Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$

Solution: The center of the incircle is $O.$ Denote the points in which the incircle meets $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ and $\overline{DA}$ as $W,$ $X,$ $Y,$ and $Z,$ respectively. Next, also denote the base of the perpendicular from $P$ to $\overline{AB},$ $\overline{AD},$ $\overline{OW},$ and $\overline{OZ}$ as $M,$ $N,$ $S,$ and $T,$ respectively.
We can easily see that the radius of the circle is $\frac{25}{2}.$ Using this and Pythagorus on right $\triangle OSP$ and $\triangle OTP,$ we find that $MW = PS = 10$ and $NZ = PT = 12.$
Since $AW = AZ$ by properties of circle tangents, we can deduce by the above information that $AM = AN+2.$ Doing Pythagorus on right $\triangle AMP$ and $\triangle ANP$ we find that $a^2 = b^2 + 56$ (because $a^2+25=b^2+81.$ ) From solving the $2$ just derived equations, we find that $AM=15$ and $AN=13.$
Next, we use Pythagorus on right $\triangle AOB$ (we can see it's right because of properties of rhombuses.) We get \[AB^2 = AO^2 + BO^2.\] We know $AB = AW + WB = 25 + WB.$ By Pythagorus on $\triangle AWO$ and $\triangle BWO,$ we also know $AO^2 = 25^2+\left(\frac{25}{2}\right)^2$ and $BO^2=WB^2+\left(\frac{25}{2}\right)^2.$ Substituting these in, we have \[25^2 + 50WB + WB^2 = 25^2+\left(\frac{25}{2}\right)^2+\left(\frac{25}{2}\right)^2+WB^2.\] Solving for $WB,$ we get $WB = \frac{25}{4}.$ Now we find that each side of the rhombus $=AB=25+\frac{25}{4}=\frac{125}{4}.$ The perimeter of the rhombus would be that times $4.$ Our final answer is \[\frac{125}{4}\cdot4=\boxed{125}.\]

Answer:  125
Problem: Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$

Solution: Notation is shown on diagram, $RT \perp AD, FG \perp AB, E = AD \cap \omega, E' = FG \cap AD.$ $RT = 9 + 16 = 25 = FG$ as hights of rhombus. \[RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,\] \[PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.\] \[PE =  \sqrt{RP^2 + RE^2} = 15 \implies E = E'.\] \[\sin \alpha = \frac {RE}{PE} =  \frac {GF}{AD} \implies AD = \frac {15 \cdot 25}{12} = \frac {125}{4}.\] The perimeter of $ABCD$ is $\frac{125}{4}\cdot4=\boxed{125}.$

Answer:  125
Problem: Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$

Solution: $p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$ , with $m\neq 2$
In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$ , with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$ ), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).
In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$ , and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $328$ polynomials.
The total is $\boxed{738}$

Answer:  738
Problem: There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)

Solution: Define $\left\{ x \right\} = x - \left\lfloor x \right\rfloor$
First, we bound $U$
We establish an upper bound of $U$ . We have \begin{align*} U & \leq \sum_{n=1}^{2023} \frac{n^2 - na}{5} \\ & = \frac{1}{5} \sum_{n=1}^{2023} n^2 - \frac{a}{5} \sum_{n=1}^{2023} n \\ & = \frac{1012 \cdot 2023}{5} \left( 1349 - a \right) \\ & \triangleq UB . \end{align*}
We establish a lower bound of $U$ . We have \begin{align*} U & =  \sum_{n=1}^{2023} \left(  \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\}  \right) \\ & = \sum_{n=1}^{2023} \frac{n^2 - na}{5} - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = UB -  \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & \geq UB -  \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} . \end{align*}
We notice that if $5 | n$ , then $\frac{n^2 - na}{5} \in \Bbb Z$ .
Thus, \begin{align*} U & \geq UB -  \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} \\ & \geq UB -  \sum_{n=1}^{2023} \mathbf 1 \left\{ 5 \nmid n \right\} \\ & = UB - \left( 2023 - \left\lfloor \frac{2023}{5} \right\rfloor \right) \\ & = UB - 1619 \\ & \triangleq LB . \end{align*}
Because $U \in \left[ - 1000, 1000 \right]$ and $UB - LB = 1619 < \left( 1000 - \left( - 1000 \right) \right)$ , we must have either $UB \in \left[ - 1000, 1000 \right]$ or $LB \in \left[ - 1000, 1000 \right]$
For $UB \in \left[ - 1000, 1000 \right]$ , we get a unique $a = 1349$ .
For $LB \in \left[ - 1000, 1000 \right]$ , there is no feasible $a$
Therefore, $a = 1349$ . Thus $UB = 0$
Next, we compute $U$
Let $n = 5 q + r$ , where $r = {\rm Rem} \ \left( n, 5 \right)$
We have \begin{align*} \left\{ \frac{n^2 - na}{5} \right\} & = \left\{ \frac{\left( 5 q + r \right)^2 - \left( 5 q + r \right)\left( 1350 - 1 \right)}{5} \right\} \\ & = \left\{ 5 q^2 + 2 q r - \left( 5 q + r \right) 270 + q + \frac{r^2 + r}{5}  \right\} \\ & = \left\{\frac{r^2 + r}{5}  \right\} \\ & = \left\{ \begin{array}{ll} 0 & \mbox{ if } r = 0, 4 \\ \frac{2}{5} & \mbox{ if } r = 1, 3 \\ \frac{1}{5} & \mbox{ if } r = 2 \end{array} \right. . \end{align*}
Therefore, \begin{align*} U & =  \sum_{n=1}^{2023} \left(  \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\}  \right) \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{q=0}^{404} \sum_{r=0}^4 \left\{\frac{r^2 + r}{5}  \right\} + \left\{ \frac{0^2 - 0 \cdot a}{5} \right\} + \left\{ \frac{2024^2 - 2024a}{5} \right\} \\ & = - \sum_{q=0}^{404} \left( 0 + 0 + \frac{2}{5} + \frac{2}{5} + \frac{1}{5} \right) + 0 + 0 \\ & = - 405 . \end{align*}
Therefore, $a + U = 1349 - 405 = \boxed{944}$

Answer:  944
Problem: There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)

Solution: We define $U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}$ . Since for any real number $x$ $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$ , we have $U \le U' \le U + 2023$ . Now, since $-1000 \le U \le 1000$ , we have $-1000 \le U' \le 3023$
Now, we can solve for $U'$ in terms of $a$ . We have: \begin{align*} U' &= \sum^{2023}_{n=1} {\frac{n^2-na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5} - \frac{na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5}} - \sum^{2023}_{n=1} {\frac{na}{5}} \\ &= \frac{\sum^{2023}_{n=1} {{n^2}} - \sum^{2023}_{n=1} {na}}{5} \\ &= \frac{\frac{2023(2023+1)(2023 \cdot 2 + 1)}{6} - \frac{a \cdot 2023(2023+1)}{2} }{5} \\ &= \frac{2023(2024)(4047-3a)}{30} \\ \end{align*} So, we have $U' = \frac{2023(2024)(4047-3a)}{30}$ , and $-1000 \le U' \le 3023$ , so we have $-1000 \le \frac{2023(2024)(4047-3a)}{30} \le 3023$ , or $-30000 \le 2023(2024)(4047-3a) \le 90690$ . Now, $2023 \cdot 2024$ is much bigger than $90690$ or $30000$ , and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$ , or $a = 1349$ , and $U' = 0$
Now, we can find $U - U'$ . We have: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor} - \sum^{2023}_{n=1} {\frac{n^2-1349n}{5}} \\ &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \end{align*}. Now, if $n^2-1349n \equiv 0 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = 0$ , and if $n^2-1349n \equiv 1 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = -\frac{1}{5}$ , and so on. Testing with $n \equiv 0,1,2,3,4, \text{ (mod 5)}$ , we get $n^2-1349n \equiv 0,2,1,2,0 \text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$ , we get: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \\ &= 404 \cdot 0 - 405 \cdot \frac{2}{5} - 405 \cdot \frac{1}{5} - 405 \cdot \frac{2}{5} - 404 \cdot t0 \\ &= -405(\frac{2}{5}+\frac{1}{5}+\frac{2}{5}) \\ &= -405 \end{align*} Since $U' = 0$ , this gives $U = -405$ , and we have $a + U = 1349-405 = \boxed{944}$

Answer:  944
Problem: There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)

Solution: We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$ ) from the numerator of $\frac{n^2-na}{5}$ . For example, $\left\lfloor \frac{7}{5} \right\rfloor = \frac{7-2}{5} = 1$
Note that the congruence of $n^2 - na$ (mod $5$ ) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$ ) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$ ).
For example, if $a \equiv 1$ (mod $5$ ), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \equiv 1,2,3,4,0$ (mod $5$ ). Let the sum of these elements be $s$ . Note that for each “loop” of the numerator (mod $5$ ), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$ ) throughout the sum, as $5 \cdot 404=2020$ . Then
$U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2}  -404s \right)}{5}$
Where $n=2023$ . Note that since $5 \cdot 404=2020$ , this is an approximation for $U$ because the equation disregards the remainder (mod $5$ ) when $n=2021, 2022$ , and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000<U<1000$ , so we can proceed with this approximation to solve for $a$
Factoring our approximation gives $U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$
We set $a= \frac{(2n+1)}{3} = 1349$ to make $\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$ , accordingly minimizing $|U|$ , yielding $U \approx  \frac{-404s}{5}$
If $a$ increases or decreases by $1$ , then $U$ changes by $\frac {(n)(n+1)}{2 \cdot 5} = \frac {2023 \cdot 2024}{10}$ which clearly breaks the inequality on $U$ . Therefore $a=1349 \equiv 4$ (mod $5$ ) giving the set of remainders $(2,1,2,0,0)$ , so $s=5$ and our approximation yields $U \approx -404$ . However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \frac{(2+1+2)}{5} = - 405$ , giving an answer of $1349-405= \boxed{944}$

Answer:  944
Problem: There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)

Solution: Consider the integral \[\int_{0}^{2023} \dfrac{n^2-na}{5} \, dn.\] We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated(if you know calculus) to be \[\dfrac{1}{15}2023^3-\dfrac{a}{10}2023^2=2023^2\left(\dfrac{2023}{15}-\dfrac{a}{10}\right).\] Because we want this to be as close to $0$ as possible, we find that $a$ should equal $1349.$ Then, evaluating the sum becomes trivial. Set \[U'=\sum_{n=1}^{2023}\dfrac{n^2-1349n}{5}\] and \[U''=\sum_{n=1}^{2023}\{\dfrac{n^2-1349n}{5}\}.\] Then $U=U'-U''.$ We can evaluate $U'$ to be $0$ and $U''$ to be $-405$ (using some basic number theory). Thus, $U=-405$ and the answer is \[1349+(-405)=\boxed{944}.\] BS2012

Answer:  944
Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that \[f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.\]
We see that $f(1) = 2$ $f(2) = 3$ , and $f(n) = f(n-1) + f(n-2)$ . This is because if the element $n$ is included in our subset, then there are $f(n-2)$ possibilities for the rest of the elements (because $n-1$ cannot be used), and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(n)$ is the $n+1$ th Fibonacci number.
This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = \boxed{235}$

Answer:  235
Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: Denote by $N_1 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset contains exactly one pair of consecutive integers.
Denote by $N_0 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset does not contain any consecutive integers.
Denote by $a$ the smallest number in set $S$
First, we compute $N_1 \left( m \right)$
Consider $m \geq 3$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 1 \right)$
Case 2: A subset contains $a$ but does not contain $a + 1$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 2 \right)$
Case 3: A subset contains $a$ and $a + 1$
To have exactly one pair of consecutive integers, this subset cannot have $a + 2$ , and cannot have consecutive integers in $\left\{ a+3, a+4, \cdots , a + m - 1 \right\}$
Thus, the number of subsets that has exactly one pair of consecutive integers is $N_0 \left( m - 3 \right)$
Therefore, for $m \geq 3$ \[N_1 \left( m \right) = N_1 \left( m - 1 \right) + N_1 \left( m - 2 \right) + N_0 \left( m - 3 \right) .\]
For $m = 1$ , we have $N_1 \left( 1 \right) = 0$ .
For $m = 2$ , we have $N_1 \left( 2 \right) = 1$
Second, we compute $N_0 \left( m \right)$
Consider $m \geq 2$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has no consecutive integers is $N_0 \left( m - 1 \right)$
Case 2: A subset contains $a$
To avoid having consecutive integers, the subset cannot have $a + 1$
Thus, the number of subsets that has no consecutive integers is $N_0 \left( m - 2 \right)$
Therefore, for $m \geq 2$ \[N_0 \left( m \right) = N_0 \left( m - 1 \right) + N_0 \left( m - 2 \right)  .\]
For $m = 0$ , we have $N_0 \left( 0 \right) = 1$ .
For $m = 1$ , we have $N_0 \left( 1 \right) = 2$
By solving the recursive equations above, we get $N_1 \left( 10 \right) = \boxed{235}$

Answer:  235
Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: Let $a_n$ be the number of subsets of the set $\{1,2,3,\ldots,n\}$ such that there exists exactly 1 pair of consecutive elements. 
Let $b_n$ be the number of subsets of the set $\{1, 2, 3\ldots, n\}$ such that there doesn't exist any pair of consecutive elements. 
First, lets see how we can construct $a_n.$ For each subset $S$ counted in $a_n,$ either:
1. $\{n-1, n\}\subseteq S,$ 2. $n\not\in S$ , or
3. $n-1 \not\in S$ and $n\in S.$ The first case counts $b_{n-3}$ subsets (as $n-1$ cannot be included and the rest cannot have any consecutive elements), The second counts $a_{n-1},$ and the third counts $a_{n-2}.$ Thus, \[a_n = a_{n-1} + a_{n-2} + b_{n-3}.\] Next, Lets try to construct $b_n.$ For each subset $T$ counted in $b_n,$ either:
1. $n \not\in T,$ or 
2. $n \in T.$ The first case counts $b_{n-1}$ subsets and the second counts $b_{n-2}.$ Thus, \[b_n = b_{n-1} + b_{n-2}.\] Since $b_1 = 2$ and $b_2 = 3,$ we have that $b_n = F_{n+1},$ so $a_n = a_{n-1} + a_{n-2} + F_{n-2}.$ (The $F_i$ is the $i$ th Fibonacci number). From here, we can construct a table of the values of $a_n$ until $n = 10.$ By listing out possibilities, we can solve for our first 3 values.
\[\begin{array}{r|l} n & a_n \\ \hline 1 & 0 \\ 2 & 1\\ 3 & 2\\ 4 & 2 + 1 + F_2 = 5\\5 & 5 + 2 + F_3 = 10\\6 & 10 + 5 + F_4 = 20 \\ 7 & 20 + 10 + F_5 = 38\\ 8 & 38 + 20 + F_6 = 71 \\ 9& 71 + 38 + F_7 = 130\\ 10 & 130 + 71 + F_8 = 235 \end{array}\] Our answer is $a_{10} = \boxed{235}.$ ~AtharvNaphade

Answer:  235
Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: Note: This is a very common stars and bars application.
Casework on number of terms, let the number of terms be $n$ . We can come up with a generalized formula for the number  of subsets with n terms.
Let $d_1, d_2, ..., d_{n-1}$ be the differences between the n terms. For example, in the set {2, 3, 6}, $d_1 = 1; d_2 = 3; d_3 = 5$
Let the range of the set be k for now, $d_1 + ... + d_{n-1} = k$ . We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let $d_1 = 1$ $1 + d_2 + ... + d_{n-1} = k$
To ensure the other $d_2, d_3, ..., d_{n-1}$ are greater than 1 such that no two other terms are consecutive or the same, let $D_2 = d_2 + 1; D_3 = d_3 + 1; ...$ $(n-2) + 1 + D_2 + ... + D_{n-1} = k$ where $D_2, D_3, ..., D_{n-1}$ are positive integers.
Finally, we add in $D_0$ , the distance between 0 and the first term of the set, and $D_n$ , the distance between the last term and 11. This way, the "distance" from 0 to 11 is "bridged" by $D_0$ , k, and $D_N$ \[D_0 + k + D_n = D_0 + ((n-1) + D_2 + ... + D_{n-1}) + D_N = 11\] \[D_0 + D_2 + D_3 + ... + D_n = 12 - n\] There are n positive terms , by Balls and Urns , there are ${(12-n)-1 \choose (n)-1} = {11-n \choose n-1}$ ways of doing this. However, recall that there were $(n-1)$ ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence $(n-1){11-n \choose n-1}$ for $n > 2$
Case 1: Two terms $n = 2$ , so $(2-1){11-2 \choose 2-1} = 1{9 \choose 1} = 9$
Case 2: Three terms $n = 3$ , so $(3-1){11-3 \choose 3-1} = 2{8 \choose 2} = 56$
Case 3: Four terms $n = 4$ , so $(4-1){11-4 \choose 4-1} = 3{7 \choose 3} = 105$
Case 4: Five terms $n = 5$ , so $(5-1){11-5 \choose 5-1} = 4{6 \choose 4} = 60$
Case 5: Six terms $n = 6$ , so $(6-1){11-6 \choose 6-1} = 5{5 \choose 5} = 5$
We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is $9+56+105+60+5=\boxed{235}$

Answer:  235
Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: Note that there are $F_{n+2}$ subsets of a set of $n$ consecutive integers that contains no two consecutive integers. (This can be proven by induction.)
Now, notice that if we take $i$ and $i+1$ as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, $i-1$ and $i+2$ cannot be chosen, and since $i-2$ and $i+3$ are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set $\{1, 2, ... , i-2 \}$ is consecutive with any element of the set $\{ i+3, i+4, ... , 10 \}$
Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has $i-2$ consecutive integer elements and the second set has $8-i$ consecutive integer elements, we know that this is $F_i F_{10-i}.$
Summing this over for all $1 \leq i \leq 9$ yields \[\sum_{i=1}^9 F_i F_{10-i} = F_1F_9 + F_2F_8 + F_3F_7 + F_4F_6 + F_5 F_5 + F_6 F_4 + F_7 F_3 + F_8 F_2 + F_9 F_1 = 2(F_1F_9 + F_2F_8 + F_3F_7+F_4F_6) + F_5^2 = 2(1 \cdot 34 + 1 \cdot 21 + 2 \cdot 13 + 3 \cdot 8) + 5^2 = 2 \cdot 105+25 = \boxed{235}.\]

Answer:  235
Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: The problem is the same as laying out a line of  polynomoes to cover spots $0,1,...10$ : 1 triomino ( $RGG$ ), $n$ dominoes ( $RG$ ), and $8-2n$ monominoes ( $R$ ).  The $G$ spots cover the members of the subset.
The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10.
There are 5 ways to choose polyomino sets, and many ways to order each set:
$R + RG + RGG =$ Polyominoes $\rightarrow$ Orderings
$0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = ~~~5$
$2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = ~60$
$4 + 2 + 1 = 7 \rightarrow 7! / 4!2!1! = 105$
$6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = ~56$
$8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = ~~~9$
The sum is $\boxed{235}$

Answer:  235
Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: Let $dp(i)$ be the number of subsets of a set $i$ consecutive integers such that the maximum value in the set is $i$ and there exists exactly one pair of consecutive integers. Define $dp2(i)$ similarly, but without any pair of consecutive integers. The base cases are $dp2(1)=dp2(2)=1$ $dp(1)=0$ , and $dp(2)=1$
The transitions are:
\[dp2(i)=\sum_{j=1}^{i-2}(dp(j))+1\]
\[dp(i)=dp2(i-1)+\sum_{j=1}^{i-2}dp(j)\]
Note that $dp2$ is the Fibonacci numbers.
\[\begin{array}{rcl} i & dp & dp2\\ \hline 1 & 0 & 1 \\ 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 3 & 3 \\ 5 & 5 & 5 \\ 6 & 10 & 8 \\ 7 & 18 & 13 \\ 8 & 33 & 21 \\ 9 & 59 & 34 \\ 10 & 105 & 55 \\ \end{array}\]
Summing over $dp$ yields $1+1+3+5+10+18+33+59+105=\boxed{235}$

Answer:  235