Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$

Solution: It is well known that if $|z|=1$ then $\bar{z}=\frac{1}{z}$ . Therefore, we have that the desired expression is equal to \[\left(z^1+z^4+z^9+...+z^{144}\right)\left(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144}\right)\] We know that $z=e^{\frac{i\pi}{4}}$ so $\bar{z}=e^{\frac{i7\pi}{4}}$ . Then, by De Moivre's Theorem, we have \[\left(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi}\right)\left(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi}\right)\] which can easily be computed as $\boxed{36}$

Answer:  36
Problem: Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$

Solution: We first calculate that $z^4 = -1$ . After a bit of calculation for the other even powers of $z$ , we realize that they cancel out add up to zero. Now we can simplify the expression to $\left(z^{1^2} + z^{3^2} + ... + z^{11^2}\right)\left(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}}\right)$ . Then, we calculate the first few odd powers of $z$ . We notice that $z^1 = z^9$ , so the values cycle after every 8th power. Since all of the odd squares are a multiple of $8$ away from each other, $z^1 = z^9 = z^{25} = ... = z^{121}$ , so $z^{1^2} + z^{3^2} + ... + z^{11^2} = 6z^{1^2}$ , and $\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}} = \frac{6}{z^{1^2}}$ . When multiplied together, we get $6 \cdot 6 = \boxed{36}$ as our answer.

Answer: C 36
Problem: Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$

Solution: $z=\mathrm{cis }(\pi/4)$
Perfect squares mod 8: $1,4,1,0,1,4,1,0,1,4,1,0$
$1/z=\overline{z}=\mathrm{cis }(7\pi/4)$
$6\mathrm{cis }(\pi/4)\cdot 6\mathrm{cis }(7\pi/4)=\boxed{36}$

Answer:  36
Problem: Circles $\omega$ and $\gamma$ , both centered at $O$ , have radii $20$ and $17$ , respectively. Equilateral triangle $ABC$ , whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$ , has vertex $A$ on $\omega$ , and the line containing side $\overline{BC}$ is tangent to $\gamma$ . Segments $\overline{AO}$ and $\overline{BC}$ intersect at $P$ , and $\dfrac{BP}{CP} = 3$ . Then $AB$ can be written in the form $\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}$ for positive integers $m$ $n$ $p$ $q$ with $\text{gcd}(m,n) = \text{gcd}(p,q) = 1$ . What is $m+n+p+q$ $\phantom{}$
$\textbf{(A) } 42 \qquad \textbf{(B) }86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 114 \qquad \textbf{(E) } 130$

Solution: [asy] size(20cm); draw(circle((0,0), 20)); label("$\omega$", (0,0), 4.05*20*dir(149)*20/21); draw(circle((0,0), 17)); label("$\gamma$", (0,0), 4.05*17*dir(149)*20/21); dot((0,0)); label("$O$", (0,0), E); pair aa = (-20, 0); dot(aa); label("$A$", aa, W); draw((-20,0)--(0,0)); real a = (-20 + (80/sqrt(13) - 34/sqrt(3))*(sqrt(13)/sqrt(12))*(sqrt(3)/2)); real ans = (80/sqrt(13) - 34/sqrt(3)); dot((a,0)); label("$P$", (a, 0), dir(290)*0.58); pair s = ((12*a + 0)/13, 0-sqrt(12)*a/13); dot(s); label("$S$", s, dir(135)); pair c = (a + 1/4*ans*1/sqrt(13), 0 + 1/4*ans*sqrt(12)/sqrt(13)); dot(c); label("$C$", c, dir(110)); pair m = (a - 1/4*ans*1/sqrt(13), 0 - 1/4*ans*sqrt(12)/sqrt(13)); dot(m); label("$M$", m, dir(285)); pair b = (a - 3/4*ans*1/sqrt(13), 0 - 3/4*ans*sqrt(12)/sqrt(13)); dot(b); label("$B$", b, S); draw(b--s); draw(s--(0,0)); draw(aa--b); draw(aa--c); draw(aa--m); markscalefactor=0.1; draw(rightanglemark(s,m,aa,3.4)); draw(rightanglemark((0,0),s,m,3.4)); [/asy]
Let $S$ be the point of tangency between $\overline{BC}$ and $\gamma$ , and $M$ be the midpoint of $\overline{BC}$ . Note that $AM \perp BS$ and $OS \perp BS$ . This implies that $\angle OAM \cong \angle AOS$ , and $\angle AMP \cong \angle OSP$ . Thus, $\triangle PMA \sim \triangle PSO$
If we let $s$ be the side length of $\triangle ABC$ , then it follows that $AM = \frac{\sqrt{3}}{2}s$ and $PM = \frac{s}{4}$ . This implies that $AP = \frac{\sqrt{13}}{4}s$ , so $\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}$ . Furthermore, $\frac{AM + SO}{AO} = \frac{AM}{AP}$ (because $\triangle PMA \sim \triangle PSO$ ) so this gives us the equation \[\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}\] to solve for the side length $s$ , or $AB$ . Thus, \[\frac{\sqrt{39}}{2}s + 17\sqrt{13} = 40\sqrt{3}\] \[\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}\] \[s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB\] The problem asks for $m + n + p + q = 80 + 13 + 34 + 3 = \boxed{130}$

Answer: E 130
Problem: Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a  \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$
$\textbf{(A) } 8 \qquad  \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution: By definition, the recursion becomes $a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}$ .  By the change of base formula, this reduces to $a_n = n^{\log_{n-1}(a_{n-1})}$ .  Thus, we have $\log_n(a_n) = \log_{n-1}(a_{n-1})$ .  Thus, for each positive integer $m \geq 3$ , the value of $\log_m(a_m)$ must be some constant value $k$
We now compute $k$ from $a_3$ .  It is given that $a_3 = 3\,\heartsuit\,2 = 3^{\frac1{\log_7(2)}}$ , so $k = \log_3(a_3) = \log_3\left(3^{\frac1{\log_7(2)}}\right) = \frac1{\log_7(2)} = \log_2(7)$
Now, we must have $\log_{2019}(a_{2019}) = k = \log_2(7)$ . At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with.
$\frac{\log{a_{2019}}}{\log{2019}} = \frac{\log{7}}{\log{2}}\implies \frac{\log{a_{2019}}}{\log{7}} = \frac{\log{2019}}{\log{2}}\implies \log_7(a_{2019}) =\log_2(2019)$
We conclude that $\log_7(a_{2019}) = \log_2(2019) \approx \boxed{11}$

Answer: D 11
Problem: Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a  \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$
$\textbf{(A) } 8 \qquad  \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution: Using the recursive definition, $a_4 = (4  \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)$ or $a_4 = (4^{m})^{k}$ where $m = \frac{1}{\log_{7}(3)}$ and $k = \log_{7}(3^{\frac{1}{\log_{7}(2)}})$ . Using logarithm rules, we can remove the exponent of the 3 so that $k = \frac{\log_{7}(3)}{\log_{7}(2)}$ . Therefore, $a_4 = 4^{\frac{1}{\log_{7}(2)}}$ , which is $4  \, \heartsuit \, 2$
We claim that $a_n = n  \, \heartsuit \, 2$ for all $n \geq 3$ . We can prove this through induction.
Clearly, the base case where $n = 3$ holds.
$a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1)  \, \heartsuit \, 2)$
This can be simplified as $a_n = (n^{\log_{n-1}(7)}) \, \diamondsuit \, ((n-1)^{\log_{2}(7)})$
Applying the diamond operation, we can simplify $a_n = n^h$ where $h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}$ . By using logarithm rules to remove the exponent of $\log_{7}(n-1)$ and after cancelling, $h = \frac{1}{\log_{7}(2)}$
Therefore, $a_n = n^{\frac{1}{\log_{7}(2)}} = n  \, \heartsuit \, 2$ for all $n \geq 3$ , completing the induction.
We have $a_{2019} = 2019^{\log_{2}(7)}$ . Taking $\log_{2019}$ of both sides gives us ${\log_{2019}(a_{2019})} = {\log_{2}(7)}$ . Then, by changing to base $7$ and after cancellation, we arrive at ${\log_{7}(a_{2019})} = {\log_{2}(2019)}$ . Because $2^{11} = 2048$ and $2^{10} = 1024$ , our answer is $\boxed{11}$

Answer: D 11
Problem: Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a  \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$
$\textbf{(A) } 8 \qquad  \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution: We are given that \[a_n=(n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] \[a_n=(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Since we are asked to find $\log_7(a_{2019})$ , we directly apply \[\log_7(a_n)=\log_7(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Using the property that $\log_ab^c=c\log_ab$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7(n^{\frac{1}{\log_7(n-1)}}))\] Now using the property that $\frac{1}{\log_ab}=\log_ba$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7n^{\log_{n-1}7})\] Once again applying the first property yields \[\log_7(a_n)=(\log_7a_{n-1})(\log_{n-1}7)(\log_7n)\] Rearranging the expression, \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7a_{n-1})\]
Now expressing $\log_7a_{n-1}$ in a similar expression as $\log_7a_n$
\[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7a_{n-2})\] \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7n-2)(\log_{n-3}7)...(\log_74)(\log_37)(\log_7a_3)\]
Because of the fact that $(\log_ab)(\log_ba)=1$ , we can cancel out the terms to get
\[\log_7(a_n)=(\log_7n)(\log_37)(\log_7a_3)\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_7(3^{\frac{1}{\log_72}}))\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_27)(\log_73)\] \[\log_7(a_n)=(\log_27)(\log_7n)\]
Using the Chain Rule for Logarithm, $(\log_ab)(\log_bc)=\log_ac$ , yields
\[\log_7(a_n)=(\log_2n)\] Finally, substituting in $n=2019$ , we have \[\log_7(a_{2019})=(\log_22019)\] \[\log_7(a_{2019})\approx11\boxed{11}\]

Answer: D 11
Problem: Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$ , and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$ . What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?
$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

Solution: For all nonnegative integers $n$ , let $\angle C_nA_nB_n=x_n$ $\angle A_nB_nC_n=y_n$ , and $\angle B_nC_nA_n=z_n$
Note that quadrilateral $A_0B_0A_1B_1$ is cyclic since $\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ$ ; thus, $\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0$ . By a similar argument, $\angle A_0A_1C_1=\angle A_0C_0C_1=90^\circ-x_0$ . Thus, $x_1=\angle A_0A_1B_1+\angle A_0A_1C_1=180^\circ-2x_0$ . By a similar argument, $y_1=180^\circ-2y_0$ and $z_1=180^\circ-2z_0$
Therefore, for any positive integer $n$ , we have $x_n=180^\circ-2x_{n-1}$ (identical recurrence relations can be derived for $y_n$ and $z_n$ ). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to $n$ (and the coefficient of $x_0$ is $(-2)^n$ ). Hence, we let $x_n=pq^n+r+(-2)^nx_0$ . We will solve for $p$ $q$ , and $r$ by iterating the recurrence to obtain $x_1=180^\circ-2x_0$ $x_2=4x_0-180^\circ$ , and $x_3=540-8x_0$ . Letting $n=1,2,3$ respectively, we have \begin{align} pq+r&=180 \\ pq^2+r&=-180 \\ pq^3+r&=540 \end{align}
Subtracting $(1)$ from $(3)$ , we have $pq(q^2-1)=360$ , and subtracting $(1)$ from $(2)$ gives $pq(q-1)=-360$ . Dividing these two equations gives $q+1=-1$ , so $q=-2$ . Substituting back, we get $p=-60$ and $r=60$
We will now prove that for all positive integers $n$ $x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60$ via induction. Clearly the base case of $n=1$ holds, so it is left to prove that $x_{n+1}=(-2)^{n+1}(x_0-60)+60$ assuming our inductive hypothesis holds for $n$ . Using the recurrence relation, we have \begin{align*} x_{n+1}&=180-2x_n \\ &=180-2((-2)^n(x_0-60)+60) \\ &=(-2)^{n+1}(x_0-60)+60 \end{align*}
Our induction is complete, so for all positive integers $n$ $x_n=(-2)^n(x_0-60)+60$ . Identical equalities hold for $y_n$ and $z_n$
The problem asks for the smallest $n$ such that either $x_n$ $y_n$ , or $z_n$ is greater than $90^\circ$ . WLOG, let $x_0=60^\circ$ $y_0=59.999^\circ$ , and $z_0=60.001^\circ$ . Thus, $x_n=60^\circ$ for all $n$ $y_n=-(-2)^n(0.001)+60$ , and $z_n=(-2)^n(0.001)+60$ . Solving for the smallest possible value of $n$ in each sequence, we find that $n=15$ gives $y_n>90^\circ$ . Therefore, the answer is $\boxed{15}$

Answer: E 15
Problem: Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$ , and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$ . What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?
$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

Solution: We note that the problem seems quite complicated, but since it is an AMC 12, the difference between the largest angle of $\triangle A_nB_nC_n$ and $60^{\circ}$ (we call this quantity S) most likely reduces to a simpler problem like some repeating sequence. The only obvious sequence (for the answer choices) is a geometric sequence with an integer common ratio. From here, we get that $S_n=0.001S^n$ and we need $S_n>30$ . The first power of two greater than $30,000$ is $2^{15}$ thus our answer is $\boxed{15}$

Answer: E 15
Problem: Let $f(x) = x^{2}(1-x)^{2}$ . What is the value of the sum \[f \left(\frac{1}{2019} \right)-f  \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]
$\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1$

Solution: First, note that $f(x) = f(1-x)$ . We can see this since \[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\] Using this result, we regroup the terms accordingly: \[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\] \[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\] Now it is clear that all the terms will cancel out (the series telescopes), so the answer is $\boxed{0}$

Answer: A 0
Problem: For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$
$\textbf{(A) } 57 \qquad\textbf{(B) } 59 \qquad\textbf{(C) } 61 \qquad\textbf{(D) } 62 \qquad\textbf{(E) } 63$

Solution: For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need \[\log_2{x} + \log_4{x} > 3\] \[\log_2{x} + 3 > \log_4{x}\] \[\log_4{x} + 3 > \log_2{x}.\] The second inequality is redundant, as it's always less restrictive than the last inequality.
Let's raise the first inequality to the power of $4$ . This gives $4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow \left(2^2\right)^{\log_2{x}} \cdot x > 64 \Rightarrow x^2 \cdot x > 64$ . Thus, $x > 4$
Doing the same for the second inequality gives $4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64$ (where we are allowed to divide both sides by $x$ since $x$ must be positive in order for the logarithms given in the problem statement to even have real values).
Combining our results, $x$ is an integer strictly between $4$ and $64$ , so the number of possible values of $x$ is $64 - 4 - 1 = \boxed{59}$

Answer: B 59
Problem: For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$
$\textbf{(A) } 57 \qquad\textbf{(B) } 59 \qquad\textbf{(C) } 61 \qquad\textbf{(D) } 62 \qquad\textbf{(E) } 63$

Solution: Using the triangle inequality, you get $\log_2{x}+\log_4{x} > 3$ . Solving for $x$ , you get $x > 4$ . Now we need an upper-bound for $x$ and since we're dealing with bases of $2$ and $4$ , we're looking for answer choices close to a power of $2$ and $4$ . All the answer choices seem to be around $64$ , and plugging that into the inequality $3+\log_4{x} > \log_2{x}$ we see $64$ is the correct number. Now we have $64 > x > 4$ and the number of integers in between is $64-4-1 = \boxed{59}$ --OGBooger

Answer: B 59
Problem: The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A);  } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution: Note that of the $12$ cities, $6$ of them ( $2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$ ). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the start and end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). It is not hard to find that there is only one configuration satisfying these conditions:
Note: $\text{X}$ s represent unused edges.
[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A);  } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E);  label("X", (0, 1.5)); label("X", (1.5, 2)); label("X", (1.5, 0)); label("X", (3, 0.5)); label("O", (1,1), NE); label("O", (2,1), NE); [/asy]
Observe that at each of the $2$ cities marked with an $\text{O}$ on a path, there are $2$ possibilities: we can either continue straight and cross back over the path later, or we can make a left turn, then turn right when we approach the junction again. This gives us a total of $2\cdot 2 = \boxed{4}$ valid paths.

Answer: E 4
Problem: The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A);  } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution: Observe that only the two central vertices can be visited twice. Since the path is of length 13, we need to repeat a vertex. Using casework on each vertex, we can find there are two paths that go through each central vertex twice, for an answer of $\boxed{4}.$

Answer:  4
Problem: The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A);  } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution: Looking at the answer choices we see that it would probably be easy enough to count the different paths. After some experimenting, we find that we must cross one of the middle vertexes twice, making counting easier.
We finally find that there are $\boxed{4}$ paths that satisfy the conditions. Luckily, $4$ is the largest answer choice so we know it must be correct.

Answer:  4