AMC10 2025 A

At $2{:}30$, Andy is $8$ miles ahead. For every hour that they both travel, Betsy gains $4$ miles on Andy. Therefore, it will take her $2$ more hours to be the same distance from the starting point as Andy, which occurs at $\boxed{\textbf{(E) } 4{:}30}$.

We are given $0.2(10) = 2$ pounds of cashews in the first box. Denote the pounds of nuts in the second nut mix as $x.$ \[5 + 0.2x = 0.4(10 + x)\] \[0.2x = 1\] \[x = 5\] Thus, we have $5$ pounds of the second mix. \[0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) }4}\]

You can split the problem into two cases: Case $1$: The two sides with equal length are both smaller than $2025$, which means that they range from $1013$ to $2024$. There are $1012$ such cases. Case $2$: There are two sides of length $2025$, so the last side must be in the range $1$ to $2025$. There are $2025$ such cases. Keep in mind, an equilateral triangle also counts as an isosceles triangle, since it has at least 2 sides of equal length. Therefore, the total number of cases is $1012 + 2025 = \boxed{\text{(D) }3037}$

When Ash joins a team, the team's average is pulled towards his age. Let $A$ be Ash's age and $N$ be the number of people on the student team. This means that there are $15-N$ people in the teacher team. Let us write an expression for the change in the average for each team. The students originally had an average of $12$, which became $14$ when Ash joined, so there was an increase of $2$. The term $A-12$ represents how much older Ash is compared to the average of the students'. If we divide this by $N+1$, which is the number of people on the student team when Ash joins, we get the average change per team member once Ash is added. Therefore, \[\frac{A-12}{N+1} = 2.\] Similarly, for teachers, the average was originally $55$, which decreased by $3$ to become $52$ when Ash joined. Intuitively, $55-A$ represents how much younger Ash is than the average age of the teachers. Dividing this by the expression $(15-N)+1$, which is the new total number of people on the teacher team, represents the average change per team member once Ash joins. We can write the equation \[\frac{55-A}{16-N} = 3.\] To solve the system, multiply equation (1) by $N+1$, and similarly multiply equation (2) by $16-N$. Then add the equations together, canceling $A$, leaving equation $43=50-N$. From this we get $N=7$ and $A= \boxed{28}.$

Group the numbers by their hill pattern: $(12)(1232)(123432)(12345432)...$ The maximums of each hill occur at terms $n = 2, 5, 10, 17...$ These terms correspond to maximums of $2, 3, 4,...$ Let $a$ be the maximum at term $N$. Since the sum of the first $x$ odd numbers is $x^2$ we have $1 + (a-1)^2 = N.$ So for example, if a = $4,$ then $N = 10,$ telling us that the peak of the hill with maximum $4$ occurs at the $10$th term. Now, we know $2025 = 45^2,$ so let $a = 46.$ Then $N = 2026,$ so the $2026$th term is $46.$ Then the $2025$th term must be $\boxed{\text{(E) }45}.$

Assume you have a diagram in front of you. Because each angle of the triangle is trisected, we have 9 $20^\circ$ angles. Using a side of the triangle as a base, we have an isosceles triangle with two $20^\circ$ angles. Using this we can show that the third angle is $140^\circ$. Following that, we use the principle of vertical angles to show that one angle of the hexagon is $140^\circ$. And with rotational symmetry, three. The average of all 6 angles has to be $120^\circ$, so the answer is $\boxed{\textbf{(C) }100}$

Use synthetic division to find that the remainder of $x^{3}+x^{2}+ax+b$ is $a+b+2$ when divided by $x-1$ and $2a+b+12$ when divided by $x-2$. Now, we solve \[\begin{cases} a+b+2=4 \\ 2a+b+12 = 6 \\ \end{cases}\] This ends up being $a=-8$, $b=10$, so $b-a=10-(-8)=18

We first number all the statements: 1) At least one of these statements is true. 2) At least two of these statements are true. 3) At least two of these statements are false. 4) At least one of these statements is false. We can immediately see that statement 4 must be true, as it would contradict itself if it were false. Similarly, statement 1 must be true, as all the other statements must be false if it were false, which is contradictory because statement 4 is true. Since both 1 and 4 are true, statement 2 has to be true. Therefore, statement 3 is the only false statement, making the answer $\boxed{\text{(B) }1}$.

Substitute $1 - a$ for $x$ and set this expression equal to $25.$ The problem boils down to finding how many real roots \[100(1-a)^3 - 300(1-a)^2 + 200(1-a) = 25\] has. We further simplify this expression and create a function $f(x):$ \[f(x) = -100a^3 + 100a - 25\] Using Descarte's Rule of Signs we get: Sign changes for $f(x)$ (possible number of positive roots): 2 \[f(-x) = +100a^3 - 100a - 25\] Sign changes for $f(-x)$ (possible number of negative roots): 1 Possibilities for roots: 1) $2$ positive roots, $1$ negative root 2) $0$ positive roots, $1$ negative root, $2$ imaginary roots So which one is it? We know if the function changes sign between an interval, then a root exists in that interval (Intermediate Value Theorem). From $a = 0$ to $\frac{1}{2},$ the function changes sign because $f(0) = -25$ while $f(\frac{1}{2}) = +\frac{25}{2}$, so a positive root exists. This eliminates the second possibility, implying that there must be $2$ positive and $1$ negative roots. So the answer is $2 + 1 = \boxed{\textbf{(C) } 3}.$

Notice that the size of the smaller semicircle is not specified, and there is no additional information that hints at any specific size for it. Hence, we can shrink the small semicircle until its area is arbitrarily small and negligible, leaving us with a semicircle with a diameter of $16$. The area of the semicircle is given by $\frac{\pi r^2}{2}$, so we have $r=\frac{16}{2}=8\Rightarrow$$A=\frac{\pi(8)^2}{2}=\boxed{\text{(C) }32\pi}$

Since the geometric sequence is more restrictive, we can test values for the common ratio until we find one that works. After a few tests, we find that a common ratio of $4$ results in the geometric sequence $1,4,16,64,$ so the arithmetic sequence is $1,22,43,64.$ The answer is $4+16+64+22+43=\boxed{\text{(E) }149}.$ A more generalized solution is as follows. Let the common difference of the arithmetic sequence be $d$, and the common ratio of the geometric sequence be $r.$ Hence, the two sequences are $1,1+d,1+2d,1+3d$ and $1,r,r^2,r^3.$ Since $z=1+3d=r^3,$ the arithmetic sequence is $1,1+d,1+2d,r^3.$ Since $d=\dfrac{1+3d-1}{3}=\dfrac{r^3-1}{3}$ is a positive integer, we seek the smallest $r\neq1$ such that $r^3-1=(r-1)(r^2+r+1)$ is divisble by $3,$ so the smallest $r$ is $4$. The rest follows like above.

The only two digits that are neither prime nor even are $1$ and $9$. We split this problem into cases based on the number of $2$s. This is because $2$ is both a prime number and an even number. Case 1: For this case, there are no $2$s. There are $4$ choices for where the even digit goes, and $3$ choices for what the even digit is. There are then $3$ choices for where the prime digit goes, and $3$ choices for what the prime number digit is. The last two spots have $2$ choices each, either $1$ or $9$. This gives a total of $4\cdot 3^3 \cdot 2^2 = 432$ options for this case. Case 2: For this case, there is one $2$, with it being the only even prime. There are $4$ choices for where $2$ goes, and $2$ choices for the other three digits each. This case gives a total of $2^3\cdot 4 = 32$ options. Hence, the answer is $432 + 32 = \boxed{\textbf{(D) }464}$

Let the side length of the largest square be $a,$ so it has area $a^2.$ Hence, the second-largest square has area $a^2k^2,$ the third-largest has $a^2k^4,$ and so on. It follows that the total shaded area is \[a^2-a^2k^2+a^2k^4-a^2k^6+...=a^2(1-k^2+k^4-k^6+...)=a^2\dfrac{1}{1+k^2}.\] The ratio of the area of the shaded region to that of the original square is then \[\dfrac{a^2\frac{1}{1+k^2}}{a^2}=\dfrac{1}{1+k^2}=\dfrac{64}{100}\] \[\implies 64+64k^2=100\implies k^2=\dfrac{36}{64}\implies k=\boxed{\text{(D) }\dfrac{3}{4}}.\] We can just let $a=1$ because the question deals with ratios, meaning that there wouldn't be a loss of generality if we let the side length equal some value, getting the same answer $\boxed{D}$.

Pair two students together and put them adjacent on any two seats. There are 6 ways to do this. Considering one of these cases (they are all the same), there are 4 seats left, in which we wish to arrange the teachers together. So pair the teachers together and put them adjacent on any two seats not already occupied by two of the students. There are 3 ways to do this. For all 6 cases, there are 6×3=18 favorable outcomes. The number of ways to arrange the 2 students and 2 teachers is $\binom{6}{2} \times \binom{4}{2} = 90$. Our probability is 1890= $\boxed{\textbf{(B) } \frac 15}$

Because $ABEF$ is a rectangle, $\angle ABC=90°$. We are given that $\angle ADE=90°$, and since $\angle ECD=\angle ACB$ by vertical angles, $\triangle ECD \sim \triangle ACB$. Let $AC=x$. By the Pythagorean Theorem, $CB=\sqrt{x^2-1}$. Since $AF=BE=7$, $EC=7-\sqrt{x^2-1}$. Because $AC=x$ and $AD=5$, $CD=5-x$. By similar triangles, \[\frac{7-\sqrt{x^2-1}}{x}=\frac{5-x}{\sqrt{x^2-1}}\]. Cross-multiplying, we get that \[7\sqrt{x^2-1}-x^2+1=5x-x^2\], so \[7\sqrt{x^2-1}=5x-1\]. We square both sides, and this is simply a quadratic in $x$: \[24x^2+10x-50=0\], which has a positive root $x=\frac{5}{4}$. Since $AB=1$, we can plug this into the Pythagorean Theorem, with $\frac{5}{4}$ being the hypotenuse, to get $BC=\frac{3}{4}$, and ${1}\cdot \frac{\frac{3}{4}}{2}$ to equal $[ABC]= \boxed{\textbf{(A)} \frac{3}{8}}$

We have three coins and three jars. Each coin is placed independently and randomly into one of the jars. Let $M$ be the maximum number of coins in any jar. We want to compute the expected value of $M$. Step 1: Count total outcomes Each coin has $3$ choices, so the total number of equally likely placements is $3^3 = 27$. Step 2: Casework on the maximum number of coins Case 1: $M = 1$. This occurs when each jar has exactly one coin. There are $3! = 6$ assignments of coins to jars. Hence, $\Pr(M=1) = \frac{6}{27} = \frac{2}{9}$. Case 2: $M = 3$. This occurs when all three coins fall into the same jar. There are $3$ jars to choose from, so $\Pr(M=3) = \frac{3}{27} = \frac{1}{9}$. Case 3: $M = 2$. This occurs when one jar has $2$ coins, another jar has $1$ coin, and the last jar has $0$ coins. We can choose which jar gets $2$ coins in $3$ ways, which jar gets $1$ coin in $2$ ways, and which $2$ coins out of the $3$ go into the jar with two coins, so we multiply by $\dbinom{3}{2}$, which is just $3$ (note we don't have to do this for the earlier cases because for case $2$, all $3$ coins go into one jar, and for case $1$, the factorial already accounts for that). Therefore, there are $3^2 \cdot 2 = 18$ outcomes. Thus, $\Pr(M=2) = \frac{18}{27} = \frac{2}{3}$. Step 3: Compute the expected value The expected value of $M$ is $\mathbb{E}[M] = 1 \cdot \frac{2}{9} + 2 \cdot \frac{2}{3} + 3 \cdot \frac{1}{9}$. Converting everything to ninths, we have $\mathbb{E}[M] = \frac{2}{9} + \frac{12}{9} + \frac{3}{9} = \frac{17}{9}$. Hence, the expected number of coins in the jar with the most coins is$\boxed{\text{(D) }\frac{17}{9}}$ . $Pr$=Probability of $\mathbb{E}$=Expected value As described in the solution, there are $3^3=27$ ways of distributing the coins into the $3$ jars. Because there are $6$ ways for M=1 and $3$ ways for M=3, there are $27 - 6 - 3 = 18$ ways for $M=2$.

The problem statement implies that $N$ divides both $273436-16=273420$ and $272760-15=272745$. We want to find $N > 16$ that satisfies both of these conditions. Hence, we can just find the greatest common divisor of the two numbers. $\gcd(273420,272745)=\gcd(675,272745)=\gcd(675,45)=45$ by the Euclidean Algorithm, so the answer is $\boxed{\text{(E) }4}.$ Note: If an integer $a$ is congruent to an integer $b$ modulo a positive integer $c$, denoted by $a \equiv b \pmod{c}$, this means that $c$ divides the difference $a-b$

We will need to determine the sum of the reciprocals of the roots. To find the sum of the reciprocals of the roots $p,q$ of the quadratic $ax^2+bx+c$, we use Vieta's formulas. Recall that $p+q = -b/a$ and $pq = c/a$. Therefore, \[\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{\frac{-b}{a}}{\frac{c}{a}} = \frac{-b}{a} \cdot \frac{a}{c} = \frac{-b}{c},\] which doesn't depend on $a$. The sum of the reciprocals of the roots of the quadratic $x^2-4x-3$ is $\frac{-(-4)}{-3} = -4/3.$ The same is true for every quadratic in the form $ax^2-4x-3$. The sum of all the reciprocals of the roots of $ax^2+bx+c$ is $2025 \cdot \left(-\frac{4}{3}\right).$ Because we have $2025$ quadratics, there are $2 \cdot 2025 = 4050$ total roots. Our answer is $\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} =\boxed{(B)\ -\dfrac{3}{2}}$.

Consider the polynomial $f(x) = -x^2+3x+1.$ When we multiply this polynomial by $x+1,$ we are essentially doing the operation given in the problem (When we multiply $p(x)$ by $x+1,$ a term of degree $d$ in the yielded expression is the sum of $1\cdot(\text{degree d})$ and $x\cdot(\text{degree d-1})$ in $p(x)$ This effect is visible in Pascal's Triangle). So, if we let the coefficients of $f(x)$ be the zero row of the array, then the $n^{th}$ row is just the coefficients of $f(x)(x+1)^n.$ The next thing to note is that the sum of the coefficients in any polynomial $p(x)$ is just $p(1).$ Therefore, the sum of the entries in the $n^{th}$ row of the array is $f(1)(1+1)^n=3\cdot2^n.$ Letting this equal $12288,$ we get $n=12.$ We are looking for the $3^{rd}$ term in the $12^{th}$ row. The $12^{th}$ row is given by the coefficients of $f(x)(x+1)^{12}=(-x^2+3x+1)(x+1)^{12}.$ Since the degree of the resulting expression is $14,$ the third term in the row is just the coefficient of $x^{12}$ in the expression, which is $-\dbinom{12}{10}+3\dbinom{12}{11}+1=\boxed{\textbf{(A) }-29}.$

Let the silo center be $O$, let the point MacDonald is situated at be $A$, and let the point $20$ meters west of the silo center be $B$. $ABO$ is then a right triangle with side lengths $15, 20,$ and $25$. Let the point $20$ meters east of the silo center be $C$, and let the point McGregor is at be $D$ with $CD=g>0$. Also let $AD$ be tangent to circle $O$ at $E$. Extend $BC$ and $AD$ to meet at point $F$. This creates $3$ similar triangles, $\triangle ABF\sim \triangle DCF \sim \triangle OEF$. Let the distance between point $C$ and $F$ be $x$. The similarity ratio between triangles $ABF$ and $DCF$ is then $\frac{longer\;leg}{shorter\;leg} = \frac{40+x}{15} = \frac{x}{g}$ This is currently unsolvable so we bring in triangle $OEF$. The hypotenuse of triangle $OEF$ is $OF=20+x$ and its shorter leg is the radius of the silo $=10$. We can then establish a second similarity relationship between triangles $OEF$ and $ABF$ with $\frac{shorter\; leg}{hypotenuse}=\frac{10}{20+x}=\frac{15}{AF}$ Now we find the hypotenuse of $ABF$ in terms of $x$ using the Pythagorean theorem. $AF^2=15^2+(40+x)^2$. Which simplifies to $AF^2=225+1600+80x+x^2=1825+80x+x^2$ So $AF=\sqrt{x^2+80x+1825}$ Plugging back in we get $\frac{10}{20+x}=\frac{15}{\sqrt{x^2+80x+1825}}$. Now we can begin to break this down by multiplying both sides by both denominators. $10(\sqrt{x^2+80x+1825})=15(20+x)$ Dividing both sides by $5$ then squaring yields, $4x^2+320x+7300=9x^2+360x+3600$ This furthermore simplifies to $5x^2+40x-3700=0$ At which point we can divide off a $5$ and then apply the quadratic formula on $x^2+8x-740=0$ which we take the positive root of. \[x = \frac{-8+\sqrt{64+2960}}{2} = \frac{-8+\sqrt{3024}}{2} =\frac{-8+\sqrt{144 \cdot 21}}{2}.\] Simplifying yields that $x=6\sqrt{21}-4$ Then to solve for $g$ we simply plug $6\sqrt{21}-4$ back into the first similarity ratio to get $\frac{36+6\sqrt{21}}{15}=\frac{6\sqrt{21}-4}{g}$ Multiply both sides by $15g$ and dividing by $36+6\sqrt{21}$ will let us solve for $g=\frac{15(6\sqrt{21}-4)}{36+6\sqrt{21}}$ and after rationalizing the denominator we get $\frac{20\sqrt{21}-75}{3}$. $20+21+75+3=\boxed{\textbf{(A)}~119}$

Let our subset be $\{11,12,13,...,20\}.$ If we add any element from the set $\{1,2,3,...,10\}$ to our current subset, we will have to remove at least one element from our subset. Hence, the maximum size of our subset is $\boxed{\text{(C) }10}$.

Descartes' Circle Formula (curvatures $k_i = \frac{1}{r_i}$) \[k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1k_2 + k_2k_3 + k_3k_1}.\] For radii 1, 2, 3 we have \[k_1 = 1,\quad k_2 = \frac{1}{2},\quad k_3 = \frac{1}{3}.\] Compute the sum and the square-root term \[k_1+k_2+k_3 = \frac{11}{6},\qquad k_1k_2+k_2k_3+k_3k_1 = 1.\] Therefore \[k_4 = \frac{11}{6} \pm 2.\] Choose the plus sign for the small circle tangent externally to the three given circles \[k_4 = \frac{11}{6} + 2 = \frac{23}{6}, \qquad r_4 = \frac{1}{k_4} = \boxed{\textbf{(B) }\frac{6}{23}}.\]

Let $CD \perp AB$ with foot $D$. Right triangles $ACD$ and $BCD$ give $AC^2 = AD^2+CD^2$, $BC^2 = BD^2+CD^2$, $AC^2-BC^2 = AD^2-BD^2 =(AD-BD)(AD+BD)$. Since $AD+BD = AB = 80$ and $AC^2-BC^2 = 75^2-45^2 = 3600$, we get the equation $3600 = 80(AD-BD)$. This equation simplifies to $45 = AD - BD$. We can solve the system of equations $AD + BD = 80$ and $AD - BD = 45$ easily via elimination, and we get $AD = \frac{125}{2}$, $BD = \frac{35}{2}$. $CD^2 = AC^2-AD^2 = 75^2-\left(\frac{125}{2}\right)^2 = \frac{6875}{4}$, $CD = \frac{25\sqrt{11}}{2}$. By Angle Bisector Theorem, $\frac{DP}{PC} = \frac{DB}{BC} = \frac{\frac{35}{2}}{45} = \frac{7}{18}$, $PC = CD-DP$ thus, $18DP = 7(CD-DP)$, $25DP = 7CD$, $DP = \left(\frac{7}{25}\right)CD = \left(\frac{7}{25}\right)\left(\frac{25\sqrt{11}}{2}\right) = \frac{7\sqrt{11}}{2}$. $BP^2 = BD^2+DP^2 = \left(\frac{35}{2}\right)^2+\left(\frac{7\sqrt{11}}{2}\right)^2 = \frac{1225}{4}+\frac{49(11)}{4} = \frac{1764}{4} = 441$, thus $BP = \boxed{\text{(D) }21}.$

To satisfy the conditions, a $\textit{fair}$ integer must have no digit be a local minimum. Let's say we have $n$ distinct digits, with each digit being a number from $1$ to $9$. To create a $\textit{fair}$ integer, we begin by placing the largest digit. For the second-largest digit, we can either place this digit to the right or to the left of the string already created. We have these $2$ options for the third-largest digit, and so on. Therefore, there are $2^{n-1}$ valid permutations to create a $\textit{fair}$ integer. We must also choose which digits will be in the permutation. If you are creating an $n$-digit long $\textit{fair}$ integer, there are $9\choose{n}$ ways to pick which digits will be in the number. Therefore, for each $n \in \{1,2,\dots, 9\}$, the number of fair integers of length $n$ is: \[\binom{9}{n} \cdot 2^{n-1}.\] Summing over all $n$: \[\sum_{n=1}^9{\binom{9}{n} \cdot 2^{n-1}}=\frac{1}{2}\left(\sum_{n=0}^9{\binom{9}{n}}2^n -1\right)=\frac{1}{2}\left((1+2)^9 -1 \right) = \frac{1}{2}(19682) = \boxed{9841}.\] Note that the Binomial Theorem was used to equate \[\sum_{n=0}^9{\binom{9}{n}}2^n = (1+2)^9.\]

Say WLOG that $AB$ is the top side of the square, and the square is of side length 1. Let us say that the midpoint of $AB$ is $M$, while the midpoint of $CD$ is $Q$. Drawing a vertical line to split the square in half, we notice that if $P$ is to the left of the line, $AP < BP$, and if P is to the right of the line, $AP > BP$. Also, drawing a quarter circle of radius 1 from point $A$, we can split the area into points P for which $AP < AB$ and $AP > AB$. Because of our constraints, there are 2 cases: Case 1: $AB > AP > BP$ In this case, $P$ will be to the right of the vertical line and inside of the quarter circle. Let us say that the intersection of the vertical line and quarter circle is $N$. The distance from $N$ to $AD$ is 1/2, and we can say that $\angle BAN$ is $60^\circ$. Sector $BAN$ of circle $A$ would therefore have an area of $\frac{\pi}{6}$. Because $\triangle AMN$ is a 30-60-90 triangle, the area of $AMN$ is $\frac{\sqrt{3}}{8}$. The probability of case 1 happening should then be $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$. Case 2: $AB < AP < BP$ In this case, $P$ will be to the left of the vertical line and outside of the quarter circle. Knowing that the quarter circle's area is $\frac{\pi}{4}$, we can subtract the probability of Case 1 happening to get the chance that $P$ is on the left of the vertical line and in circle $A$. Doing this would give $\frac{\pi}{12}+\frac{\sqrt{3}}{8}$. To get the probability of Case 2 happening, we can subtract this from the area of rectangle $AMQD$. This would give us $\frac{1}{2}-\frac{\pi}{12}-\frac{\sqrt{3}}{8}$. Adding both cases, we get the total probability as $\frac{1}{2}+\frac{\pi}{12}-\frac{\sqrt{3}}{4} = \frac{6+\pi-3\sqrt{3}}{12}$. Formatting this gives us $6+1+3+3+12 = \boxed{\textbf{(A) } 25}$.