Navvye/Olympiad-Problems-CSV huggingface dataset:
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: Let $s$ be the numbers of elements in a subset. First we examine the second condition. No elements less than $s$ can be put in a subset of size $s$ , therefore the "lowest" element that can go into the subset is $s$ , whereas the "highest" element that can go into the subset is $15$ . This is a total of $15-s+1$ or $16-s$ possible elements.
Now we consider the first condition. No consecutive elements are allowed. This means that if an element $e$ is put into the subset, both $e-1$ and $e+1$ are no longer possibilities. Assume that all subsets are ordered from least to greatest (we are looking for the number of combinations, so we can order these combinations however we want). Then the first element will be $s$ (as a reminder, the lowest possible element in a subset is $s$ ), the second element will be at least $s+2$ , and so on. After $s$ elements are chosen, we will have skipped $s-1$ elements (these are the elements that were "eliminated" as they were consecutive). Therefore, we ignore exactly $s-1$ elements (if we ignore more or less elements, then $s$ changes) Since we must ignore $s-1$ elements, we can simply remove those beforehand. ( $16-s)-(s-1) = 17-2s$ possible elements
Now we look for the bounds of $s$ . We are looking for non-empty subsets, so $s\ge1$ . If $s$ is too large, there will not be enough non-consecutive terms between $s$ and $15$ . Specifically, the highest element in a subset using "optimal" selection will be $s+2(s-1)$ or $3s-2$ . If $3s-1>15$ , that means s is too large. Therefore $3s-2\le15$ ; solving for $s$ yields $s\le5$ . Now we know that $1 \le s \le 5$
We want to know the number of ways to arrange $17-2s$ "balls" into $s$ identical "boxes" with at most $1$ ball per box, for $s=1$ to $s=5$ . This is equivalent to $\sum_{s=1}^{5}{17-2s \choose s} = 405$ , or $\boxed{405}$

Answer: E 405
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: We will split the problem into cases, and maybe one could then generalize this to arbitrary $n$
$\bold{Case 1}:$ $n=15, k=1$ ). Then this is easy. We have $\binom{15}{1}=15$
$\bold{Case 2}:$ $n=14, k=2$ ). Now we have something tricky. To get a good grasp on this case, let us consider the smallest element; $2$ , in the first spot. We then have $15-1=14$ numbers left. However, we cannot have the digit $3$ . Hence we have to choose $2$ numbers from $14-1=13$ integers left. This can be done in $\binom{13}{2}=78$ ways.
$\bold{Case 3}:$ $n=13, k=3$ ). As in the last case, we can use the smallest element to get a good grasp on this case. Put the digit $3$ in the first spot. Then there are $11$ integers left for the second spot. But in total, we have to avoid $2$ elements for each subset of cardinality $3$ ${a,b,c}$ . So we have to choose $3$ elements from $13-2=11$ elements, which can be done in $\binom{11}{3}=165$ ways.
$\bold{Case 4}:$ $n=12, k=4$ ). As in the previous case, we have to avoid $3$ numbers, and choose $4$ from them. Which is choosing $4$ elements from $12-3=9$ elements, which can be done in $\binom{9}{4}=126$ ways.
$\bold{Case 5}:$ $n=11, k=4$ ). Now you are accustomed to the strategy. We remove $4$ integers leaving $\binom{7}{5}=21$ ways.
Adding up all of the cases yields $405$ , as in $\boxed{405}$

Answer: E 405
Problem: A football game was played between two teams, the Cougars and the Panthers.  The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points.  How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$

Solution: If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 $\Rightarrow \boxed{10}$

Answer: A 10
Problem: A football game was played between two teams, the Cougars and the Panthers.  The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points.  How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$

Solution: Let the Panthers' score be $x$ . The Cougars then scored $x+14$ . Since the teams combined scored $34$ , we get $x+x+14=34 \\ \rightarrow 2x+14=34 \\ \rightarrow 2x=20 \\ \rightarrow x = 10$
and the answer is $\boxed{10}$

Answer: A 10
Problem: Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade.  There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar.  Water contains no calories.  How many calories are in 200 grams of her lemonade?
$\text {(A) } 129 \qquad \text {(B) } 137 \qquad \text {(C) } 174 \qquad \text {(D) } 223 \qquad \text {(E) } 411$

Solution: Francesca makes a total of $100+100+400=600$ grams of lemonade, and in those $600$ grams, there are $25$ calories from the lemon juice and $386$ calories from the sugar, for a total of $25+386=411$ calories per $600$ grams. We want to know how many calories there are in $200=600/3$ grams, so we just divide $411$ by $3$ to get $137\implies\boxed{137}$

Answer: B 137
Problem: Mr. and Mrs. Lopez have two children.  When they get into their family car, two people sit in the front, and the other two sit in the back.  Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat.  How many seating arrangements are possible?
$\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \text {(D) } 24 \qquad \text {(E) } 48$

Solution: There are only two possible occupants for the driver's seat. After the driver is chosen, any of the remaining three people can sit in the front, and there are two arrangements for the other two people in the back. Thus, there are $2\cdot 3\cdot 2 = \boxed{12}$ possible seating arrangements. ~ aopsav (Credit to AoPS Alcumus)

Answer:  12
Problem: How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$

Solution: Let the integer have digits $a$ $b$ , and $c$ , read left to right. Because $1 \leq a<b<c$ , none of the digits can be zero and $c$ cannot be 2. If $c=4$ , then $a$ and $b$ must each be chosen from the digits 1, 2, and 3. Therefore there are $\binom{3}{2}=3$ choices for $a$ and $b$ , and for each choice there is one acceptable order. Similarly, for $c=6$ and $c=8$ there are, respectively, $\binom{5}{2}=10$ and $\binom{7}{2}=21$ choices for $a$ and $b$ . Thus there are altogether $3+10+21=\boxed{34}$ such integers.

Answer:  34
Problem: How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$

Solution: Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.
If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit.
If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,
and so on.
So, the answer is $3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B$

Answer:  34
Problem: How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$

Solution: The last digit is 4, 6, or 8.
If the last digit is $x$ , the possibilities for the first two digits correspond to 2-element subsets of $\{1,2,\dots,x-1\}$
Thus the answer is ${3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = \boxed{34}$

Answer:  34
Problem: How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$

Solution: The answer must be half of a triangular number (evens and decreasing/increasing) so $\boxed{34}$ or the letter B.
-

Answer:  34
Problem: How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$

Solution: Casework:
For the sake of simplicity, we are going to call the number $\overline{abc}$
1. If $a=1$
a. $c=2$ . No such number exists.
b. $c=4$ $b=2, 3$
c. $c=6$ $b=2, 3, 4, 5$
d. $c=8$ $b=2, 3, 4, 5, 6, 7$
2. If $a=2$ : continue as above.
We can count up that there are 34 such integers, so select $\boxed{34}$

Answer: B 34
Problem: In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
$\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$

Solution: If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: \[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5\] Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Then, the first side has length $3*7=21$ , the second side has length $7$ , the third side has length $15$ , and so the perimeter is $21+7+15=43 \Rightarrow \boxed{43}$

Answer: A 43
Problem: The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$ -intercept $(0,-p)$ , where $p\ne 0$ . What is $b$
$\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p$

Solution: A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$ . Because the $y$ -intercept is $(0,-p)$ and $p\ne 0$ , it follows that $a=-2/p$ . Thus \[y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,\] so $\boxed{4}$

Answer:  4
Problem: An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathrm{(B)}\ 121 \qquad \mathrm{(C)}\ 221 \qquad \mathrm{(D)}\ 230 \qquad \mathrm{(E)}\ 231$

Solution: Let the starting point be $(0,0)$ . After $10$ steps we can only be in locations $(x,y)$ where $|x|+|y|\leq 10$ . Additionally, each step changes the parity of exactly one coordinate. Hence after $10$ steps we can only be in locations $(x,y)$ where $x+y$ is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable.
Once we pick $x\in\{-10,\dots,10\}$ , we have $11-|x|$ valid choices for $y$ , giving a total of $\boxed{121}$ possible positions.

Answer:  121
Problem: An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathrm{(B)}\ 121 \qquad \mathrm{(C)}\ 221 \qquad \mathrm{(D)}\ 230 \qquad \mathrm{(E)}\ 231$

Solution: $10$ moves results in a lot of possible endpoints, so we try small cases first.
If the object only makes $1$ move, it is obvious that there are only 4 possible points that the object can move to.
If the object makes $2$ moves, it can move to $(0, 2)$ $(1, 1)$ $(2, 0)$ $(1, -1)$ $(0, -2)$ $(-1, -1)$ $(-2, 0)$ as well as $(0, 0)$ , for a total of $9$ moves.
If the object makes $3$ moves, it can end up at $(0, 3)$ $(2, 1)$ $(1, 2)$ $(3, 0)$ $(2, -1)$ $(1, -2)$ $(0, -3)$ $(-1, -2)$ $(-2, -1)$ $(0, -3)$ etc. for a total of $16$ moves.
At this point we can guess that for n moves, there are $(n + 1)^2$ different endpoints. Thus, for 10 moves, there are $11^2 = 121$ endpoints, and the answer is $\boxed{121}$

Answer: B 121
Problem: Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 \qquad \mathrm{(D)}\ \frac 15  \qquad \mathrm{(E)}\ \frac 14$

Solution: Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$
The equation $\lfloor\log_{10}x\rfloor = k$ can be rewritten as $10^k \leq x < 10^{k+1}$ . The second one gives us $10^k \leq 4x < 10^{k+1}$ . Combining these, we get that both hold at the same time if and only if $10^k \leq x < \frac{10^{k+1}}4$
Hence for each integer $k$ we get an interval of values for which $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ . These intervals are obviously pairwise disjoint.
For any $k\geq 0$ the corresponding interval is disjoint with $(0,1)$ , so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$ . Hence our answer is the sum of the lengths of the intervals for $k<0$
For a fixed $k$ the length of the interval $\left[ 10^k, \frac{10^{k+1}}4 \right)$ is $\frac 32\cdot 10^k$
This means that our result is $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{16}$

Answer:  16
Problem: Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 \qquad \mathrm{(D)}\ \frac 15  \qquad \mathrm{(E)}\ \frac 14$

Solution: The largest value for $x$ is $10^{0}$ . If $x > 10^{-1}$ , then $\lfloor\log_{10}4x\rfloor$ doesn't fulfill the condition unless $10^{-2} \leq x < 0.25 * 10^{-1}$ . The same holds when you get smaller, because $x = 0.25 * 10^{n}$ for $n \leq 0$ is the lowest value such that $4x$ becomes a higher power of 10.
Recognize that this is a geometric sequence. The probability of choosing $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-1$ is $(9/10)* (15/90) =15/100$ , because there is a 90 percent chance of choosing $x > 10^{-1}$ , and only values of $x$ between $10^{-1}$ and $0.25*10^{0}$ work in this case. Then, for $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-2$ , you have $(1/10) * ((9/10) *(15/90))$ . This is a geometric series with ratio $1/10$ . Using $a/(1-r)$ for the sum of an infinite geometric sequence, we get $(15/100)/(1-(1/10)) =  \boxed{16}$

Answer:  16
Problem: Suppose $a$ $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492  \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 501$

Solution: The power of $10$ for any factorial is given by the well-known algorithm \[\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots\] It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher powers of $5$ . As we list out the powers of 5, it is clear that $5^{4}=625$ is less than 2006 and $5^{5}=3125$ is greater. Therefore, set $a$ and $b$ to be 624. Thus, c is $2006-(624\cdot 2)=758$ . Applying the algorithm, we see that our answer is $152+152+188= \boxed{492}$

Answer:  492
Problem: Isosceles $\triangle ABC$ has a right angle at $C$ .  Point $P$ is inside $\triangle ABC$ , such that $PA=11$ $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers.  What is $a+b$
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]
$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$

Solution: [asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); MP("\alpha",C,5*dir(80),f); MP("90^\circ-\alpha",C,3*dir(30),f); MP("s",(A+C)/2,plain.S,f); MP("s",(B+C)/2,plain.W,f); [/asy] Using the Law of Cosines on $\triangle PBC$ , we have:
\begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*}
Using the Law of Cosines on $\triangle PAC$ , we have: \begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*}
Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$ \begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2  \end{align*}
Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$ .  Thus, $a+b=85+42=\boxed{127}$

Answer:  127
Problem: Isosceles $\triangle ABC$ has a right angle at $C$ .  Point $P$ is inside $\triangle ABC$ , such that $PA=11$ $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers.  What is $a+b$
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]
$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$

Solution: Let point $P$ have coordinates $(x,y)$ and $C$ have coordinates $(0,0).$ Then, $A$ has $(s,0)$ and $B$ has $(0,s)$
By distance formula, we have \[x^2+y^2=36 \tag{1}.\] \[x^2+(y-s)^2=49 \tag{2}.\] \[(x-s)^2+y^2=121 \tag{3}\]
Expanding $(2)$ and $(3)$ gives \[x^2+y^2-2ys+s^2=49,\] and \[x^2+y^2-2sx+s^2=121,\] respectively. Then, using equation $(1)$ , we have that \[-2ys+s^2=49-36=13,\] and \[-2sx+s^2=121-36=85.\]
Then, solving for $x$ and $y$ gives \[x=\frac{85-s^2}{-2s}=\frac{s^2-85}{2s},\] and \[y=\frac{13-s^2}{-2s}=\frac{s^2-13}{2s}.\] Plugging these values of $x$ and $y$ into equation $(1)$ yields \[\left(\frac{s^2-85}{2s}\right)^2+\left(\frac{s^2-13}{2s}\right)^2=36.\]
We multiply both sides by $4s^2$ and expand, yielding the equation \[s^4-170s^2+7225+s^4-26s^2+169=144s^2.\] Simplifying gives the equation \[2s^4-340s^2+7394=0.\] Solving this quadratic gives $s^2=85\pm 42\sqrt{2}.$ Now, if this were the actual test, we stop here, noting that the question tells us $a$ and $b$ are positive, so $s^2$ must be $85+42\sqrt{2}$ , and our solution is $127$
However, here is why $s^2$ cannot be $85-42\sqrt{2}$
If $s^2=85-42\sqrt{2}$ , using $1.4$ as an approximation for $\sqrt{2}$ $s^2\approx 85-42\cdot 1.4=85-58.8=26.2$ , and $s$ is slightly greater than $5$ . Also note that this implies that $AB\approx 7$
Note that at least one of $\angle BPA$ $\angle APC$ and $\angle BPC$ must be obtuse, since they sum to $360^{\circ}$ . Then, note the well known fact that if $\angle A$ is the largest angle in $\Delta ABC$ $BC$ must be the largest side. However, combined with the first fact, implies that either $AB$ is the largest side of $\Delta ABP$ $BC$ is the largest side of $\Delta BPC$ , or $AC$ is the largest side of $\Delta APC$ . By our approximations, this cannot possibly be true, even if we are generous with our margin of error, so $s^2$ cannot equal $85-42\sqrt{2}$ , and $s^2=85+42\sqrt{2}$
The answer is $85+42=\boxed{127}$

Answer:  127
Problem: line passes through $A\ (1,1)$ and $B\ (100,1000)$ . How many other points with integer coordinates are on the line and strictly between $A$ and $B$
$(\mathrm {A}) \ 0 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 9$

Solution: The slope of the line is $\frac{1000-1}{100-1}=\frac{111}{11},$ so all points on the line have the form $(1+11t, 1+111t)$ for some value of $t$ (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if $t$ is an integer, and the point is strictly between $A$ and $B$ if and only if $0<t<9$ . Thus, there are $\boxed{8}$ points with the required property.
-Paixiao

Answer:  8
Problem: In the five-sided star shown, the letters $A$ $B$ $C$ $D$ and $E$ are replaced by the
numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
numbers at the ends of the line segments $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ form an
arithmetic sequence, although not necessarily in that order. What is the middle
term of the arithmetic sequence?
[asy] draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); label("$A$",(0.5,1.54),N); label("$B$",(1,0),SE); label("$C$",(-0.31,0.95),W); label("$D$",(1.31,0.95),E); label("$E$",(0,0),SW); [/asy]
$(\mathrm {A}) \ 9 \qquad (\mathrm {B}) \ 10 \qquad (\mathrm {C})\ 11 \qquad (\mathrm {D}) \ 12 \qquad (\mathrm {E})\ 13$

Solution: Let the terms in the arithmetic sequence be $a$ $a + d$ $a + 2d$ $a + 3d$ , and $a + 4d$ . We seek the middle term $a + 2d$
These five terms are $A + B$ $B + C$ $C + D$ $D + E$ , and $E + A$ , in some order. The numbers $A$ $B$ $C$ $D$ , and $E$ are equal to 3, 5, 6, 7, and 9, in some order, so \[A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.\] Hence, the sum of the five terms is \[(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.\] But adding all five numbers, we also get $a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d$ , so \[5a + 10d = 60.\] Dividing both sides by 5, we get $a + 2d = \boxed{12}$ , which is the middle term. The answer is (D).

Answer:  12
Problem: A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005 , how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$

Solution: Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$ ; only, in base $9$ , the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$

Answer:  1462