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

Solution: For a list ${a_1, a_2, \dots, a_k}$ with $k$ terms, $2$ valid lists with $k+1$ terms can be created by $2$ ways:
1. Add $a_{k+1}$ to the end of the list, making a new list ${a_1, a_2, \dots, a_k, a_{k+1} }$
2. Increase the value of all existing terms by one, making a new list ${a_2, a_3, \dots, a_{k+1}}$ . Then add $a_1$ to the end of the list, making a new list ${a_2, a_3, \dots, a_{k+1}, a_1}$
Let $F(n)$ be the number of lists with $n$ elements, $F(n) = 2 \cdot F(n-1)$ . As $F(2) = 2$ $F(n) = 2^{n-1}$ $F(10) = 2^9 = \boxed{512}$

Answer: B 512
Problem: A trapezoid has side lengths 3, 5, 7, and 11. The sum of all the possible areas of the trapezoid can be written in the form of $r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3$ , where $r_1$ $r_2$ , and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to $r_1+r_2+r_3+n_1+n_2$
$\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65$

Solution: Name the trapezoid $ABCD$ , where $AB$ is parallel to $CD$ $AB<CD$ , and $AD<BC$ . Draw a line through $B$ parallel to $AD$ , crossing the side $CD$ at $E$ . Then $BE=AD$ $EC=DC-DE=DC-AB$ . One needs to guarantee that $BE+EC>BC$ , so there are only three possible trapezoids:
\[AB=3, BC=7, CD=11, DA=5, CE=8\] \[AB=5, BC=7, CD=11, DA=3, CE=6\] \[AB=7, BC=5, CD=11, DA=3, CE=4\]
In the first case, by Law of Cosines, $\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14$ , so $\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14$ . Therefore the area of this trapezoid is $\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}$
In the second case, $\cos(\angle BCD) = (6^2+7^2-3^2)/(2\cdot 6\cdot 7) = 19/21$ , so $\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21$ . Therefore the area of this trapezoid is $\frac{1}{2} (5+11) \cdot 7 \cdot 4\sqrt{5}/21 =\frac{32}{3}\sqrt{5}$
In the third case, $\angle BCD = 90^{\circ}$ , therefore the area of this trapezoid is $\frac{1}{2} (7+11) \cdot 3 = 27$
So $r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5$ , which rounds down to $\boxed{63}$

Answer: D 63
Problem: A trapezoid has side lengths 3, 5, 7, and 11. The sum of all the possible areas of the trapezoid can be written in the form of $r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3$ , where $r_1$ $r_2$ , and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to $r_1+r_2+r_3+n_1+n_2$
$\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65$

Solution: Let the area of the trapezoid be $S$ , the area of the triangle be $S_1$ , the area of the parallelogram be $S_2$
By Heron's Formula $S_1 = \sqrt{\frac{b+c+d-a}{2} \cdot \frac{c+d-a-b}{2} \cdot \frac{a+b+d-c}{2} \cdot \frac{b+c-a-d}{2}}$
$S_2 = \frac{S_1 \cdot 2}{c-a} \cdot a = \frac{2aS_1}{c-a}$
$S = S_1 + S_2 = S_1(1+\frac{2a}{c-a}) = S_1 \cdot \frac{c+a}{c-a} = \frac14 \cdot \frac{c+a}{c-a} \cdot \sqrt{(b+c+d-a)(c+d-a-b)(a+b+d-c)(b+c-a-d)}$
If $a = 3$ $b = 7$ $c = 11$ $d = 5$
$S = \frac14 \cdot \frac{14}{8} \cdot \sqrt{(7+11+5-3)(11+5-3-7)(3+7+5-11)(7+11-3-5)} = \frac{35\sqrt{3}}{2}$
If $a = 3$ $b = 11$ $c = 5$ $d = 7$
$S = \frac14 \cdot \frac{8}{2} \cdot \sqrt{(11+5+7-3)(5+7-3-11)(3+5+7-11)(11+5-3-7)}$ , which is impossible as $5+7-3-11 = -2$
If $a = 3$ $b = 5$ $c = 7$ $d = 11$
$S = \frac14 \cdot \frac{10}{4} \cdot \sqrt{(5+7+11-3)(7+11-3-5)(3+5+11-7)(5+7-11-3)}$ , which is impossible as $5+7-11-3 = -2$
If $a = 5$ $b = 11$ $c = 7$ $d = 3$
$S = \frac14 \cdot \frac{12}{2} \cdot \sqrt{(11+7+3-5)(7+3-5-11)(5+11+3-7)(11+7-5-3)}$ , which is impossible as $7+3-5-11 = -6$
If $a = 5$ $b = 3$ $c = 11$ $d = 7$
$S = \frac14 \cdot \frac{16}{6} \cdot \sqrt{(3+11+7-5)(11+7-5-3)(5+3+7-11)(3+11-5-7)} = \frac{32\sqrt{5}}{3}$
If $a = 7$ $b = 3$ $c = 11$ $d = 5$
$S = \frac14 \cdot \frac{18}{4} \cdot \sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27$
Thus the answer is $\frac{35}{2} + \frac{32}{3} + 27 + 3 + 5$ , which rounds down to $\boxed{63}$

Answer: D 63
Problem: Last summer $30\%$ of the birds living on Town Lake were geese, $25\%$ were swans, $10\%$ were herons, and $35\%$ were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 60$

Solution: To simplify the problem, WLOG, let us say that there were a total of $100$ birds. The number of birds that are not swans is $75$ . The number of geese is $30$ . Therefore the percentage is just $\frac{30}{75} \times 100 = 40 \Rightarrow \boxed{40}$

Answer: C 40
Problem: At a twins and triplets convention, there were $9$ sets of twins and $6$ sets of triplets, all from different families. Each twin shook hands with all the twins except his/her siblings and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?
$\textbf{(A)}\ 324 \qquad \textbf{(B)}\ 441 \qquad \textbf{(C)}\ 630 \qquad \textbf{(D)}\ 648 \qquad \textbf{(E)}\ 882$

Solution: There are $18$ total twins and $18$ total triplets. Each of the twins shakes hands with the $16$ twins not in their family and $9$ of the triplets, a total of $25$ people. Each of the triplets shakes hands with the $15$ triplets not in their family and $9$ of the twins, for a total of $24$ people. Dividing by two to accommodate the fact that each handshake was counted twice, we get a total of $\frac{1}{2}\times 18 \times (25+24) = 9 \times 49 = 441 \rightarrow \boxed{441}$

Answer: B 441
Problem: A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock $A.$ How many hours did it take the power boat to go from $A$ to $B$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 3.5 \qquad \textbf{(C)}\  4 \qquad \textbf{(D)}\ 4.5 \qquad \textbf{(E)}\ 5$

Solution: WLOG let the speed of the river be 0. This is allowed because the problem never states that the speed of the current has to have a magnitude greater than 0. In this case, when the powerboat travels from $A$ to $B$ , the raft remains at $A$ . Thus the trip from $A$ to $B$ takes the same time as the trip from $B$ to the raft. Since these times are equal and sum to $9$ hours, the trip from $A$ to $B$ must take half this time, or $4.5$ hours. The answer is thus $\boxed{4.5}$

Answer: D 4.5
Problem: A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock $A.$ How many hours did it take the power boat to go from $A$ to $B$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 3.5 \qquad \textbf{(C)}\  4 \qquad \textbf{(D)}\ 4.5 \qquad \textbf{(E)}\ 5$

Solution: What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Think of the blue arrow as the power boat and the red arrow as the raft in the following three diagrams, which represent different time intervals of the problem.
[asy] size(8cm,8cm); pair A, B; A=(-3,4); B=(3,4); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((-2.5,6.4),dir(180),blue); arrow((-2.5,6.2),dir(180),red); pair A, B; A=(-3,5); B=(3,5); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((2.6,5.4),dir(360),blue); arrow((-0.5,5.2),dir(180),red); pair A, B; A=(-3,6); B=(3,6); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((.3,4.4),dir(360),blue); arrow((1.1,4.2),dir(180),red);[/asy]
Thinking about the distance covered as their distances with respect to each other, they are $0$ distance apart in the first diagram when they haven't started to move yet, some distance $d$ apart in the second diagram when the power boat reaches $B$ , and again $0$ distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of $d$ on the way there, and again cover a distance of $d$ on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.
Let $b$ denote the speed of the power boat (only the power boat, not factoring in current) and $r$ denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from $A$ to $B$ , the boat travels at a velocity of $b+r$ , and on the way back, travels at a velocity of $-(b-r)=r-b$ , since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from $A$ to $B$ becomes $(r+b)-r=b$ , and on the way back it becomes $(r-b)-r=-b$ . Since the boat's velocities with respect to the raft are exact opposites, $b$ and $-b$ , we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.
From this, we have that the boat travels a distance $d$ at rate $b$ with respect to the raft both on the way to $B$ and on the way back. Thus, using $\dfrac{distance}{speed}=time$ , we have $\dfrac{2d}{b}=9\text{ hours}$ , and to see how long it took to travel half the distance, we have $\dfrac{d}{b}=4.5\text{ hours}\implies\boxed{4.5}$

Answer: D 4.5
Problem: A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock $A.$ How many hours did it take the power boat to go from $A$ to $B$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 3.5 \qquad \textbf{(C)}\  4 \qquad \textbf{(D)}\ 4.5 \qquad \textbf{(E)}\ 5$

Solution: Let $t$ be the time it takes the power boat to go from $A$ to $B$ in hours, $r$ be the speed of the river current (and thus also the raft), and $p$ to be the speed of the power boat with respect to the river.
Using $d = rt$ , the raft covers a distance of $9r$ , the distance from $A$ to $B$ is $(p + r)t$ , and the distance from $B$ to where the raft and power boat met up is $(9 - t)(p - r)$
Then, $9r + (9 - t)(p - r) = (p + r)t$ . Solving for $t$ , we get $t = 4.5$ , which is $\boxed{4.5}$

Answer: D 4.5
Problem: Triangle $ABC$ has side-lengths $AB = 12, BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\  33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42$

Solution: Let $O$ be the incenter of $\triangle{ABC}$ . Because $\overline{MO} \parallel \overline{BC}$ and $\overline{BO}$ is the angle bisector of $\angle{ABC}$ , we have
\[\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}\]
It then follows due to alternate interior angles and base angles of isosceles triangles that $MO = MB$ . Similarly, $NO = NC$ . The perimeter of $\triangle{AMN}$ then becomes \begin{align*} AM + MN + NA &= AM + MO + NO + NA \\ &= AM + MB + NC + NA \\ &= AB + AC \\ &= 30 \rightarrow \boxed{30}

Answer: B 30
Problem: Triangle $ABC$ has side-lengths $AB = 12, BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\  33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42$

Solution: Let $O$ be the incenter. $AO$ is the angle bisector of $\angle MAN$ . Let the angle bisector of $\angle BAC$ meets $BC$ at $P$ and the angle bisector of $\angle ABC$ meets $AC$ at $Q$ . By applying both angle bisector theorem and Menelaus' theorem,
$\frac{AO}{OP} \times \frac{BP}{BC} \times \frac{CQ}{QA} = 1$
$\frac{AO}{OP} \times \frac{12}{30} \times \frac{24}{12} = 1$
$\frac{AO}{OP}=\frac{5}{4}$
$\frac{AO}{AP}=\frac{5}{9}$
Perimeter of $\triangle AMN = \frac{12+24+18}{9} \times 5 = 30  \rightarrow \boxed{30}$

Answer: B 30
Problem: Triangle $ABC$ has side-lengths $AB = 12, BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\  33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42$

Solution: Like in other solutions, let $O$ be the incenter of $\triangle ABC$ . Let $AO$ intersect $BC$ at $D$ . By the angle bisector theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{18} = \frac{2}{3}$ . Since $BD+DC = 24$ , we have $\frac{BD}{24-BD} = \frac{2}{3}$ , so $3BD = 48 - 2BD$ , so $BD = \frac{48}{5} = 9.6$ . By the angle bisector theorem on $\triangle ABD$ , we have $\frac{DO}{OA} = \frac{BD}{BA} = \frac{4}{5} = 0.8$ , so $\frac{DA}{OA} = 1 + \frac{DO}{OA} = \frac{9}{5} = 1.8$ , so $\frac{AO}{AD} = \frac{5}{9}$ . Because $\triangle AMN \sim \triangle ABC$ , the perimeter of $\triangle AMN$ must be $\frac{5}{9} (12 + 18 + 24) = 30$ , so our answer is $\boxed{30}$

Answer: B 30
Problem: Triangle $ABC$ has side-lengths $AB = 12, BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\  33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42$

Solution: We know that the ratio of the perimeter of $\triangle AMN$ and $\triangle ABC$ is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from $A$ to $BC$ is $\frac{9\sqrt{15}}{4}$ from Herons and then $A=\frac{bh}{2}$ and also that the height from $A$ to $MN$ is simply the height from $A$ to $BC$ minus the inradius. We know the area and the semiperimeter so $r=\frac{A}{s}$ which gives us $r=\sqrt{15}$ . Now we know that the altitude from $A$ to $MN$ is $\frac{5\sqrt{15}}{4}$ so the ratios of the heights from $A$ for $\triangle AMN$ and $\triangle ABC$ is $\frac{5}{9}$ . Thus the perimeter of $\triangle AMN$ is $\frac{5}{9} \times 54 = 30$ so our answer is $\boxed{30}$

Answer: B 30
Problem: Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 2880 \qquad \textbf{(C)}\ 3120 \qquad \textbf{(D)}\ 3250 \qquad \textbf{(E)}\ 3750$

Solution: This solution essentially explains other ways of thinking about the cases stated in Solution 2.
Case 1:
${6\choose5} \cdot 5!$
5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices.
Case 2:
${6\choose4} \cdot4\cdot5\cdot3!$
4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices.
Case 3:
${6\choose3} \cdot {3\choose2} \cdot5\cdot2$
3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices.
Solving each case and adding them up gets you 3120. $\boxed{3120}$

Answer: C 3120
Problem: Suppose that $\left|x+y\right|+\left|x-y\right|=2$ . What is the maximum possible value of $x^2-6x+y^2$
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution: Plugging in some values, we see that the  graph of the equation $|x+y|+|x-y| = 2$ is a square bounded by $x= \pm 1$ and $y = \pm 1$
Notice that $x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9$ means the square of the distance from a point $(x,y)$ to point $(3,0)$ minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point $(3,0)$ , which is $(-1, \pm 1)$ . Either one, when substituting into the function, yields $\boxed{8}$

Answer: D 8
Problem: Suppose that $\left|x+y\right|+\left|x-y\right|=2$ . What is the maximum possible value of $x^2-6x+y^2$
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution: Since the equation $|x+y|+|x-y| = 2$ is dealing with absolute values, the following could be deduced: $(x+y)+(x-y)=2$ $(x+y)-(x-y)=2$ $-(x+y)+(x-y)=2$ , and $-(x+y)-(x-y)=2$ . Simplifying would give $x=1$ $y=1$ $y=-1$ , and $x=-1$ . In $x^2-6x+y^2$ , we care most about $-6x,$ since both $x^2$ and $y^2$ are non-negative. To maximize $-6x$ , though, $x$ would have to be -1. Therefore, when $x=-1$ and $y=-1$ or $y=1$ , the equation evaluates to $\boxed{8}$

Answer: D 8
Problem: At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$ . Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textbf{(D)}\ 406 \qquad \textbf{(E)}\ 1024$

Solution: We start with $2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19$ . After rearranging, we get $\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)$
Since $\lfloor\log_{2}(N-1)\rfloor$ is a positive integer, $\frac{N+19}{2}$ must be in the form of $2^{m}$ for some positive integer $m$ . From this fact, we get $N=2^{m+1}-19$
If we now check integer values of N that satisfy this condition, starting from $N=19$ , we quickly see that the first values that work for $N$ are $45$ and $109$ , that is, $2^6-19$ and $2^7 -19$ , giving values of $5$ and $6$ for $m$ , respectively. Adding up these two values for $N$ , we get $45 + 109 = 154 \rightarrow \boxed{154}$

Answer: C 154
Problem: At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$ . Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textbf{(D)}\ 406 \qquad \textbf{(E)}\ 1024$

Solution: We examine the value that $2^{1+\lfloor\log_{2}(N-1)\rfloor}$ takes over various intervals. The $\lfloor\log_{2}(N-1)\rfloor$ means it changes on each multiple of 2, like so:
2 --> 1
3 - 4 --> 2
5 - 8 --> 3
9 - 16 --> 4
From this, we see that $2^{1+\lfloor\log_{2}(N-1)\rfloor} - N$ is the difference between the next power of 2 above $2^{\lfloor\log_{2}(N-1)\rfloor}$ and $N$ . We are looking for $N$ such that this difference is 19. The first two $N$ that satisfy this are $45 = 64-19$ and $109=128-19$ for a final answer of $45 + 109 = 154 \rightarrow \boxed{154}$

Answer: C 154
Problem: At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$ . Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textbf{(D)}\ 406 \qquad \textbf{(E)}\ 1024$

Solution: Note that each $N$ is $19$ less than a power of $2$ . So, the answer will be $38$ less than the sum of $2$ powers of $2$ . Adding $38$ to each answer, we get $76$ $128$ $192$ $444$ , and $1062$ . Obviously we can take out $76$ and $1062$ . Also, $128$ will not work because two powers of two will never sum to another power of $2$ (unless they are equal, which is a contradiction to the question). So, we have $192$ and $444$ . Note that $444 = 1 + 443 = 2 + 442 = 4 + 440 = 8 + 436 = 16 + 428 = 32 + 412$ , etc. We quickly see that $444$ will not work, leaving $192$ which corresponds to $\boxed{154}$ . We can also confirm that this works because $192 = 128 + 64 = 2^7 + 2^6$

Answer: C 154
Problem: At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$ . Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textbf{(D)}\ 406 \qquad \textbf{(E)}\ 1024$

Solution: In order to fix the exponent and get rid of the logarithm term, let $N = 2^m + k + 1$ , with $0 \leq k < 2^m$ . Doing so, we see that $\lfloor \log_2{N - 1} \rfloor = m$ , which turns our given relation into \[2^m = 20 + k,\] for which the solutions of the form $(m, k)$ $(5, 12)$ and $(6, 44)$ , follow trivially. Adding up the two values of $N$ gives us $32 + 12 + 1 + 64 + 44 + 1 = 154$ , so the answer is $\boxed{154}$

Answer: C 154
Problem: Let $f(x)=ax^2+bx+c$ , where $a$ $b$ , and $c$ are integers. Suppose that $f(1)=0$ $50<f(7)<60$ $70<f(8)<80$ $5000k<f(100)<5000(k+1)$ for some integer $k$ . What is $k$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: From $f(1) = 0$ , we know that $a+b+c = 0$
From the first inequality, we get $50 < 49a+7b+c < 60$ . Subtracting $a+b+c = 0$ from this gives us $50 < 48a+6b < 60$ , and thus $\frac{25}{3} < 8a+b < 10$ . Since $8a+b$ must be an integer, it follows that $8a+b = 9$
Similarly, from the second inequality, we get $70 < 64a+8b+c < 80$ . Again subtracting $a+b+c = 0$ from this gives us $70 < 63a+7b < 80$ , or $10 < 9a+b < \frac{80}{7}$ . It follows from this that $9a+b = 11$
We now have a system of three equations: $a+b+c = 0$ $8a+b = 9$ , and $9a+b = 11$ . Solving gives us $(a, b, c) = (2, -7, 5)$ and from this we find that $f(100) = 2(100)^2-7(100)+5 = 19305$
Since $15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)$ , we find that $k = 3 \rightarrow \boxed{3}$

Answer: C 3
Problem: Let $f(x)=ax^2+bx+c$ , where $a$ $b$ , and $c$ are integers. Suppose that $f(1)=0$ $50<f(7)<60$ $70<f(8)<80$ $5000k<f(100)<5000(k+1)$ for some integer $k$ . What is $k$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: $f(x)$ is some non-monic quadratic with a root at $x=1$ . Knowing this, we'll forget their silly $a$ $b$ , and $c$ and instead write it as $f(x)=p(x-1)(x-r)$
$f(7)=6p(7-r)$ , so $f(7)$ is a multiple of 6. They say $f(7)$ is between 50 and 60, exclusive. Notice that the only multiple of 6 in that range is 54. Thus, $f(7)=6p(7-r)=54$
$f(8)=7p(8-r)$ , so $f(8)$ is a multiple of 7. They say $f(8)$ is between 70 and 80, exclusive. Notice that the only multiple of 7 in that range is 77. Thus, $f(8)=7p(8-r)=77$
Now, we solve a system of equations in two variables.
\begin{align*} 6p(7-r)&=54 \\ 7p(8-r)&=77 \\ \\ p(7-r)&=9 \\ p(8-r)&=11 \\ \\ 7p-pr&=9 \\ 8p-pr&=11 \\ \\ (8p-pr)-(7p-pr)&=11-9 \\ \\ p&=2 \\ \\ 2(7-r)&=9 \\ \\ r&=2.5 \end{align*}
$f(100)=2(100-1)(100-2.5)=19305 \implies k=3 \implies \boxed{3}$

Answer: C 3