AMC10 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.

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

We can start by assigning the values x and y for both sides. Here is the equation representing the area: $x \cdot y = 2024$ Let's write out 2024 fully factorized. $2^3 \cdot 11 \cdot 23$ Since we know that $x^2 > (x+1)(x-1)$, we want the two closest numbers possible. After some quick analysis, those two numbers are $44$ and $46$. $\\44+46=90$ Now we multiply by $2$ and get $\boxed{\textbf{(B) }180}.$ Solution by IshikaSaini.

We can factor the expression as \[7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).\] Note that $57=19\cdot3$, this expression is actually divisible by 19. The answer is $\boxed{\textbf{(A) } 0}$.

The factors of $42$ are $1, 2, 3, 6, 7, 14, 21, 42$. Multiply the unit digits to get $\boxed{\textbf{(D) } 6}$

If $\frac{a+b+c}{3} = 0$, that means $a+b+c=0$, and $(a+b+c)^2=0$. Expanding that gives \[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc\] If $\frac{a^2+b^2+c^2}{3} = 10$, then $a^2+b^2+c^2=30$. Thus, we have \[30 + 2ab + 2ac + 2bc = 0\] Arithmetic will give you that $ab + bc + ac = -15$. To find the arithmetic mean, divide that by 3, so $\frac{ab + bc + ac}{3} = \boxed{\textbf{(A) }-5}$

Let $AB = CD$ have length $b$ and let the altitude of the parallelogram perpendicular to $\overline{AD}$ have length $h$. The area of the parallelogram is $bh$ and the area of $\triangle ABE$ equals $\frac{(b/2)(h)}{2} = \frac{bh}{4}$. Thus, the area of quadrilateral $BCDE$ is $bh - \frac{bh}{4} = \frac{3bh}{4}$. We have from $AA$ that $\triangle CBF \sim \triangle AEF$. Also, $CB/AE = 2$, so the length of the altitude of $\triangle CBF$ from $F$ is twice that of $\triangle AEF$. This means that the altitude of $\triangle CBF$ is $2h/3$, so the area of $\triangle CBF$ is $\frac{(b)(2h/3)}{2} = \frac{bh}{3}$. Then, the area of quadrilateral $CDEF$ equals the area of $BCDE$ minus that of $\triangle CBF$, which is $\frac{3bh}{4} - \frac{bh}{3} = \frac{5bh}{12}$. Finally, the ratio of the area of $CDEF$ to the area of triangle $CFB$ is $\frac{\frac{5bh}{12}}{\frac{bh}{3}} = \frac{\frac{5}{12}}{\frac{1}{3}} = \frac{5}{4}$, so the answer is $\boxed{\textbf{(A) } 5:4}$.

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 think of this problem like boxes. First start with 9. We see that we can arrange the groups of people into the 9 boxes. We take 9 people of different languages and arrange them in each of the boxes. This means we have $100 - 9 \times 9 = 19$ people remaining. We then take 18 people of different language (as they can be put in a pair and still qualify) and put them in the boxes. This gives us 1 person left over. Now, this is where MAA wants you to choose 10, but one can quickly see that the last person can be put into any group of three (except a group that includes their language) and still qualify whilst maintaining all the constraints. Therefore our answer is $\boxed{\textbf{(A) } 9}$. This solution is a basic introduction/summary of the Pigeonhole Principle

Note that $\sqrt{1183}=13\sqrt7$. Since $x$ and $y$ are positive integers, and $\sqrt{x}+\sqrt{y}=\sqrt{1183}$ we can represent each value of $\sqrt{x}$ and $\sqrt{y}$ as the product of a positive integer and $\sqrt7$. Let's say that $\sqrt{x}=m\sqrt7$ and $\sqrt{y}=n\sqrt7$, where $m$ and $n$ are positive integers. This implies that \[x+y=(\sqrt{x})^2+(\sqrt{y})^2=7m^2+7n^2=7(m^2+n^2)\] and that $m+n=13$. WLOG, assume that ${m}\geq{n}$. It is not hard to see that $x+y$ reaches its minimum when $m^2+n^2$ reaches its minimum. We now apply algebraic manipulation to get that \[m^2+n^2=(m+n)^2-2mn\] Since $m+n$ is determined, we now want $mn$ to reach its maximum. Since $m$ and $n$ are positive integers, we can use the AM-GM inequality to get that: $\frac{m+n}{2}\geq{\sqrt{mn}}$. When $mn$ reaches its maximum, $\frac{m+n}{2}={\sqrt{mn}}$. This implies that $m=n=\frac{13}{2}$. However, this is not possible since $m$ and $n$ and integers. Under this constraint, we can see that $mn$ reaches its maximum when $m=7$ and $n=6$. Therefore, the minimum possible value of $x+y$ is $7(m^2+n^2)=7(7^2+6^2)=\boxed{\textbf{(B)}595}$ A similar method is to take $y=1183-26\sqrt{7x}-x^2$, then noting $x=7a^2$ and bashing to find the value of a where x is closest to y.

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).

Consider the numbers as $1,0,1,0,...,1,0$. Note that the number of odd integers is monotonously decreasing. We need to get rid of $1012$ $0$'s, so we either add or multiply four $0$s together to get $1012\rightarrow 253 \rightarrow 63+1=64 \rightarrow 16 \rightarrow 4 \rightarrow 1.$ To get rid of the final $0$, we need to consume three other $1$'s to result in one $1$. Thus the answer is $1012-2=\boxed{\textbf{(A) } 1010 }$,

Intuitive solution: We perform casework based on how many snails tie. Let's say we're dealing with the following snails: $A,B,C,D,E$. $5$ snails tied: All $5$ snails tied for $1$st place, so only $1$ way. $4$ snails tied: $A,B,C,D$ all tied, and $E$ either got $1$st or last. ${5}\choose{1}$ ways to choose who isn't involved in the tie and $2$ ways to choose if that snail gets first or last, so $10$ ways. $3$ snails tied: We have $ABC, D, E$. There are $3! = 6$ ways to determine the ranking of the $3$ groups. There are $5\choose2$ ways to determine the two snails not involved in the tie. So $6 \cdot 10 = 60$ ways. $2$ snails tied: We have $AB, C, D, E$. There are $4! = 24$ ways to determine the ranking of the $4$ groups. There are $5\choose{3}$ ways to determine the three snails not involved in the tie. So $24 \cdot 10 = 240$ ways. $1$ snail tied: This is basically just every snail for a place, so $5! = 120$ ways. The answer is $1+10+60+240+120 = \boxed{\text{(D) }431}$.

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 $\boxed{2}$.

Let's look at zero slope first. All lines of such form will be expressed in the form $y=k$, where $k$ is some real number. If $k$ is an integer, the line passes through infinitely many lattice points. One such example is $y=1$. If $k$ is not an integer, the line passes through $0$ lattice points. One such example is $y=1.1$. So we have $2$ cases. Let's now look at the second case. The line has slope $\frac{p}{q}$, where $p$ and $q$ are relatively prime integers. The line has the form $y = \frac{p}{q}x + m$. If the line passes through lattice point $(a,b)$, then the line must also pass through the lattice point $(a+kq, b+kp)$, where $a,b,k$ are all integers. Therefore, the line can pass through infinitely many lattice points but it cannot pass through exactly $1$ or $2$. The line can pass through $0$ lattice points, such as $y=x+0.5$. This contributes $2$ more cases. If the line has an irrational slope, it can never pass through more than $2$ lattice points. We prove this using contradiction. Let's say the line passes through lattice points $(a,b)$ and $(c,d)$. Then the line has slope $\frac{d-b}{c-a}$, which is rational. However, the slope of the line is irrational. Therefore, the line can pass through at most $1$ lattice point. One example of this is $y=\sqrt{2}x$. This line contributes $2$ final cases. Our answer is therefore $\boxed{6}$.

Let $A_R, A_L, B_R, B_L, C_R, C_L$ denote the shoes. There are $6$ ways to choose the first shoe. WLOG, assume it is $A_R$. We have $A_R,$ __, __, __, __, __. $~~~~~$ Case $1$: The next shoe in line is $A_L$. We have $A_R, A_L,$ __, __, __, __. Now, the next shoe in line must be either $B_L$ or $C_L$. There are $2$ ways to choose which one, but assume WLOG that it is $B_L$. We have $A_R, A_L, B_L,$ __, __, __. $~~~~~ ~~~~~$ Subcase $1$: The next shoe in line is $B_R$. We have $A_R, A_L, B_L, B_R,$ __, __. The only way to finish is $A_R, A_L, B_L, B_R, C_R, C_L$. $~~~~~ ~~~~~$ Subcase $2$: The next shoe in line is $C_L$. We have $A_R, A_L, B_L, C_L,$ __, __. The only way to finish is $A_R, A_L, B_L, C_L, C_R, B_R$. $~~~~~$ In total, this case has $(6)(2)(1 + 1) = 24$ orderings. $~~~~~$ Case $2$: The next shoe in line is either $B_R$ or $C_R$. There are $2$ ways to choose which one, but assume WLOG that it is $B_R$. We have $A_R, B_R,$ __, __, __, __. $~~~~~ ~~~~~$ Subcase $1$: The next shoe is $B_L$. We have $A_R, B_R, B_L,$ __, __, __. $~~~~~ ~~~~~ ~~~~~$ Sub-subcase $1$: The next shoe in line is $A_L$. We have $A_R, B_R, B_L, A_L,$ __, __. The only way to finish is $A_R, B_R, B_L, A_L, C_L, C_R$. $~~~~~ ~~~~~ ~~~~~$ Sub-subcase $2$: The next shoe in line is $C_L$. We have $A_R, B_R, B_L, C_L,$ __, __. The remaining shoes are $C_R$ and $A_L$, but these shoes cannot be next to each other, so this sub-subcase is impossible. $~~~~~ ~~~~~$ Subcase $2$: The next shoe is $C_R$. We have $A_R, B_R, C_R,$ __, __, __. The next shoe in line must be $C_L$, so we have $A_R, B_R, C_R, C_L,$ __, __. There are $2$ ways to finish, which are $A_R, B_R, C_R, C_L, A_L, B_L$ and $A_R, B_R, C_R, C_L, B_L, A_L$. $~~~~~$ In total, this case has $(6)(2)(1 + 2) = 36$ orderings. Our final answer is $24 + 36 = \boxed{\textbf{(A) } 60}$

Notice that the sum of radii of two circles tangent to each other will equal to the distance from center to center. Set the center of the big circle be at $(0,1).$ Since both circles are tangent to a line (in this case, $y=0$), the y-coordinates of the centers are just its radius. Hence, the center of the smaller circle is at $\left(x_2, \frac14\right)$. From the the sum of radii and distance formula, we have: \[1+\frac14 = \sqrt{x_2^2 + \left(\frac34\right)^2} \Rightarrow x_2 = 1.\] So, the coordinates of the center of the smaller circle are $(1, \frac14).$ Now, let $(x_3,r_3)$ be the coordinates of the new circle. Then, from the fact that sum of radii of this circle and the circle with radius $1$ is equal to the distance from the two centers, you have: \[\sqrt{(x_3-0)^2+(r_3-1)^2} = 1+r_3.\] Similarly, from the fact that the sum of radii of this circle and the circle with radius $\frac14$, you have: \[\sqrt{(x_3-1)^2+\left(r_3-\frac14\right)^2}= \frac14 + r_3.\] Squaring the first equation, you have: \[x_3^2+r_3^2-2r_3+1=1+2r_3+r_3^2 \Rightarrow 4r_3=x_3^2 \Rightarrow x_3=2\sqrt{r_3}.\] Squaring the second equation, you have: \[x_3^2-2x_3+1+r_3^2-\frac{r_3}{2}+\frac{1}{16}=\frac{1}{16}+\frac{r_3}{2}+r_3^2 \Rightarrow x_3^2-2x_3+1=r_3\] Plugging in from the first equation we have \[r_3-1=x_3^2-2x_3=4r_3-4\sqrt{r_3} \Rightarrow 3r_3-4\sqrt{r_3}+1=0 \Rightarrow (3\sqrt{r_3}-1)(\sqrt{r_3}-1)=0 \Rightarrow r_3=1, \frac19.\] Summing these two yields $\boxed{\frac{10}{9}}.$

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 $4$, $r=6-5+4=\boxed{\textbf{(A) }5}$.

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

First, we know that $P(2022)$ and $P(2024)$ must be integers since they are both divisible by $2$. Then let’s consider the remaining two numbers. Since they are not divisible by $2$, the result of the first term must be a certain number $+\frac{1}{2}$. The divisibility rule for $4$ states that the last two digits of the number must be divisible by $4$, and we know that the last two digits of ${2023}^2$ are the last two digits of ${23}^2$, which is $29$. And $29$ is $1$ greater than a multiple of $4$. Therefore, the result of the second term must be $\frac{1}{4}$. Similarly, the remaining two terms must each be $\frac{1}{8}$. Their sum is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1$, so $P(2023)$ and $P(2025)$ are also integers. Therefore, the answer is $\boxed{\textbf{(E) }4}$.

The $3$x$3$x$3$ block has side lengths of $3a, 3b, 3c$. The $2$x$2$x$7$ block has side lengths of $2b, 2c, 7a$. We can create the following system of equations, knowing that the new block has $1$ unit taller, deeper, and wider than the original: \[3a+1 = 2b\] \[3b+1=2c\] \[3c+1=7a\] Adding all the equations together, we get $b+c+3 = 4a$. Adding $a-3$ to both sides, we get $a+b+c = 5a-3$. The question states that $a,b,c$ are all relatively prime positive integers. Therefore, our answer must be congruent to $2 \pmod{5}$. The only answer choice satisfying this is $\boxed{E(92)}$.