ terms. This number comes out to be $1,008,016$ $\implies \boxed{1,008,016}$ (Author: David Camacho)

Answer: D 1,008,016
Problem: The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

Solution: Noticing how $y$ and $z$ are negative in the second part of the expression, let $x=a$ and $y+z = b$ . Then we get \[(a+b)^{2006} + (a-b)^{2006}\] We know that the terms that don't cancel out have even powers of $a$ and $b$ . Our expansion will be in the form of \[2a^{2006} + x_1a^{2004}b^{2} + x_2a^{2002}b^{4} + \cdots + 2b^{2006}\] Note that $b^n = (x+y)^n$ has $n+1$ terms. Furthermore, the current expression is irreducible as each term has a different $x$ power. Thus, when we write $a$ and $b$ back to their original terms, we will have $1+3+5+ \cdots + 2007 = 1004^2 = 1,008,016 \implies \boxed{1,008,016}$

Answer: D 1,008,016
Problem: How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ }  405$

Solution: This question can be solved fairly directly by casework and pattern-finding.  We give a somewhat more general attack, based on the solution to the following problem:
How many ways are there to choose $k$ elements from an ordered $n$ element set without choosing two consecutive members?
You want to choose $k$ numbers out of $n$ with no consecutive numbers. For each configuration, we can subtract $i-1$ from the $i$ -th element in your subset. This converts your configuration into a configuration with $k$ elements where the largest possible element is $n-k+1$ , with no restriction on consecutive numbers. Since this process is easily reversible, we have a bijection .
Without consideration of the second condition, we have: ${15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \choose 8}$
Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than $k$ , which translates to subtracting $k - 1$ from the "top" of each binomial coefficient .
Now we have, after we cancel all the terms ${n \choose k}$ where $n < k$ ${15 \choose 1} + {13 \choose 2} + {11 \choose 3} + {9 \choose 4} + {7 \choose 5}=  15 + 78 + 165 + 126 + 21 = \boxed{405}$

Answer: E 405