Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $K$ be the measure of the area bounded by the $x$ -axis, the line $x=8$ , and the curve defined by
\[f={(x,y)\quad |\quad y=x \text{  when  } 0 \le x \le 5, y=2x-5 \text{  when  } 5 \le x \le 8}.\]
Then $K$ is:
$\text{(A) } 21.5\quad \text{(B) } 36.4\quad \text{(C) } 36.5\quad \text{(D) } 44\quad \text{(E) less than 44 but arbitrarily close to it}$

Solution: [asy]  import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=9.2,ymin=-5.2,ymax=13.2;  pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);   /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs);  Label laxis; laxis.p=fontsize(10);  xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  draw((0,0)--(5,5)--(8,11)--(8,0)--(0,0)); dot((0,0)); dot((5,5)); dot((8,11)); dot((8,0));  [/asy]
The shape can be divided into a triangle and a trapezoid.  For the triangle, the base is $5$ and the height is $5$ , so the area is $\frac{5 \cdot 5}{2} = \frac{25}{2}$ .  For the trapezoid, the two bases are $5$ and $11$ and the height is $3$ , so the area is $\frac{3(5+11)}{2} = 24$ .  Thus, the total area is $\frac{25}{2} + 24 = \frac{73}{2} = \boxed{36.5}$

Answer: C 36.5
Problem: If it is known that $\log_2(a)+\log_2(b) \ge 6$ , then the least value that can be taken on by $a+b$ is:
$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$

Solution: We use the logarithm property of addition: \begin{align*} \log_2(a)+\log_2(b) \ge 6 &= \log_2(ab) \ge 6\\ &\Rightarrow 2^{log_2(ab)} \ge 2^6\\ &= ab \ge 64 \end{align*} Due to the Quadratic Optimization or the AM-GM Inequality , the least value is obtained when $a = b$ .
Therefore, $a = b = 8 \Rightarrow a + b = \boxed{16}$

Answer: D 16
Problem: [asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]
A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the center $M$ is:
$\text{(A) } 1\quad \text{(B) } 15\quad \text{(C) } 15\tfrac{1}{3}\quad \text{(D) } 15\tfrac{1}{2}\quad \text{(E) } 15\tfrac{3}{4}$

Solution: Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$ , where $x$ is the horizontal distance from the center and $y$ is the height.  The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center.  That means $0 = 400a + 16$ , so $a = -\frac{1}{25}$
Thus, the equation that models height based on distance from the center $y = -\frac{1}{25}x^2 + 16$ , so the height of the arch $5$ inches from the center is $-\frac{1}{25} \cdot 5^2 + 16 = \boxed{15}$ inches.

Answer: B 15
Problem: Let a sequence $\{u_n\}$ be defined by $u_1=5$ and the relationship $u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.$ If $u_n$ is expressed as a polynomial in $n$ , the algebraic sum of its coefficients is:
$\text{(A) 3} \quad \text{(B) 4} \quad \text{(C) 5} \quad \text{(D) 6} \quad \text{(E) 11}$

Solution: Note that the first differences create a linear function, so the sequence ${u_n}$ is quadratic.
The first three terms of the sequence are $5$ $8$ , and $15$ .  From there, a system of equations can be written. \[a+b+c=5\] \[4a+2b+c=8\] \[9a+3b+c=15\] Solve the system to get $a=2$ $b=-3$ , and $c=6$ .  The sum of the coefficients is $\boxed{5}$

Answer: C 5
Problem: The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$ , then $n$ equals:
$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$

Solution: The formula for the sum of the angles in any polygon is $180(n-2)$ . Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being $160$ , we can find the sum of the angles.
$a_{n}=160$
$a_{1}=160-5(n-1)$
Plugging this into the formula for finding the sum of an arithmetic sequence...
$n(\frac{160+160-5(n-1)}{2})=180(n-2)$
Simplifying, we get $n^2+7n-144$
Since we want the positive solution to the quadratic, we can easily factor and find the answer is $n=\boxed{9}$

Answer:  9
Problem: The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$ .  If $5b9$ is divisible by 9, then $a+b$ equals
$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution: If $5b9$ is divisible by $9$ , this must mean that $5 + b + 9$ is a multiple of $9$ . So, \[5 + b + 9 = 9, 18, 27, 36...\]
Because $5 + 9 = 14$ and $b$ is in between 0 and 9,
\[5 + b + 9 = 18\] \[b = 4\]
The question states that \[2a3 + 326 = 549\] so \[2a3 = 549 - 326\] \[2a3 = 223\] \[a = 2\]
\[a + b = 6\] which is answer choice $\boxed{6}$

Answer: C 6
Problem: Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$ , all logarithms to the same base and $x \not= 1$ .  If $\frac{b^2}{ac}=x^y$ , then $y$ is:
$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$

Solution: We are given: \[\frac{b^2}{ac} = x^y\]
Taking the logarithm on both sides: \[\log{\left(\frac{b^2}{ac}\right)} = \log{x^y}\]
Using the properties of logarithms: \[2\log{b} - \log{a} - \log{c} = y \log{x}\]
Substituting the values given in the problem statement: \[2q \log{x} - p \log{x} - r \log{x} = y \log{x}\]
Since $x \neq 1$ , dividing each side by $\log{x}$ we get: \[y = \boxed{2}\]

Answer: C 2
Problem: A triangle is circumscribed about a circle of radius $r$ inches.  If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:
$\text{(A)}\text{ independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2}$

Solution: The area $K$ of the triangle can be expressed in terms of its inradius $r$ and its semiperimeter $s$ as: \[K = r \times s = r \times \frac{P}{2}\]
So, $\frac{P}{K} = \boxed{2}$

Answer: C 2
Problem: If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$ , in inches, is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$

Solution: For rectangle $ABCD$ with perimeter 20, the diagonal $AC$ is given by: \[AC = \sqrt{l^2 + w^2}\] To minimize $AC$ $l$ and $w$ should be equal (i.e., the rectangle is a square). Thus, $l = w = 5$ .
So, the minimum $AC$ is: \[AC = \sqrt{5^2 + 5^2} = \boxed{50}\] ~ proloto

Answer: B 50
Problem: In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

Solution: After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$ , but we want to find the value of $AD$ . We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$ , and we have $10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6$ , and we see that $x = \boxed{166}$

Answer: E 166
Problem: In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

Solution: [asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -9.78, xmax = 9.78, ymin = -5.72, ymax = 5.72;  /* image dimensions */   draw((-1,4)--(-4.08,3.78)--(-3.1,-3.42)--(1.56,-0.22)--cycle, linewidth(2));   /* draw figures */ draw((-1,4)--(-4.08,3.78), linewidth(2));  draw((-4.08,3.78)--(-3.1,-3.42), linewidth(2));  draw((-3.1,-3.42)--(1.56,-0.22), linewidth(2));  draw((1.56,-0.22)--(-1,4), linewidth(2));  draw(circle((-2.287661623108666,0.35726272352132055), 3.863626188061437), linewidth(2));  draw((-1,4)--(-3.1,-3.42), linewidth(2));  draw((-4.08,3.78)--(1.56,-0.22), linewidth(2));   /* dots and labels */ dot((-1,4),dotstyle);  label("$A$", (-0.92,4.2), NE * labelscalefactor);  dot((-4.08,3.78),dotstyle);  label("$B$", (-4.42,4), NE * labelscalefactor);  dot((-3.1,-3.42),dotstyle);  label("$C$", (-3.4,-3.94), NE * labelscalefactor);  dot((1.56,-0.22),dotstyle);  label("$D$", (1.8,-0.4), NE * labelscalefactor);  dot((-1.566733533935139,1.9975415134291763),linewidth(4pt) + dotstyle);  label("$O$", (-1.34,1.98), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy] (Diagram not to scale)
Since $AO \cdot OC = BO \cdot OD$ $ABCD$ is cyclic through power of a point. From the given information, we see that $\triangle{AOB}\sim \triangle{DOC}$ and $\triangle{BOC} \sim \triangle{AOD}$ . Hence, we can find $CD=\frac{9}{2}$ and $AD=2 \cdot BC$ . Letting $BC$ be $x$ , we can use Ptolemy's to get \[6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}\] Since we are solving for $AD=2x=2\cdot\sqrt{\frac{83}{2}}=\sqrt{4\cdot\frac{83}{2}} = \boxed{166}$

Answer: E 166
Problem: In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

Solution: The solution says it all. Since $\angle AOD$ is supplementary to $\angle AOB$ $cos(\angle AOD) = cos(180^{\circ} - \angle AOB)=-cos(\angle AOB)$ . The law of cosines on $\triangle AOB$ gives us $cos(\angle AOB)=\frac {8^2+4^2-6^2}{(2)(8)(4)}=\frac {11}{16}$ . Again, we can use the law of cosines on $\triangle AOD$ , which gives us \[AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}\] \[=\sqrt {8^2+6^2-(2)(8)(6)cos(\angle 180^{\circ} - AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}\] \[=\sqrt{166}\] which gives us $\boxed{166}$

Answer: E 166
Problem: Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ .  To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$

Solution: [asy] draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); label("$A$",(0,10),N); label("$B$",(-9.5,-5.2),N); label("$C$",(9.5,-5.2),N);  dot((-3,0)); label("$P$",(-3,-2),N); draw((-3,0)--(0,10)); draw((-3,0)--(-8.66,-5)); draw((-3,0)--(8.66,-5));  dot((-9,7.5)); label("$P'$",(-9.2,7.5),N); draw((-9,7.5)--(0,10)); draw((-9,7.5)--(-8.66,-5)); draw((-9,7.5)--(-3,0));  [/asy]
Notice that $6^2+8^2=10^2.$ That makes us want to construct a right triangle.
Rotate $\triangle APC$ $60^{\circ}$ about A. Note that $\triangle PAC \cong \triangle P'AB$ , so \[\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.\]
Therefore, $\triangle APP'$ is equilateral, so $P'P=8$ , which means $\angle P'PB = 90^{\circ}.$
Let $\angle BP'P = \alpha .$ Notice that $\cos\alpha = \frac{8}{10}=\frac{4}{5},$ and $\sin\alpha = \frac{3}{5}.$
Applying the Law of Cosines to $\triangle APC$ (remembering $\angle APC = \angle AP'B$ ): \begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 164-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}
We want to find the area of $\triangle ABC$ , which is \[AC^2\cdot\frac{\sqrt3}{4}=25\sqrt3+36\approx\boxed{79}.\]

Answer: D 79
Problem: Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ .  To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$

Solution: Fun formula:  Given a point whose distances from the vertices of an equilateral triangle are $a$ $b$ , and $c$ , the side length of the triangle is:
\[s=\sqrt{\frac{1}{2}\left(a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+c^2a^2)-3(a^4+b^4+c^4)}\right)}\]
Given that the area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$ , the answer is:
\begin{align*} [ABC] &= \frac{\sqrt{3}}{4}\cdot\frac{1}{2}\left(a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+c^2a^2)-3(a^4+b^4+c^4)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{6\cdot 16(3^2 4^2+4^2 5^2+5^2 3^2)-3\cdot16(3^4+4^4+5^4)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96(144+400+225)-48(81+256+625)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-48\cdot962}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-96\cdot481}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot288}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot96\cdot3}\right)\\ &= 25\sqrt{3}\pm36 \approx \{6.5, \text{or } 78.5\} \end{align*}
$6.5$ is not a choice, therefore the answer is $\boxed{79}$

Answer: D 79
Problem: Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ .  To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$

Solution: Rotate $P$ and $B$ $60^{\circ}$ CCW around $A$ , becoming $X$ and $C$ .  Rotate $P$ and $C$ $60^{\circ}$ CCW around $B$ , becoming $Y$ and $A$ .  Rotate $P$ and $A$ $60^{\circ}$ CCW around $C$ , becoming $Z$ and $B$
[asy] import graph; import geometry; size(12cm);  pair A, B, C, P, X, Y, Z;  // Define the equilateral triangle ABC real a = sqrt(100+48*sqrt(3)); A = (0, 0); B = rotate(60)*A + (a, 0); C = rotate(120)*B + (a, 0);  // Define point P using given distances pair[] P_candidates = intersectionpoints(Circle(A,8), Circle(B,6)); for (pair candidate : P_candidates) {     if (length(candidate - C) < 10.1 && length(candidate - C) > 9.9) {         P = candidate;         break;     } }  // Rotate C and P about A through 60 degrees to get B and X X = rotate(60,A)*P;  // Rotate A and P about B through 60 degrees to get C and Y Y = rotate(60,B)*P;  // Rotate B and P about C through 60 degrees to get A and Z Z = rotate(60,C)*P;  // Draw the triangle and the segments draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P);  // Connect X, Y, Z to P and to the vertices of the triangle draw(X--P, dashed); draw(Y--P, dashed); draw(Z--P, dashed); draw(X--A, dashed); draw(X--C, dashed); draw(Y--A, dashed); draw(Y--B, dashed); draw(Z--B, dashed); draw(Z--C, dashed);  // Label the points label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$P$", P, NNE*2); label("$X$", X, NW); label("$Y$", Y, S); label("$Z$", Z, E);  // Add the distances label("$8$", (A+P)/2, NW); label("$6$", (B+P)/2, NE); label("$10$", (C+P)/2, N);  // Add right angle marks draw(rightanglemark(C,X,P,15)); draw(rightanglemark(P,B,Z,15)); draw(rightanglemark(A,P,Y,15)); [/asy]
Notice that since $\triangle AXC\cong\triangle APB$ $\triangle BYA\cong\triangle BPC$ , and $\triangle CZB\cong\triangle CPA$ , then
\[[AYBZCX]=2\cdot[ABC]\]
Now the area of the big hexagon is easy to compute since it's comprised of 3 equilateral triangle and 3 right triangles:
\begin{align*}[ABC] &= \frac{1}{2}[AYBZCX] = \frac{1}{2}\left(\underbrace{3\cdot\frac12\cdot6\cdot8}_{\text{3 right triangles}}+\underbrace{\frac{\sqrt{3}}{4}\left(6^2+8^2+10^2\right)}_{\text{3 equilateral triangles}}\right)\\ &= \frac{1}{2}\left(72+\frac{\sqrt{3}}{4}\cdot200)\right) = 36+25\sqrt{3}\\ &\approx \boxed{79}

Answer: D 79
Problem: Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ .  To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$

Solution: Let $s$ be the side length of $ABC.$ Notice that $s\le 14$ by the triangle inequality. This means that \[[ABC]\le\dfrac{14^2\sqrt{3}}{2}\approx 84.87.\] This automatically rules out choices $A, B,$ and $C.$ Now, we will look at if the area is $50$ . By the equilateral triangle area formula, $s$ would equal $10\sqrt{\dfrac{2}{\sqrt{3}}}\approx 10.75.$ This is very close to $10.$ If $s=10,$ $\angle APB=90$ and $\angle APC, \angle BPC<90$ by the Pythagorean theorem and Pythagorean inequalities. Thus, \[\angle APB+\angle APC+\angle BPC<270.\] $\angle APB+\angle APC+\angle BPC$ needs to be $360,$ and it probably cannot increase by more than $90$ by just adding $0.75$ to $s$ (more rigorous proof below) Thus, the only viable answer choice is $\boxed{79}.$

Answer:  79
Problem: The number of real values of $x$ that satisfy the equation \[(2^{6x+3})(4^{3x+6})=8^{4x+5}\] is:
$\text{(A)  zero} \qquad \text{(B)  one} \qquad \text{(C)  two} \qquad \text{(D)  three} \qquad \text{(E)  greater than 3}$

Solution: We know that $2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}$ .
We also know that $8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$ .
There are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$ , so the answer is $\boxed{3}$

Answer: E 3
Problem: If $F(n+1)=\frac{2F(n)+1}{2}$ for $n=1,2,\cdots$ and $F(1)=2$ , then $F(101)$ equals:
$\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53$

Solution: Notice that $\frac{2F(n)+1}{2}=F(n)+\frac{1}{2}.$
This means that for every single increment $n$ goes up from $1$ $F(n)$ will increase by $\frac{1}{2}.$ Since $101$ is $100$ increments from $1$ $F(n)$ will increase $\frac{1}{2}\times100=50.$
Since $F(1)=2,$ $F(101)$ will equal $2+50=\boxed{52}.$

Answer: D 52
Problem: The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 3 \qquad  \textbf{(E) }\ \text{more than 4}$

Solution: Solution by e_power_pi_times_i
Take the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$ . Factoring results in $(2x-5)(x-1) = 0$ , so there are $\boxed{2}$ real solutions.

Answer: C 2
Problem: The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 3 \qquad  \textbf{(E) }\ \text{more than 4}$

Solution: Notice that $a^0=1, a>0$ . So $2^0=1$ . So $2x^2-7x+5=0$ . Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$ . So this means that the equation has two real solutions. Therefore, select $\boxed{1}$

Answer: B 1
Problem: The expression $(81)^{-2^{-2}}$ has the same value as:
$\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4$

Solution: Let us recall $\text{PEMDAS}$ . We  calculate the exponent first. $(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}$ When we substitute, we get $81^{\frac{1}{4}}=\sqrt[4]{81}=\boxed{3}$

Answer: C 3
Problem: When the repeating decimal $0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:
$\textbf{(A)}\ 15 \qquad \textbf{(B) }\ 45 \qquad \textbf{(C) }\ 114 \qquad \textbf{(D) }\ 135 \qquad \textbf{(E) }\ 150$

Solution: We let $x=0.\overline{36}$ . Thus, $100x=36.\overline{36}$ . We find that $100x-x=99x=36.\overline{36}-0.\overline{36}=36$ , or $x=\frac{36}{99}=\frac{4}{11}$ . Since $4+11=15$ , the answer is $\boxed{15}$

Answer: A 15
Problem: The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$ -axis if the value of $c$ is:
$\textbf{(A)}\ - 16 \qquad  \textbf{(B) }\ - 4 \qquad  \textbf{(C) }\ 4 \qquad  \textbf{(D) }\ 8 \qquad  \textbf{(E) }\ 16$

Solution: Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of $y = x^2 - 8x + c$ to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So,
\begin{align*}     0 &= b^2-4ac\\     0 &= (-8)^2-4c\\     c &=  64\\     c &= 16\\ \end{align*}
Therefore $c=16$ , and our answer is $\boxed{16}$

Answer: E 16
Problem: Let $n$ be the number of integer values of $x$ such that $P = x^4 + 6x^3 + 11x^2 + 3x + 31$ is the square of an integer. Then $n$ is:
$\textbf{(A)}\ 4 \qquad  \textbf{(B) }\ 3 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 1 \qquad  \textbf{(E) }\ 0$

Solution: First, we wish to factor $P$ into a more manageable form. 
From the beginning of $P$ , we notice $x^4+6x^3$ , which gives us the idea to use $(x^2+3x)^2=x^4+6x^3+9x^2$
This gives us \[P=(x^2+3x)^2+2x^2+3x+31\] This is not useful, but it gives us a place to start from.
We can then try $(x^2+3x+1)=x^4+6x^3+11x^2+6x+1$ \[P=(x^2+3x+1)^2-3(x-10)\] This is much more useful, as it moves all non-linear terms inside of a squared expression.
We can then say $P=(x^2+3x+1)^2-3(x-10)=a^2$ , where $a^2$ is the square of an integer mentioned on the problem. Right from here, we can set $x=10$ , which cancels out the $3(x-10)$ , giving $(100+30+1)^2=a^2$ . This gives us one solution, $x=10$ $a=131$
We can then rearrange the expression, giving us $(x^2+3x+1)^2-a^2=3(x-10)$ . Factoring using difference of squares, we obtain \[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\]
We can then state that when $x$ is greater than $3$ and less than $-9$ $x^2+3x+1$ will be greater than $|3x-10|$ . This is obtained by setting $x^2+3x+1>|3x-10|$ and then solving the inequality. We can then conclude that $x^2+3x+1+|a|>|3x-10|$
Next, we claim that $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$ when $x \neq 10$ . We can prove this by first noting that since $x$ and $a$ are integers, $x^2+3x+1-|a|$ is an integer. Next, we shall assume that $x^2+3x+1-|a|=0$ . Solving this and plugging back into the original equation, we obtain $(2|a|)(0)=3(x-10)$ . Solving we obtain $x=10$ , which is a contraction to $x \neq 10$ . Therefore, $x^2+3x+1-|a| \neq 0$ and $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$
Finally, we can go back to the equation \[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\] We note that since $(x^2+3x+1+|a|)$ is larger than $3(x-10)$ , in order for there to be solutions, $(x^2+3x+1-|a|)$ must be in the range $(-1,1)$ . However, this contradicts what was proven earlier, so when $x \neq 10$ and $x < -9$ or $x > 3$ , there are no solutions for $x$
Now, all that remains to be checked are values of $x$ between $-9$ and $3$ . Using brute force and checking each value individually, we can assert that there are no such solutions for $x$ , leaving us with only $1$ solution, $x=10$ . Therefore, the answer is $\boxed{1}$

Answer: D 1