AMC12 2024 B

If the person is the 1013th from the left, that means there is 1012 people to their left. If the person is the 1010th from the right, that means there is 1009 people to their right. Therefore, there are $1012 + 1 + 1009 = \boxed{\textbf{(B) } 2022}$ people in line.

$10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!$ $6! \cdot 7! = 720 \cdot 7!$ Therefore, the equation is equal to $720 \cdot 7! - 720 \cdot 7! = (B)\ 0$ [ONLY FOR CERTAIN CHINESE TESTPAPERS] $0 - 5! = (A)\ -120$

$\pi = 3.14159\dots$ is slightly less than $\dfrac{22}{7} = 3.\overline{142857}$. So $7\pi \approx 21.9$ The inequality expands to be $-21.9 \le 2x \le 21.9$. We find that $x$ can take the integer values between $-10$ and $10$ inclusive. There are $\boxed{\text{E. }21}$ such values. Note that if you did not know whether $\pi$ was greater than or less than $\dfrac{22}{7}$, then you might perform casework. In the case that $\pi > \dfrac{22}{7}$, the valid solutions are between $-11$ and $11$ inclusive: $23$ solutions. Since, $23$ is not an answer choice, we can be confident that $\pi < \dfrac{22}{7}$, and that $\boxed{\text{E. } 21}$ is the correct answer. Test advice: If you are in the test and do not know if $\frac{22}{7}$ is bigger or smaller than $\pi$, you can use the extremely sophisticated method of just dividing $\dfrac{22}{7}$ via long division. Once you get to $3.142$ you realise that you don't need to divide further since $\pi = 3.1416$ when rounded to 4 decimal places.Therefore, you do not include $11$ and $-11$ and the answer is 21.

Consider the triangular array of numbers: \[1\] \[2, 3\] \[4, 5, 6\] \[7, 8, 9, 10\] \[11, 12, 13, 14, 15\] \[\vdots\]. The numbers in a row congruent to $1 \bmod{5}$ will be in bucket A. Similarly, the numbers in a row congruent to $2, 3, 4, 0 \bmod{5}$ will be in buckets B, C, D, and E respectively. Note that the $n^\text{th}$ row ends with the $n^\text{th}$ triangle number, $\frac{n(n+1)}{2}$. We must find values of $n$ that make $\frac{n(n+1)}{2}$ close to $2024$. \[\frac{n(n+1)}{2} \approx 2024\] \[n(n+1) \approx 4048\] \[n^2 \approx 4048\] \[n \approx 63\] Trying $n = 63$ we find that $\frac{n(n+1)}{2} = 2016$. Since $2016$ will be the last ball in row $63$, ball $2024$ will be in row $64$. Since $64 \equiv 4 \bmod{5}$, ball $2024$ will be placed in bucket $\boxed{\text{D. } D}$.

Recall that the sum of the first $n$ odd numbers is $n^2$. Thus \[1 + 3 + 5 + 7+ \dots + 97 + 99 = 50^2 = 2500.\] If we want to minimize the number of sign flips to make the number negative, we must flip the signs corresponding to the values with largest absolute value. This will result in the inequality \[1 + 3 + 5 +\dots + (2n - 3) + (2n - 1) - (2n + 1) - (2n + 3)-\dots - 97 - 99 < 0.\] The positive section of the sum will contribute $n^2$, and the negative section will contribute $-(2500-n^2) = (n^2 - 2500)$. The inequality simplifies to \[n^2 + (n^2 - 2500) < 0\] \[2n^2 < 2500\] \[n^2 < 1250\] The greatest positive value of $n$ satisfying the inequality is $n = 35$, corresponding to $35$ positive numbers, and $\boxed{\text{B. } 15}$ negatives. ALEX

Generally, the number of digits of number $n$ in base $b$ is \[\lfloor \log_b n \rfloor + 1.\] In this question, it is $\lfloor \log_{5} (5\times 10^{13})\rfloor+1$. Note that \begin{align*} \log_{5} (5\times 10^{13}) &= 1+\frac{13}{\log_{10} 5} \\ &\approx 1+\frac{13}{0.7} \\ &\approx 19.5 \end{align*} Hence, our answer is $\fbox{\textbf{(B)} 20}$

We know that $WX = 4$, $WZ = 8$, so $YZ = 4$ and $YX = 8$. Since $\angle WMA = 90^\circ$, triangles $WXM$ and $MYA$ are similar. Therefore, $\frac{WX}{MY} = \frac{XM}{YA}$, which gives $\frac{4}{8 - XM} = \frac{XM}{4 - ZA}$. We also know that the areas of triangles $WXM$ and $WAZ$ are equal, so $WX \cdot XM = WZ \cdot ZA$, which implies $4 \cdot XM = 8 \cdot ZA$. Substituting this into the previous equation, we get $\frac{4}{8 - 2ZA} = \frac{2ZA}{4 - ZA}$, yielding $ZA = 1$ and $XM = 2$. Thus, \[\triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(C) }15}\]

We have log2⁡x⋅log3⁡x=2(log2⁡x+log3⁡x)1=2(log2⁡x+log3⁡x)log2⁡x⋅log3⁡x1=2(1log3⁡x+1log2⁡x)1=2(logx⁡3+logx⁡2)logx⁡6=12x12=6x=36 so $\boxed{\textbf{(C) }36}$

Inequalities of the form $|x|+|y| \le 8$ are well-known and correspond to a square in space with centre at origin and vertices at $(8, 0)$, $(-8, 0)$, $(0, 8)$, $(0, -8)$. The diagonal length of this square is clearly $16$, so it has an area of \[\frac{1}{2} \cdot 16 \cdot 16 = 128\] Now, \[(x^2 + y^2 - 25)^2 \le 49\] Converting to polar form, \[r^2 - 25 \le 7 \implies r \le \sqrt{32},\] and \[r^2 - 25 \ge -7\implies r\ge \sqrt{18}.\] The intersection of these inequalities is the circular region $T$ for which every circle in $T$ has a radius between $\sqrt{18}$ and $\sqrt{32}$, inclusive. The area of such a region is thus $\pi(32-18)=14\pi.$ The requested probability is therefore $\frac{14\pi}{128} = \frac{7\pi}{64},$ yielding $(m,n)=(7,64).$ We have $m+n=7+64=\boxed{\textbf{(B)}\ 71}.$

$\textbf{First Case}$ We start off by knowing that there must exist an ordered pair (0,y,z) and a big term 7 such that the range is satisfied and the median could also be satisfied. Because x≤y≤z, we say x=0. Then, we get the sum 24.8. We need 24.8+y+z9∈Z. This means that we need the nearest multiple of 9, which is 36, and we get y+z=11.2. We need one of these to be the median, WLOG say y. Then, y∈{4,5}. So if y=4, we get z=7.2, and if y=5, we get 6.2. We see that if z=6.2, the range will still remain as 7, and therefore the one ordered triple (0,5,6.2) satisfies this. We know that the median can be y, so we have y∈{4,5}. We do casework on 4 and 5. $\textbf{Case 1}$ Say y=4. Then, because we already know there exists an ordered triple where z is not the largest, there must exist at least one ordered triple where z is, so we say z−x=7. We also know that the mean must be divisible by 9. Quickly summing each number up we get 24.8. The next number divisible by 9 is 36. We add y to get 28.8. We then know x+z=36−28.8=7.2. The smallest values give x=0.1 and z=7.1. This satisfies our constraints. We don't want any errors, so we check the next sum, which is 45. We get then 7.2=x+z and z−x=7. This gives x=4.6 and z=11.6. This is a contradiction, as x≤y≤z, and the values incrementally become too large, and therefore we only have one ordered triple for this case, which is (0.1,4,7.1) $\textbf{Case 2}$ We now check y=5. This makes our sum 29.8. The nearest multiple of 9 is 36, again. This gives us 6.2=x+z and z−x=7. Solving gives us -0.4 and 6.6. This however doesn't work, as 6.6 is not the greatest value. We decide to check the next multiple of 9, which is 45. This gives us 15.2=x+z and z−x=7. This gives us x=4.1 and z=11.1. This also doesn't work, as x becomes incrementally larger, and therefore there are no ordered triples that satisfy this. $\textbf{Last Case}$ We have one more case, and that is x is not the first number, but z is the last, meaning z−1=7, and z=8. This means that 24.8+x+y+z9∈Z. So, if z=8, we get 32.8. We again see that we need the nearest multiple of 9 which is 36, and we get 3.2=x+y. We see that x and y still need to be a integer median and 3.2 is too small, so 36 cannot work. We try 45, and see 12.2=x+y. After some trial and error, we see that x=6 and y=6.2, giving us the ordered triple (6,6.2,8). Continuing as before shows us that x again incrementally increases, and therefore we have only one ordered triple. We have $\boxed{\textbf{(C) }3}$ ordered triples. These are (0.1,4,7.1), (0,5,6.2), and (6,6.2,8).

Add up $x_1$ with $x_{89}$, $x_2$ with $x_{88}$, and $x_i$ with $x_{90-i}$. Notice \[x_i+x_{90-i}=\sin^2(i^{\circ})+\sin^2((90-i)^{\circ})=\sin^2(i^{\circ})+\cos^2(i^{\circ})=1\] by the Pythagorean identity. Since we can pair up $1$ with $89$ and keep going until $44$ with $46$, we get \[x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+\left(\frac{\sqrt{2}}{2}\right)^2+1^2=\frac{91}{2}\] Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$

By making a rough estimate of where $z$, $z^2$, and $z^3$ are on the complex plane, we can draw a pretty accurate diagram (like above.) Here, points $Z_1$, $Z_2$, and $Z_3$ lie at the coordinates of $z$, $z^2$, and $z^3$ respectively, and $O$ is the origin. We're given $|z|=2$, so $|z^2|=|z|^2=4$ and $|z^3|=|z|^3 = 8$. This gives us $OZ_1=2$, $OZ_2=4$, and $OZ_3=8$. Additionally, we know that $\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}$ (since every power of $z$ rotates around the origin by the same angle.) We set these angles equal to $\theta$. We have that [OZ1Z2Z3]=[OZ1Z2]+[OZ2Z3]=12⋅2⋅4sin⁡θ+12⋅4⋅8sin⁡θ=4sin⁡θ+16sin⁡θ=20sin⁡θ Since this is equal to $15$, we have $20\sin\theta=15$, so $\sin\theta=\frac{3}{4}$. Thus, $\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}$.

Adding up the first and second equation, we get: \begin{align*} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2x^2 - 16x + 2y^2 - 4y \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \\ &= 2(x - 4)^2 + 2(y - 1)^2 - 34 \end{align*} All squared values must be greater than or equal to $0$. As we are aiming for the minimum value, we set the two squared terms to be $0$. This leads to $\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}$

First note that the Euler's totient function of $125$ is $100$. We can set up two cases, which depend on whether a number is relatively prime to $125.$ If $n$ is relatively prime to $125$, then $n^{100} \equiv 1 \pmod{125}$ because of Euler's Totient Theorem. If $n$ is not relatively prime to $125$, it must have a factor of $5$. Express $n$ as $5m$, where $m$ is some integer. Then $n^{100} \equiv (5m)^{100} \equiv 5^{100}\cdot m^{100} \equiv 125 \cdot 5^{97} \cdot m^{100} \equiv 0 \pmod{125}$. Therefore, $n^{100}$ can only be congruent to $0$ or $1 \pmod{125}$. Our answer is $\boxed{2}$.

We rewrite: $A(0,1)$ $B(\log _{2} 3, 2)$ $C(\log _{2} 7, 3)$. From here we setup Shoelace Theorem and obtain: $\frac{1}{2}(2(\log _{2} 3) - log _{2} 7)$. Following log properties and simplifying gives $\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}$.

There are ${16 \choose 4}$ ways to choose the first committee, ${12 \choose 4}$ ways to choose the second, ${8 \choose 4}$ for the third, and ${4 \choose 4}=1$ for the fourth. Since the committees are indistinguishable, we need to divide the product by $4!$. Thus the $16$ people can be grouped in \[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}=\frac{16!}{(4!)^5}\] ways. In each committee, there are $4 \cdot 3=12$ ways to choose the chairperson and secretary, so $12^4$ ways for all $4$ committees. Therefore, there are \[\frac{16!}{(4!)^5}12^4\] total possibilities. Since $16!$ contains $6$ factors of $3$, $(4!)^5$ contains $5$, and $12^4$ contains $4$, $r=6-5+4=\boxed{\textbf{(A) }5}$.

Since $-10 \le a,b \le 10$, there are 21 integers to choose from, and $P(21,2) = 21 \times 20 = 420$ equally likely ordered pairs $(a,b)$. Applying Vieta's formulas, $x_1 \cdot x_2 \cdot x_3 = -6$ $x_1 + x_2+ x_3 = -a$ $x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b$ Cases: (1) $(x_1,x_2,x_3) = (-1,1,6) , b = -1, a=-6$ valid (2) $(x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0$ valid (3) $(x_1,x_2,x_3) = (1,-2,3) , b = -5, a=-2$ valid (4) $(x_1,x_2,x_3) = (-1,2,3) , b = 1, a=-4$ valid (5) $(x_1,x_2,x_3) = (-1,-2,-3) , b = 11$ invalid the total event space is $21 \cdot (21- 1)$ (choice of select a times choice of selecting b given no-replacement) hence, our answer is $\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}$

The first $20$ terms are $F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765$ So the answer is $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319}$. - Do not do this unless you have no other option or no time to find a smarter way

Let O be circumcenter of the equilateral triangle Easily get $OF = \frac{14\sqrt{3}}{3}$ $2 \cdot (\triangle(OFC) + \triangle(OCE)) = OF^2 \cdot \sin(\theta) + OF^2 \cdot \sin(120 - \theta)$ \[= \frac{14^2 \cdot 3}{9} ( \sin(\theta) + \sin(120 - \theta) )\] \[= \frac{196}{3} ( \sin(\theta) + \sin(120 - \theta) )\] \[= 2 \cdot {\frac{1}{3} } \cdot(ADBECF) = 2\cdot \frac{91\sqrt{3}}{3}\] \[\sin(\theta) + \sin(120 - \theta) = \frac{13\sqrt{3}}{14}\] \[\sin(\theta) + \frac{ \sqrt{3}}{2}\cos( \theta) +\frac{ \sqrt{1}}{2}\sin( \theta) = \frac{13\sqrt{3}}{14}\] \[\sqrt{3} \sin( \theta) + \cos( \theta) = \frac{13 }{7}\] \[\cos( \theta) = \frac{13 }{7} - \sqrt{3} \sin( \theta)\] \[\frac{169 }{49} - \frac{26\sqrt{3} }{7} \sin( \theta) + 4 \sin( \theta)^2 = 1\] \[\sin( \theta) = \frac{5\sqrt{3} }{14} or \frac{4\sqrt{3} }{7}\] $\frac{4\sqrt{3} }{7}$ is invalid given $\theta \leq 60^\circ$ , $\sin(\theta ) < \sin( 60^\circ ) = \frac{\sqrt{3} }{2} = \frac{\sqrt{3} \cdot 3.5}{7}$ \[\cos( \theta) = \frac{11 }{14}\] \[\tan( \theta) = \frac{5\sqrt{3} }{11} \boxed{B }\]

Let the midpoint of $BC$ be $M$, and let the length $BM = CM = a$. We know there are limits to the value of $x$, and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length $BC$ to $AC$ and $AB$, and doesn't contain any information about the median. Therefore we're going to have to write the side $BC$ in terms of $x$ and then use the triangle inequality to find bounds on $x$. We use Stewart's theorem to relate $BC$ to the median $AM$: $man + dad = bmb + cnc$. In this case $m = \frac{a}2$, $n=\frac{a}2$, $a = m+n$, $d = x$, $b = 42$, $c = 40$. Therefore we get the equation $2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2$ $2a^2 + 2x^2 = 42^2 + 40^2$. Notice that since $20-21-29$ is a pythagorean triple, this means $2a^2 + 2x^2 = 58^2$. \[\implies a^2 = \frac{58^2}{2}-x^2\] \[\implies a = \sqrt{\frac{58^2}{2}-x^2}\] By triangle inequality, $2a+40>42 \implies a>1$ and $40+42>2a \implies a<41$ Let's tackle the first inequality: \[\sqrt{\frac{58^2}{2}-x^2}>1 \implies x^2 < \frac{58^2}{2}-1\] \[\implies x^2 < \frac{40^2+42^2}{2}-1 \implies x^2<41^2\] Here we use the property that $\frac{x^2+(x+2)^2}{2}-1 = (x+1)^2$. Therefore in this case, $x<41$. For the second inequality, \[\sqrt{\frac{58^2}{2}-x^2} < 41 \implies x^2 > \frac{58^2}{2}-41^2\] \[\implies x^2 > \frac{58^2}{2}-1 + 1 - 41^2\] \[\implies x^2 > 41^2 + 1 - 41^2 \implies x^2 > 1 \implies x > 1\] Therefore we have $1<x<41$, so the domain of $f(x)$ is $(1,41)$. The area of this triangle is $\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)$. The maximum value of the area occurs when the triangle is right, i.e. $\theta = 90^{\circ}$. Then the area is $\frac{1}{2} \cdot 40 \cdot 42 = 840$. The length of the median of a right triangle is half the length of it's hypotenuse, which squared is $40^2+42^2 = 58^2$. Thus the length of $x$ is $29$. Our final answer is $1+41+840+29 = \boxed{\textbf{911 } (C)}$

Let $\alpha$ and $\beta$ be the smallest angles of the $3-4-5$ and $5-12-13$ triangles respectively. We have \[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\] Then \[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\] Let $\theta$ be the smallest angle of the third triangle. Consider \[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\] In order for this to be undefined, we need \[1-\frac{56}{33}\cdot\tan{\theta}=0\] so \[\tan{\theta}=\frac{33}{56}\] Hence the base side lengths of the third triangle are $33$ and $56$. By the Pythagorean Theorem, the hypotenuse of the third triangle is $65$, so the perimeter is $33+56+65=\boxed{\textbf{(C) }154}$.

Let $AB=c$, $BC=a$, $AC=b$. According to the law of sines, \[\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}=2\cos \angle A\] According to the law of cosines, \[\cos \angle A=\frac{b^2+c^2-a^2}{2bc}\] Hence, \[\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}\] This simplifies to $b^2=a(a+c)$. We want to find the positive integer solution $(a, b, c)$ to this equation such that $a, b, c$ forms a triangle, and $a+b+c$ is minimized. We proceed by casework on the value of $b$. Remember that $a<a+c$. Case $1$: $b=1$ Clearly, this case yields no valid solutions. Case $2$: $b=2$ For this case, we must have $a=1$ and $c=3$. However, $(1, 2, 3)$ does not form a triangle. Hence this case yields no valid solutions. Case $3$: $b=3$ For this case, we must have $a=1$ and $c=8$. However, $(1, 3, 8)$ does not form a triangle. Hence this case yields no valid solutions. Case $4$: $b=4$ For this case, $a=1$ and $c=15$, or $a=2$ and $c=6$. As one can check, this case also yields no valid solutions Case $5$: $b=5$ For this case, we must have $a=1$ and $c=24$. There are no valid solutions Case $6$: $b=6$ For this case, $a=2$ and $c=16$, or $a=4$ and $c=5$, or $a=3$ and $c=9$. The only valid solution for this case is $(4, 6, 5)$, which yields a perimeter of $15$. When $b\ge 7$, it is easy to see that $a+c>7$. Hence $a+b+c>14$, which means $a+b+c\ge15$. Therefore, the answer is $\boxed{\textbf{(C) }15}$

To find the height of the pyramid, we need the length from the center of the octagon (denote as $I$) to its vertices and the length of AV. From symmetry, we know that $\overline{AV} = \overline{DV}$, therefore $\triangle{AVD}$ is a 45-45-90 triangle. Denote $\overline{AV}$ as $x$ so that $\overline{AD} = x\sqrt{2}$. Doing some geometry on the isosceles trapezoid $ABCD$ (we know this from the fact that it is a regular octagon) reveals that $\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}$ and $\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2$. To find the length $\overline{IA}$, we cut the octagon into 8 triangles, each with a smallest angle of 45 degrees. Using the law of cosines on $\triangle{AIB}$ we find that ${\overline{IA}}^2=(2+\sqrt{2})/2$. Finally, using the pythagorean theorem, we can find that ${\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}$ which is answer choice $\boxed{B}$.

First we derive the relationship between the inradius of a triangle $r$, and its three altitudes $a, b, c$. Using an area argument, we can get the following well known result \[\left(\frac{AB+BC+AC}{2}\right)r=A\] where $AB, BC, AC$ are the side lengths of $\triangle ABC$, and $A$ is the triangle's area. Substituting $A=\frac{1}{2}\cdot AB\cdot c$ into the above we get \[\frac{r}{c}=\frac{AB}{AB+BC+AC}\] Similarly, we can get \[\frac{r}{b}=\frac{AC}{AB+BC+AC}\] \[\frac{r}{a}=\frac{BC}{AB+BC+AC}\] Hence, 1r=1a+1b+1c Note that there exists a unique, non-degenerate triangle with altitudes $a, b, c$ if and only if $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are the side lengths of a non-degenerate triangle, i.e., $\frac{1}{b}+\frac{1}{c}>\frac{1}{a}$. With this in mind, it remains to find all positive integer solutions $(r, a, b, c)$ to the above such that $\frac{1}{b}+\frac{1}{c}>\frac{1}{a}$, and $a\le b\le c\le 9$. We do this by doing casework on the value of $r$. Since $r$ is a positive integer, $r\ge 1$. Since $a\le b\le c\le 9$, $\frac{1}{r}\ge \frac{1}{3}$, so $r\le3$. The only possible values for $r$ are 1, 2, and 3. Case 1: $r=1$ For this case, we can't have $a\ge 4$, since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ would be too small. When $a=3$, we must have $b=c=3$. When $a\le2$, we would have $\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}$, which doesn't work. Hence this case only yields one valid solution $(1, 3, 3, 3)$ Case 2: $r=2$ For this case, we can't have $a\ge 7$, for the same reason as in Case 1. When $a=6$, we must have $b=c=6$. When $a=5$, we must have $b=5, c=10$ or $b=10, c=5$. Regardless, $10$ appears, so it is not a valid solution. When $a\le4$, $\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}$. Hence, this case also only yields one valid solution $(2, 6, 6, 6)$ Case 3: $r=3$ The only possible solution is $(3, 9, 9, 9)$, and clearly it is a valid solution. Hence the only valid solutions are $(1, 3, 3, 3), (2, 6, 6, 6), (3, 9, 9, 9)$, and our answer is $\boxed{\textbf{(B) }3}$

Let $a$ be the number of balls that are both striped and red, $x$ be the number of balls that are striped but blue, and $y$ be the number of balls that are red but dotted. Then there must be $6-a-x-y$ balls that are dotted and blue. Let $A$ be the event, "the ball Frida selects is red", and $B$ be the event, "the ball Frida selects is striped". $A$ and $B$ are independent if and only if $\frac{a}{6}=\frac{a+x}{6}\cdot\frac{a+y}{6}$, i.e., \[6a=(a+x)(a+y)\] Before we continue, I'd like to clarify that the sample space $S$, under which we are computing probability that $A$ and $B$ are independent, consists of $4^6$ total outcomes, each equally likely. Notice that once given $a, x, y$, there are ${6\choose a}{6-a \choose x}{6-a-x \choose y}=\frac{6!}{a!x!y!(6-a-x-y)!}$ corresponding outcomes. Now, it remains to find all nonnegative integer solutions $(a, x, y)$ to the above equation such that $a+x+y<6$, add up all the corresponding outcomes, and then divide by $4^6$. We can find the solutions relatively easily by doing casework on $a=0, 1, 2, 3, 4, 5, 6$. Long story short, one finds the solutions to be \[(0, 0, 0), (0, 0, 1)\cdots (0, 0, 6)\] \[(0, 1, 0), (0, 2, 0)\cdots (0, 6, 0)\] \[(1, 5, 0), (1, 0, 5), (1, 2, 1), (1, 1, 2)\] \[(2, 4, 0), (2, 0, 4) (2, 2, 1), (2, 1, 2)\] \[(3, 3, 0), (3, 0, 3)\] \[(4, 2, 0), (4, 0, 2)\] \[(5, 1, 0), (5, 0, 1)\] \[(6, 0, 0)\] There are little tricks we can use to make the process of adding up the corresponding outcomes faster. First, we notice that the first, second, and last row of solutions contributes $2^7=128$ outcomes. Then we can take advantage of the symmetry between $x, y$ to add up the rest of the outcomes. In the end we get \[128+2(6+180+15+180+20+15+6)=972\] Hence, the probability is $\frac{972}{4^6}=\frac{243}{2^{10}}$. Therefore, the answer is $\boxed{\textbf{(A) } 243}$