AMC12 2025 B

$\frac{350}{20}$ is $17.5$. We round down to find the maximum he can make (he cannot brew half of a cup), so the answer is $\boxed{\text{(B) }17}$.

By a modulo $10$ argument, the ones digits repeat with period $10$ in the following order: \[1,4,9,6,5,6,9,4,1,0\] The sum of the numbers can be verified to be $45$. There are $202$ periods that occur from $1$ to $2025$, and there are five extra numbers, those being $1,4,9,6,5$, corresponding to $2021,2022,2023,2024,2025$. The sum of these numbers is $25$. Hence, the total is \[202\cdot 45+25=9090+25=\boxed{\textbf{(D)}~9115}\]

We find that $i(i-3)=-1-3i$ and $(i-1)(i-2)=-1-3i+2=1-3i$, so i(i−1)(i−2)(i−3)=i(i−3)⋅(i−1)(i−2)=(−1−3i)(1−3i)=−(1+3i)(1−3i)=−(12−(3i)2)=−(1+9)=(D) −10

By definition of bases, $\underline{a}~\underline{b}$ base seven is equal to $7a+b$, and $\underline{b}~\underline{a}$ base nine is equal to $9b+a$. Therefore, we must have $7a+b=9b+a$, or $6a=8b$, or $3a=4b$. But in base seven, we must have $a,b<7$. Testing cases yields $a=4,b=3$ as the only solution. Their sum is $\boxed{\textbf{(A)}~7}$. The first equation comes from the following idea. In base $10$, a two-digit number can be represented as $10$ times the tens digit plus the units digit, or $10^1 \cdot a + 10^0 \cdot b$. If we insert the base numbers into this expression for $\underline{a}~\underline{b}$ and $\underline{b}~\underline{a}$, we get $7^1 \cdot a + b = 9^1 \cdot b + a$. The rest of the solution is above.

Let $x+y = a$. Then we have the equation \[35x + 22a = 400.\] Because the other two terms are divisible by $5$, $22a$ must be divisible by $5$ too. Specifically, $a$ is divisible by $5$. Let $a=5b$ and substitute: \[35x + 110b = 400\] \[7x + 22b = 80.\] After some analysis, we find that if $b=3$, $x = 2$. In fact, this is the only solution with positive integer solutions. Therefore, $a = x+y = 5 \cdot 3 = \boxed{\textbf{(E) }15}$.

The problem is essentially telling us that $\underline A~\underline B~\underline B.\underline B~\underline A$ is divisible by $36$, so it is divisible by $9$. Then, by the divisibility by $9$ condition, \[A+B+B+B+A=2A+3B\equiv 0\pmod{9}\] This means that $A$ must be divisible by $3$, or otherwise $2A+3B$ would not be divisible by $3$ and thus $9$. Since $A\ne0$, we must have one of $A=3,6,9$. But $A$ must be even or else the entire number would not be even and thus would not be divisible by $4$. Hence $A=6$. Then, $2\cdot 6+3B\equiv 0\pmod{9}$, so $4+B\equiv 0\pmod{3}$, and thus $B\equiv 2\pmod{3}$. This yields $B=2,5,8$, of which $B=5$ is the only number that allows $\underline A~\underline B~\underline B.\underline B~\underline A$ to be divisible by $4$. Thus the answer is $6+5=\boxed{\textbf{(C)}~11}$.

By logarithm rules, write ∑n=2255log2⁡(1+1n)(log2⁡n)(log2⁡(n+1))=∑n=2255log2⁡(n+1n)(log2⁡n)(log2⁡(n+1))=∑n=2255log2⁡(n+1)−log2⁡n(log2⁡n)(log2⁡(n+1))=∑n=2255(1log2⁡n−1log2⁡(n+1)) But this telescopes to $\frac{1}{\log_{2}2}-\frac{1}{\log_{2}256}=\frac{1}{1}-\frac{1}{8}=\boxed{\textbf{(C)}~\frac{7}{8}}$.

Because every coefficient of our polynomial $f(x)$ is an integer, the conjugate of $4+\sqrt{5}$, $4-\sqrt{5}$ is also a root. Let the third and final root be $r_3$. Thanks to Vieta's Formulas, we know that the sum of the roots is \[(r_3) + (4+\sqrt{5}) + (4 - \sqrt{5}) = 5\] \[r_3 = -3.\] We now calculate $f(1) = (1+3)(1-4-\sqrt{5})(1-4+\sqrt{5})$. $(1-4-\sqrt{3})(1-4+\sqrt{3}) = (-3-\sqrt{5})(-3+\sqrt{5}) = 9 - 5 = 4.$ So $f(1) = 4 \cdot 4 = 16.$ Hence $f(1) = 16 = 1-5 + a + b$. Therefore, $a + b = \boxed{20}$.

We wish to find $6^{6^6}\pmod{100}$ and find the tens digit. By the Chinese Remainder Theorem, it suffices to find $6^{6^6}\pmod{4}$ and $6^{6^6}\pmod{25}$ and find the solution to the system. We know that $6^{6^6}\equiv 0\pmod{4}$, so we just need to find the remainder modulo $25$. By Euler’s Totient Theorem, $6^{6^6}\equiv 6^r\pmod{25}$, where $r\equiv 6^6\pmod{\varphi(25)=20}$. Hence we need to find $6^6\pmod{20}$. Now we again use the Chinese Remainder Theorem. We know that $6^6\equiv 0\pmod{4}$ and $6^6\equiv 1^6\equiv 1\pmod{5}$. Testing cases yields $6^6\equiv 16\pmod{20}$. Therefore, we know that $6^{6^6}\equiv 6^{16}\pmod{25}$. Now we can simplify: \[6^{6^6}\equiv 6^{16}\equiv (6^4)^4\equiv 1296^4\equiv (-4)^4\equiv 256\equiv 6\pmod{25}\] Solving the system $6^{6^6}\equiv 6\pmod{25}$ and $6^{6^6}\equiv 0\pmod{4}$ yields $6^{6^6}\equiv 56\pmod{100}$. Hence the answer is $\boxed{\textbf{(C)}~5}$.

Without loss of generality, let $\triangle ABC$ have side-lengths $AB=2, BC=2\sqrt{3},$ and $AC=4.$ Let $D$ be the foot of the perpendicular from $B$ to $\overline{AC}, \ E$ be the midpoint of $\overline{AB},$ and $F$ be the intersection of $\overline{CE}$ and $\overline{BD}.$ Note that $\triangle ADB$ and $\triangle BDC$ are both $30^\circ{-}60^\circ{-}90^\circ$ triangles. From the side-length ratio, we get $AD=1$ and $DC=3.$ We obtain the following diagram: From here, we will proceed with mass points. Throughout this solution, we will use $W_P$ to denote the weight of point $P.$ Let $W_C=1.$ Since $3AD=DC,$ it follows that $W_A=3$ and $W_D=W_C+W_A=4.$ Since $AE=EB$ and $W_A=3,$ it follows that $W_B=3.$ Now we focus on $\overline{BD}:$ Since $W_B=3$ and $W_D=4,$ we have $\frac{DF}{FB}=\frac xy=\frac34.$ Therefore, the answer is \[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]

We will arrange the nine athletes in a line, with the first three being in the blue group, the next three being in the red group, and the last three being in the green group. Suppose the three tallest athletes are named $A, B, C$, in some order. We have $3!$ ways to choose which group each of these athletes can be in. We then have $3 \cdot 3 \cdot 3$ ways to determine which of the three positions in that group they can take. From here, we must distribute the remaining six athletes in the remaining six spaces, which can be done in $6!$ ways. There are therefore $6 \cdot 27 \cdot 6!$ favorable outcomes. There are also $9!$ total ways to arrange the athletes with no restrictions. Our answer is $\frac{6 \cdot 27 \cdot 6!}{9!} = \frac{6 \cdot 27}{9 \cdot 8 \cdot 7} = \boxed{\textbf{(C) }\frac{9}{28}}.$

The area cleaned by the wiper follows a thickened curve with vertical ends, where the curve is a $60^\circ$ arc with radius $3$. Since the sides are vertical, by Cavalieri's Principle, the area is equivalent to a rectangle with side lengths $3.5$ and the distance between the two vertical ends. Let $B'$ be the result of point $B$ at its leftmost point after it has swept $60^\circ$. Then $ABB'$ is equilateral so $BB' = AB = 3$. So the area is $3.5 \cdot 3 = \boxed{\textbf{(C) } 10.50}$. Note: if you aren't aware of Cavalieri's Principle, you can still recognize this by slicing the top curved part of the shape and moving it to the bottom, creating a rectangle with the same area as the original shape.

Create an arbitrary six slice circular diagram. The first slice has 3 options, and the second has 2, third has 2, fourth has 2, fifth has 2, and the sixth has only 1 color to be chosen. Understand that if X is the set containing all possible cases of the circle, then X has a correlation to another set, call it K, which contains all the colors. This can be quite easily sought as a rotation of one element in X that accurately maps to another element in X is considered the same. In simpler terms, to account for overlapping cases in X, one must divide by 2 (as two colorings that are rotated are considered different, and we don't want that as it defeats the purpose of "unique"). Then, multiplying out we have 3×24=48. Dividing by 2 to account for overlap in X, we get 48/2 or $\boxed{\textbf{(D) } 24}$

Since the mean of the first three terms is $2025$, their sum is $2025 * 3 = 6075$. Then, incorporating the fourth term makes the mean $2025-1=2024$, so their sum must be $2024 * 4 = 8096$, implying that the 4th term is $8096-6075=2021$ Doing the same for the 5th term, $2023 * 5= 10115$, 5th term is $10115-8096=2019$ Algebraically, we can model this pattern for the $k$th term = x as $(k-1)(2029-k) + x = (k)(2028-k)$ $2029k-2029-k^2+k+x=2028k-k^2$ $x=2029-2k$ So the maximum $k\le n$ for which $x$ is positive is 1014, giving our answer $n=\boxed{\textbf{(B) }1014}$

Extend the edges pointing downwards to converge at a point $A$ to form a square pyramid. Consider 3 square pyramids, the large one formed by the top vertices of the original figure and $A$, the middle one formed by the medians running through the sides of the original figure and point $A$, and the smaller one formed by the bottom vertices of the original figure and point $A$. Note that all pyramids are similar since they are all essentially scaled by a certain factor. The median length is $\frac{3+1}{2}=2$ Using side length to volume ratios, find that the volumes must have ratios $1:8:27$ Then, you get that the ratio of the volume thus filled to the volume that we must fill is equivalent to $8-1:27-8 = 7:19$. Thus, it will take $\frac{19}{7}$ more time to fill the remaining volume giving us an answer of $\frac{19}{7} * 35 = \boxed{\textbf{(D) }95}$

When divided by $60$, $2025$ yields $33$, with a remainder of $45$. Since $33 \equiv 9 \pmod{24}$, the clock is now at $9:45$. At this time, the hour is $75\%$ over, so the hour hand is also $75\%$ of the way between $9$ and $10$. There are $360/12 = 30 ^\circ$ degrees between $9$ and $10$, and since the hour hand is $75\%$ of the way, it is $30^\circ \cdot \frac{3}{4} = \frac{45}{2}^\circ$ from the minute hand, which is pointing directly at $9$. We now just need to find $\tan{(45/2)}$. Let this be $x$. Using the tangent addition formula, we get that \[\tan{45} = \frac{x + x}{1 - x\cdot x}\] \[1 = \frac{2x}{1-x^2}\] \[x^2+2x-1 = 0.\] Using the Quadratic Formula, and knowing $x$ must be positive since $0 < x < 90$, we get $x = \tan{(45/2)}= \boxed{\sqrt{2}-1}$ Note: One could solve this quickly using the *tangent half angle identity*, which is $\tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1+\cos(\theta)} = \frac{1-\cos(\theta)}{\sin(\theta)}$ ~math660

Denote $1=\text{red}$, $2=\text{blue}$, $3=\text{yellow}$. We need $1\to 2\to 3\to 1$. WLOG place $1$ in the center $(0,0)$, $2$ on the left edge $(-1,0)$, $3$ on the top-left corner $(-1,1)$. \[\begin{bmatrix} 3 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\] $3$ must see $1$, so the top edge $(0,1)$ must also have $1$. Then, $1$ must see $2$, so the top-right corner $(1,1)$ becomes $2$, which must see $3$, so the right edge $(1,0)$ must have $3$. \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix}\] Now this bottom-right corner can vary either as $1$, $2$ or $3$. Cases on $(1,-1)$: If $1$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{bmatrix}\] but the 3 needs a 1 and does not have it, so there are $0$ cases. If $2$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ b & a & 2 \end{bmatrix}\] if $a=1$, $b$ can be $1$ or $3$. If $a=2$, $b=3$, but the 3 needs a 1 and can't get it. If $a=3$, $b=1,2$, so there are $4$ cases in total. If $3$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 2 & 1 & 3 \end{bmatrix}\] Since the 2 in the bottom left corner does not have a 3 nearby, there are $0$ cases. WLOG the center fixes a factor of $3$, so the answer is $4\cdot 3=\boxed{\textbf{(C) } 12}$.

Let the probability of a win be $p$ and the probability of a loss be $q$. If the first game is a win, then we must find the expected number of further games to get a loss, which will be $\frac{1}{q}$. In addition to the first game played, there will be $1+\frac{1}{q}$ games played. Therefore, the expected number of games needed to get a win on game $1$ and then a loss is $(p)\left(1+\frac{1}{q}\right).$ Similarly, if the first game is a loss, we need to the find the expected number of further games to get a win, which will be $\frac{1}{p}$. There will be a total of $1+\frac{1}{p}$ games played. Therefore, the expected number of games needed to get a loss on game $1$ and then a win is $(q)\left(1+\frac{1}{p}\right).$ The answer is \[(p)\left(1+\frac{1}{q}\right) + (q)\left(1+\frac{1}{p}\right)\] \[\left(\frac{1}{3}\right) \left(1+\frac{3}{2}\right) + \left(\frac{2}{3}\right)(1+3) = \frac{5}{6} + \frac{8}{3} = \boxed{\frac{7}{2}}.\]

Let's say $n$ is one of the numbers that got written twice in the same box. Suppose it is at column $x$ and row $y$. We will write an expression for $n$ in terms of $x$ and $y$ in two ways: from Horace's perspective and Vera's perspective. From Horace's perspective, $n = (y-1)(141) + x$. This is because there are $y-1$ full rows before $n$, and we then need $x$ more numbers to reach $n$. Similarly, Vera will say $n = (x-1)(91) + y$. We now have the Diophantine equation \[(y-1)(141) + x = (x-1)(91)+y\] \[141y-141+x = 91x-91+y\] \[140y=90x+50\] \[14y = 9x + 5\] One such solution is, of course, $x=y=1$. We find further solutions by adding $14$ to $x$ and $9$ to $y$. For example, the second solution is $(15,10)$. This continues until $(141,91)$ is reached. There are $\boxed{11}$ ordered pairs in this list.

We will solve this using states. Let $P_n$ be the probability of reaching $4$, given that you start from $n$. We want to find $P_0$. Of course, $P_4 = 1$. We also know that \[P_3 = \frac{1}{4}P_4 + \frac{1}{4}P_2\] \[P_2 = \frac{1}{4}P_3 + \frac{1}{4}P_1\] \[P_1 = \frac{1}{4}P_2 + \frac{1}{4}P_0\] \[P_0 = \frac{1}{2}P_1.\] Solving the system, we find that $P_1 = \frac{2}{97}$ and $P_0 = \boxed{\frac{1}{97}}.$

Let the angles between the $8$ and $9$ sides be $\theta_1$ and $\theta_2$. Relating the two triangles' areas, we get \[\frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_1} = \frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_2}\] \[\sin{\theta_1} = \sin{\theta_2}\] For this to be true, $\theta_1 + \theta_2 = 180.$ We will now apply Law of Cosines to get an expression for $s_1$ and $s_2$ in terms of $\theta_1$ and $\theta_2$. \[s_1^2 = 8^2 + 9^2 - (2)(8)(9)\cos(\theta_1)\] \[s_2^2 = 8^2 + 9^2 - (2)(8)(9)\cos(180-\theta_1).\] Noting that $\cos{\theta_1} = -\cos{(180 - \theta_1)}$, we can add the two equations to get that \[s_1^2 + s_2^2 = 290.\] We now need two perfect squares that add to $290$. After some analysis, we find $121 + 169 = 290$. Therefore, $s_1 = 11$ and $s_2 = 13$, so our answer is $11 + 13 = \boxed{24}.$ Note that the $\cos{\theta_1} = 1/6.$ Also, note that $289 + 1 = 290$, but $17$ and $1$ are not valid side lengths due to the Triangle Inequality.

Because we can factor $z$ from each vertex, the area of the triangle is equal to $|z|^2$ multiplied by the area of the triangle with vertices $2, 1+i, 1-i$. The vertical side of that triangle has length $2$ and the altitude to that side has length $1$, so the area is $1$. We now just need the maximum possible value of $|z|^2$. The equation $|4z-2| = 1$ can be rewritten as $|z-1/2| = 1/4$. Therefore, it exists as a circle with radius $1/4$ and center at $(1/2, 0)$ on the complex plane. The maximum possible magnitude of a complex number on this circle occurs with $\frac{3}{4}$, with magnitude $\frac{3}{4}$. We can justify this because the line connecting the origin and the complex number $\frac{3}{4}$ goes through the center of the circle. Our answer is then $\left(\frac{3}{4}\right)^2 = \boxed{\frac{9}{16}}.$

Notice that there are $z$ distinct remainders modulo $z$. However, if we let $R(x)$ denote the remainder when $2025x$ is divided by $z$, we know that by the problem condition, \[R(0)<R(1)<R(2)<\cdots<R(z-1)\] But there are only $z$ distinct remainders, and each of the $z$ terms above must correspond to a distinct remainder, so we must have $R(k)=k$ for all $k=0,1,\dots,z-1$. Then $2025k\equiv k\pmod{z}$, so $2024k\equiv 0\pmod{z}$. Since $k$ can vary, this is equivalent to $2024\equiv 0\pmod{z}$, or $z\mid 2024$. Therefore, the values $z$ that satisfy the property are the factors of $2024$, so we simply need to find the sum of the factors of $2024$ and subtract $1$ at the end since $z\ne 1$. But $2024=2^3\cdot 11\cdot 23$, so the sum of the factors minus $1$ is \[(1+2+4+8)(1+11)(1+23)-1=15\cdot12\cdot24-1=4320-1=\boxed{\textbf{(E)}~4319 }.\]

Let $f(x)=\sin(20\pi x)$ and $g(x)=\log_{20}(x)$. Note that $g$ passes through $\left(\frac{1}{20},-1\right)$ and $(1,0)$ and $(20,1)$; these are the extrema and midpoint of $f$. We want to find the number of intersections of $f$ and $g$. Let an occurrence of sine passing under the $x$-axis a down-dip, and similarly define an up-dip. We find that the period of $f$ is $\frac{1}{10}$, so between $x=\frac{1}{20}$ and $x=1$ the number of periods is $9.5$. The first period indeed counts, so effectively we have $10$ down-dips in this interval. Each down-dip contributes $2$ to the total, so we have $20$ total intersections. From $x=1$ to $x=20$, there are $190$ periods, each of which also contributes $2$ to the total due to the up-dips. Therefore, this case contributes $380$ points to the total. But $(1,0)$ is counted in both cases, so the total is $20+380-1=\boxed{\textbf{(D) } 399}$.

Let $\triangle ABC$ and center $O$ be such that $OA=1, OB=2, OC=3$. Suppose the length of the side of $ABC$ is $s$. Noticing that $OA + OB = OC$, we suspect that $OACB$ is a cyclic quadrilateral. If it was, we could apply Ptolemy's Theorem, which would say that \[CB \cdot OA + OB \cdot AC = OC \cdot AB\] \[s + 2s = 3s.\] Because Ptolemy's is true, $OACB$ is cyclic. Because it's cyclic, $\angle AOB = 180 - \angle BCA = 120$. Applying Law of Cosines on $AOB$, we get \[s^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos{120}\] \[s^2 = 1 + 4 +2 = \boxed{7}.\]