problem_id stringlengths 32 32 | link stringlengths 75 84 | problem stringlengths 14 5.33k | solution stringlengths 15 6.63k | letter stringclasses 5
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d296defa043a641f71ebe6408e5de095 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_21 | Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ i... | Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftri... | C | 55 |
d296defa043a641f71ebe6408e5de095 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_21 | Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ i... | Let the length $AB=x, BC=y.$ Then, we have \begin{align*} (y+2x)^2\cdot\frac{\sqrt 3}{4}&=324\sqrt3, \\ (x+2y)^2\cdot\frac{\sqrt 3}{4}&=192\sqrt3. \end{align*} We get \begin{align*} y+2x&=36, \\ x+2y&=16\sqrt3. \end{align*} We want $3x+3y,$ and it follows that \[3x+3y=(y+2x)+(x+2y)=36+16\sqrt3.\] Finally, the answer is... | C | 55 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the... | Suppose the roommate took sheets $a$ through $b$ , or equivalently, page numbers $2a-1$ through $2b$ . Because there are $(2b-2a+2)$ numbers taken, \[\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.\] The first possible solution that comes to mi... | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the... | Suppose the smallest page number borrowed is $k,$ and $n$ pages are borrowed. It follows that the largest page number borrowed is $k+n-1.$
We have the following preconditions:
Together, we have \begin{align*} \frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&=19 \\ 1275-\frac{(2k+n-1)n}{2}&=19(50-n) \\ 2550-(2k+n-1)n&=38(50-n) \\ ... | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the... | Let $n$ be the number of sheets borrowed, with an average page number $k+25.5$ . The remaining $25-n$ sheets have an average page number of $19$ which is less than $25.5$ , the average page number of all $50$ pages, therefore $k>0$ . Since the borrowed sheets start with an odd page number and end with an even page numb... | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the... | Let $(2k-1)-2n$ be pages be borrowed, the sum of the page numbers on those pages is $(2n+2k+1)(n-k)$ while the sum of the rest pages is $1275-(2n+2k+1)(n-k)$ and we know the average of the rest is $\frac{1275-(2n+2k+1)}{50-2n+2k}$ which equals to $19$ ; multiply this out we got $950-38(n-k)=1275-(2n+2k+1)(n-k)$ and we ... | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the... | Let $c$ be the number of consecutive sheets Hiram’s roommate borrows, and let $b$ be the number of sheets preceding the $c$ borrowed sheets (i.e. if the friend borrows sheets $3$ $4$ , and $5$ , then $c=3$ and $b=2$ ).
The sum of the page numbers up till $b$ sheets is $1+2+3+\cdots + 2b=\frac{2b\cdot(2b+1)}{2} = b(2b+1... | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the... | The sum of all the page numbers is \[1+2+3+\cdots+50 = 1275.\] If we add the page numbers on each sheet, we get this sequence: \[3, 7, 11, \ldots, 99.\] So we can write the sum of the numbers on the first sheet that the roommate borrowed as $4n+3$ for some nonnegative integer, $n$ . If the roommate borrowed $k$ sheets,... | B | 13 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B... | Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid.
WLOG assume that A is in the center. \[\begin{tabular}{ c c c } ? & ? & ? \\ ? & A & ? \\ ? & ? & ? \end{tabular}\] In this configurat... | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B... | Without the loss of generality, we fix the top-left square with a red chip. We apply casework to its two adjacent chips:
Case (1): The top-center and center-left chips have different colors. [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cy... | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B... | We consider all possible configurations of the red chips for which rotations matter: [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, red); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fi... | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B... | $(1) \quad$ $\begin{tabular}{ c c c } R & G & ? \\ B & R & ? \\ ? & ? & ? \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } R & G & B \\ B & R & G \\ R & G & B \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } R & G & R \\ B & R & B \\ G & B & G \end{tabular}$ $\quad 3 \c... | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B... | Case (1) : We have a permutation of R, B, and G as all of the rows. There are $3!$ ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, $\frac{3!}{e} \approx 2$ , so there are two possi... | E | 36 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | Since $3\pi\approx9.42$ , we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\boxed{19}$ | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | $|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$ . Since $\pi$ is approximately $3.14$ $3\pi$ is approximately $9.42$ . We are trying to solve for $-9.42<x<9.42$ , where $x\in\mathbb{Z}$ . Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$ , for $x\in\mathbb{Z}$ . The number of integer values of $x$ is $9-(-9)+1=19$ . Therefore, the ans... | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | $3\pi \approx 9.4.$ There are two cases here.
When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$
When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$ . Dividing by $-1$ and flipping the sign, we get $x>-9.4.$
From case 1 and 2, we know that $-9.4 < x < 9.4$ . Since $x$ is an integer, we must have $x$ between $-9$ and $9$... | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | Looking at the problem, we see that instead of directly saying $x$ , we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions: $9+9=18.$ But what about zero? Even though z... | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | There are an odd number of integer solutions $x$ to this inequality since if any non-zero integer $x$ satisfies this inequality, then so does $-x,$ and we must also account for $0,$ which gives us the desired. Then, the answer is either $\textbf{(A)}$ or $\textbf{(D)},$ and since $3 \pi > 3 \cdot 3 > 9,$ the answer is ... | D | 19 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)... | Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, s... | C | 8 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)... | We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$ . Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{8}$ . ~samrocksnature | C | 8 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)... | Since there are an equal number of juniors and seniors on the debate team, suppose there are $x$ juniors and $x$ seniors. This number represents $25\% =\frac{1}{4}$ of the juniors and $10\%= \frac{1}{10}$ of the seniors, which tells us that there are $4x$ juniors and $10x$ seniors. There are $28$ juniors and seniors in... | C | 8 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)... | The amount of juniors must be a multiple of $4$ , since exactly $\frac{1}{4}$ of the students are on the debate team. Thus, we can immediately see that $\boxed{8}$ | C | 8 |
baf43db6a59c5d1245b61ddc16d9f589 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_4 | At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
$\textbf{(A)} ~23 \qquad\textb... | There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\tfrac{64}{2}=\boxed{32}$ ~Punxsutawney Phil | B | 32 |
5466a87961a99ab15febde232a84df47 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_5 | The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give $24$ , while the other two multiply to $30$ . What is the sum of the ages of Jonie's four cousins?
$\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad... | First look at the two cousins' ages that multiply to $24$ . Since the ages must be single-digit, the ages must either be $3 \text{ and } 8$ or $4 \text{ and } 6.$
Next, look at the two cousins' ages that multiply to $30$ . Since the ages must be single-digit, the only ages that work are $5 \text{ and } 6.$ Remembering ... | B | 22 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of ... | Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$ . The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$ . Therefore, the answer is $\boxed{76}$ | C | 76 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of ... | Suppose the morning class has $m$ students and the afternoon class has $a$ students. We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline & & & \\ [-2.5ex] \textbf{Morning} & m & 84 & 84m \\ \hline & & & \\ [-2.5ex] \textbf{Afternoo... | C | 76 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of ... | Of the average, $\frac{3}{3+4}=\frac{3}{7}$ of the scores came from the morning class and $\frac{4}{7}$ came from the afternoon class. The average is $\frac{3}{7}\cdot 84+\frac{4}{7}\cdot 70=\boxed{76}.$ | C | 76 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of ... | WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{76}.$ | C | 76 |
b829d9c71c7c89f24fa97941134e7dac | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_7 | In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$ . Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$
$\textbf{(A) }24\pi \qquad \textbf{(B) ... | Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ We construct the circles one by one:
The diagram below shows one possible configuration of the four circles: [asy] /* diagram made by samrocksnature, edited by MRENTHUSIASM */ pair A=(10,0); pair B=(-10,0); draw(A--B); filldra... | D | 65 |
4787a0177733cbd1e33350d998204a63 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_8 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row f... | In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $D$ and $E,$ respectively. Note that the numbers in the yellow cells are consecutive odd perfect squares, as we can prove by induction. [asy] /* Made by MRENTHUSIASM */... | A | 367 |
4787a0177733cbd1e33350d998204a63 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_8 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row f... | In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $C$ and $G,$ respectively. [asy] /* Made by MRENTHUSIASM */ size(11.5cm); fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,... | A | 367 |
4787a0177733cbd1e33350d998204a63 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_8 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row f... | From the full diagram below, the answer is $210+157=\boxed{367} [/asy] This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options. | A | 367 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ... | The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$
By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$... | D | 7 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ... | Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\cdot{i}$ .
A $90^\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\circ$ , and then translating the origin back to the point $(1,... | D | 7 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ... | Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$ .
If we let the original point be $(x, y)$ , then we can use that the starting point is $(1,5)$ to obtain two vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ .
We know that two vectors are perpendicular... | D | 7 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ... | Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$
Let the original point be $\langle x,y \rangle.$ From point $(1,5),$ we form the vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ that extend out from the initial point. If they are perpendicular, we know ... | D | 7 |
7def323ed742d23088be16a2c3447700 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_10 | An inverted cone with base radius $12 \mathrm{cm}$ and height $18 \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24 \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ... | The volume of a cone is $\frac{1}{3} \cdot\pi \cdot r^2 \cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi$
The volume of a cylinder is $\pi \cdot r^2 \cdot h$ so the volume of the water in the c... | A | 1.5 |
7def323ed742d23088be16a2c3447700 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_10 | An inverted cone with base radius $12 \mathrm{cm}$ and height $18 \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24 \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ... | The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has $\frac{1}{3}$ of the volume of the cylinder, and so the height is divided by $3$ . Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since $2^2 ... | A | 1.5 |
d597cbb2c0b40ad7fb8f0bc71fc68697 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_11 | Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perim... | Let the side lengths of the rectangular pan be $m$ and $n$ . It follows that $(m-2)(n-2) = \frac{mn}{2}$ , since half of the brownie pieces are in the interior. This gives $2(m-2)(n-2) = mn \iff mn - 4m - 4n + 8 = 0$ . Adding 8 to both sides and applying Simon's Favorite Factoring Trick , we obtain $(m-4)(n-4) = 8$ . S... | D | 60 |
d597cbb2c0b40ad7fb8f0bc71fc68697 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_11 | Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perim... | Obviously, no side of the rectangular pan can have less than $5$ brownies beside it. We let one side of the pan have $5$ brownies, and let the number of brownies on its adjacent side be $x$ . Therefore, $5x=2\cdot3(x-2)$ , and solving yields $x=12$ and there are $5\cdot12=60$ brownies in the pan. $64$ is the only choic... | D | 60 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\te... | We can start by setting up an equation to convert $\underline{32d}$ base $n$ to base 10. To convert this to base 10, it would be $3{n}^2+2n+d.$ Because it is equal to 263, we can set this equation to 263. Finally, subtract $d$ from both sides to get $3{n}^2+2n = 263-d$
We can also set up equations to convert $\underlin... | B | 11 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\te... | $32d$ is greater than $263$ when both are interpreted in base 10, so $n$ is less than $10$ . Some trial and error gives $n=9$ $263$ in base 9 is $322$ , so the answer is $9+2=\boxed{11}$ | B | 11 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\te... | We have \[3n^2 + 2n + d = 263\] \[3n^2 + 2n + 4 = 6^3 + 6^2 + 6d + 1\] Subtracting the 2nd from the 1st equation we get \begin{align*} d-4 &= 263 - (216 + 36 + 6d + 1) \\ &= 263 - 253 - 6d \\ &= 10 - 6d \end{align*} Thus we have $d=2.$ Substituting into the first, we have $3n^2 + 2n + 2 = 263 \Rightarrow 3n^2 + 2n - 2... | B | 11 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\te... | Find that $d=2$ using one of the methods above. Then we have that $3n^2 + 2n = 261$ . We know that $n$ is an integer, so we can solve the equation $n(2n+3)=261$ (this is guaranteed to have a solution if we did this correctly). The prime factorization of $261$ is $3^2 \cdot 29$ , so the corresponding factor pairs are $(... | B | 11 |
4258c79f8131051481eacc756d986a1b | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_14 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | [asy] size(8cm); pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9); draw(O--A--B); draw(O--R); draw(O--L); label("$A$", A, NE); label("$B$", B, N); label("$R$", R, NE); label("$L$", L, NE); label("$O$", O, S); label("$d$", O--A, W); label("$2d$", A--B, W); label("$r$", O--R, S); label("$r$", O--L, NW); ... | B | 6 |
4258c79f8131051481eacc756d986a1b | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_14 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | [asy] real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label("... | B | 6 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | We square $x+\frac{1}{x}=\sqrt5$ to get $x^2+2+\frac{1}{x^2}=5$ . We subtract 2 on both sides for $x^2+\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\frac{1}{x^4}=9$ so $x^4+\frac{1}{x^4}=7$ . We can factor out $x^7$ from our original expression of $x^{11}-7x^7+x^3$ to get that it is equal to $x^7(x^4-7+\frac{... | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | Multiplying both sides by $x$ and using the quadratic formula, we get $\frac{\sqrt{5} \pm 1}{2}$ . We can assume that it is $\frac{\sqrt{5}+1}{2}$ , and notice that this is also a solution the equation $x^2-x-1=0$ , i.e. we have $x^2=x+1$ . Repeatedly using this on the given (you can also just note Fibonacci numbers), ... | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | We can immediately note that the exponents of $x^{11}-7x^7+x^3$ are an arithmetic sequence, so they are symmetric around the middle term. So, $x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})$ . We can see that since $x+\frac{1}{x} = \sqrt{5}$ $x^2+2+\frac{1}{x^2} = 5$ and therefore $x^2+\frac{1}{x^2} = 3$ . Continuing from ... | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | We begin by multiplying $x+\frac{1}{x} = \sqrt{5}$ by $x$ , resulting in $x^2+1 = \sqrt{5}x$ . Now we see this equation: $x^{11}-7x^{7}+x^3$ . The terms all have $x^3$ in common, so we can factor that out, and what we're looking for becomes $x^3(x^8-7x^4+1)$ . Looking back to our original equation, we have $x^2+1 = \sq... | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | The equation we are given is $x+\tfrac{1}{x}=\sqrt{5}...$ Yuck. Fractions and radicals! We multiply both sides by $x,$ square, and re-arrange to get \[x^2+1=\sqrt{5}x \implies x^4+2x^2+1=5x^2 \implies x^4-3x^2+1=0.\] Now, let us consider the expression we wish to acquire. Factoring out $x^3,$ we have \[x^3\left(x^8-7x^... | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | Multiplying by x and solving, we get that $x = \frac{\sqrt{5} \pm 1}{2}.$ Note that whether or not we take $x = \frac{\sqrt{5} + 1}{2}$ or we take $\frac{\sqrt{5} - 1}{2},$ our answer has to be the same. Thus, we take $x = \frac{\sqrt{5} - 1}{2} \approx 0.62$ . Since this number is small, taking it to high powers like ... | B | 0 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)... | The divisibility rule of $15$ is that the number must be congruent to $0$ mod $3$ and congruent to $0$ mod $5$ . Being divisible by $5$ means that it must end with a $5$ or a $0$ . We can rule out the case when the number ends with a $0$ immediately because the only integer that is uphill and ends with a $0$ is $0$ whi... | C | 6 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)... | First, note how the number must end in either $5$ or $0$ in order to satisfying being divisible by $15$ . However, the number can't end in $0$ because it's not strictly greater than the previous digits. Thus, our number must end in $5$ . We do casework on the number of digits.
Case 1: $1$ digit. No numbers work, so $0$... | C | 6 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)... | Like solution 2, we can proceed by using casework. A number is divisible by $15$ if is divisible by $3$ and $5.$ In this case, the units digit must be $5,$ otherwise no number can be formed.
Case 1: sum of digits = 6
There is only one number, $15.$
Case 2: sum of digits = 9
There are two numbers: $45$ and $135.$
Case 3... | C | 6 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)... | An integer is divisible by $15$ if it is divisible by $3$ and $5$ . Divisibility by $5$ means ending in $0$ or $5$ , but since no digit is less than $0$ , the only uphill integer ending in $0$ could be $0$ , which is not positive. This means the integer must end in $5$
All uphill integers ending in $5$ are formed by pi... | C | 6 |
9971bd3158bbac13c80c951631a3e3de | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_17 | Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,2,3, \dots,10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- $11,$ Oscar-- $4,$ Aditi-- $7,$ Tyrone-- $16,$ Kim-- $17.$ Which ... | By logical deduction, we consider the scores from lowest to highest: \begin{align*} \text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \\ &\implies \text{Aditi is given cards 2 and 5.} \\ &\implies \text{Ravon is given cards 4 and 7.} && (\bigstar) \\ &\implies \text{Tyrone is given cards 6 and 1... | C | 4. |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then... | Let the lowest value be $L$ and the highest $G$ , and let the sum be $Z$ and the amount of numbers $n$ . We have $\frac{Z-G}{n-1}=32$ $\frac{Z-L-G}{n-2}=35$ $\frac{Z-L}{n-1}=40$ , and $G=L+72$ . Clearing denominators gives $Z-G=32n-32$ $Z-L-G=35n-70$ , and $Z-L=40n-40$ . We use $G=L+72$ to turn the first equation into ... | D | 36.8 |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then... | Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$ , and $k$ be the number of elements in S.
Then, $S=x+y+z$
First, when the greatest integer is removed, $\frac{S-x}{k-1}=32$
When the smallest integer is also removed, $\frac{S-x-y}{k-2}=35$
When the greatest i... | D | 36.8 |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then... | We should plug in $36.2$ and assume everything is true except the $35$ part. We then calculate that part and end up with $35.75$ . We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$ . The answer is then $\boxed{36.8}$ . You can work backwards beca... | D | 36.8 |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then... | Let $S = \{a_1, a_2, a_3, \hdots, a_n\}$ with $a_1 < a_2 < a_3 < \hdots < a_n.$ We are given the following: \[{\begin{cases} \sum_{i=1}^{n-1} a_i = 32(n-1) = 32n-32, \\ \sum_{i=2}^n a_i = 40(n-1) = 40n-40, \\ \sum_{i=2}^{n-1} a_i = 35(n-2) = 35n-70, \\ a_n-a_1 = 72 \implies a_1 + 72 = a_n. \end{cases}}\] Subtracting th... | D | 36.8 |
4693b8f76efa4eac7ea2286add2491ed | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_20 | The figure is constructed from $11$ line segments, each of which has length $2$ . The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$ , where $m$ and $n$ are positive integers. What is $m + n ?$ [asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.4713... | [asy] /* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F... | D | 23 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perime... | We can set the point on $CD$ where the fold occurs as point $F$ . Then, we can set $FD$ as $x$ , and $CF$ as $1-x$ because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for $x$ , we get,
\[x^2 + \left(\frac{1}{3}\right)^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x ... | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perime... | Let the line we're reflecting over be $\ell$ , and let the points where it hits $AB$ and $CD$ , be $M$ and $N$ , respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line $\... | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perime... | Assume that E is the midpoint of $\overline{AB}$ . Then, $\overline{AE}=\frac{1}{2}$ and since $C'D=\frac{1}{3}$ $\overline{AC'}=\frac{2}{3}$ . By the Pythagorean Theorem, $\overline{EC'}=\frac{5}{6}$ . It easily follows that our desired perimeter is $\boxed{2}$ ~samrocksnature | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perime... | As described in Solution 1, we can find that $DF=\frac{4}{9}$ , and $C'F = \frac{5}{9}.$
Then, we can find we can find the length of $\overline{AE}$ by expressing the length of $\overline{EF}$ in two different ways, in terms of $AE$ . If let $AE = a$ , by the Pythagorean Theorem we have that $EC = \sqrt{a^2 + \left(\fr... | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perime... | Draw a perpendicular line from $\overline{AB}$ at $E$ , and let it intersect $\overline{DC}$ at $E'$ . The angle between $\overline{AB}$ and $\overline{EE'}$ is $2\theta$ , where $\theta$ is the angle between the fold and a line perpendicular to $\overline{AD}$ . The slope of the fold is $\frac{1}{3}$ because it is per... | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perime... | It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means: \[C' F + DF = CF + DF = CD = AD = 3 C'D \implies C' D : DF : C' F = 3 : 4 : 5.\]
For a similar triangle, the ratio of the perimeter to the larger leg is ... | A | 2 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac... | Let our denominator be $(5!)^3$ , so we consider all possible distributions.
We use PIE (Principle of Inclusion and Exclusion) to count the successful ones.
When we have at $1$ box with all $3$ balls the same color in that box, there are $_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3$ ways for the distributions to occur (... | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac... | As In Solution 1, the probability is \[\frac{\binom{5}{1}\cdot 5\cdot (4!)^3 - \binom{5}{2}\cdot 5\cdot 4\cdot (3!)^3 + \binom{5}{3}\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - \binom{5}{4}\cdot 5\cdot 4\cdot 3\cdot 2\cdot (1!)^3 + \binom{5}{5}\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{(5!)^3}\] \[= \frac{5\cdot 5\cdot (4!)^3 - 10\cd... | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac... | Use complementary counting.
Denote $T_n$ as the total number of ways to put $n$ colors to $n$ boxes by 3 people out of which $f_n$ ways are such that no box has uniform color. Notice $T_n = (n!)^3$ . From this setup we see the question is asking for $1-\frac{f_5}{(5!)^3}$ . To find $f_5$ we want to exclude the cases o... | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac... | WLOG fix which block Ang places into each box. (This can also be thought of as labeling each box by the color of Ang's block.) There are then $(5!)^2$ total possibilities.
As in Solution 1 , we use PIE. With $1$ box of uniform color, there are ${}_{5} C _{1} \cdot (4!)^2$ possibilities ( ${}_{5} C _{1}$ for selecting o... | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac... | $!n$ denotes the number of derangements of $n$ elements, i.e. the number of permutations where no element appears in its original position. Recall the recursive formula $!0 = 1, !1 = 0, !n = (n-1)(!(n-1)+{!}(n-2))$
We will consider the number of candidate boxes (ones that currently remain uniform-color) after each pers... | D | 471 |
b1b6cc5abf05ac743aa96a4df5c90c24 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_23 | A square with side length $8$ is colored white except for $4$ black isosceles right triangular regions with legs of length $2$ in each corner of the square and a black diamond with side length $2\sqrt{2}$ in the center of the square, as shown in the diagram. A circular coin with diameter $1$ is dropped onto the square ... | To find the probability, we look at the $\frac{\text{success region}}{\text{total possible region}}$ . For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least $\frac{1}{2}$ , as it's the radius of the coin. This implies the ... | C | 68 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move... | We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position.
First we note that symmetrical positions are P-positions, as the second player can win by mirrorin... | B | 6,2,1 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move... | $(6,1,1)$ can be turned into $(2,2,1,1)$ by Arjun, which is symmetric, so Beth will lose.
$(6,3,1)$ can be turned into $(3,1,3,1)$ by Arjun, which is symmetric, so Beth will lose.
$(6,2,2)$ can be turned into $(2,2,2,2)$ by Arjun, which is symmetric, so Beth will lose.
$(6,3,2)$ can be turned into $(3,2,3,2)$ by Arjun,... | B | 6,2,1 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move... | Consider the following, much simpler game.
Arjun and Beth can each either take 1 or 2 bricks from the right-hand-side of a continuous row of initially $n$ bricks.
It is easy to see that for $n$ a multiple of 3, Beth can "mirror" whatever Arjun plays: if he takes 1, she takes 2, and if he takes 2, she takes 1. With this... | B | 6,2,1 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move... | We can start by guessing and checking our solutions. Let's start with $A$ $(6,1,1)$ . Arjun goes first, and he has a winning strategy. His strategy is to cut the 6 block in half, so whatever Beth does, Arjun will copy. If we see this in practice, Arjun has made the block in to $(2,2,1,1)$ . If Beth takes 1, he will tak... | B | 6,2,1 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positi... | I know that I want about $\frac{2}{3}$ of the box of integer coordinates above my line. There are a total of 30 integer coordinates in the desired range for each axis which gives a total of 900 lattice points. I estimate that the slope, m, is $\frac{2}{3}$ . Now, although there is probably an easier solution, I would t... | E | 85 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positi... | As the procedure shown in the Solution 1, the lower bound of $m$ is $\frac{2}{3}.$ Here I give a more logic way to show how to find the upper bound of $m.$ Denote $N=\sum_{x=1}^{30}(\lfloor mx \rfloor)$ as the number of lattice points in $S$
$N = \lfloor m \rfloor+\lfloor 2m \rfloor+\lfloor 3m \rfloor+\cdots+\lfloor 30... | E | 85 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positi... | It's easier to calculate the number of lattice points inside a rectangle with vertices $(0,0)$ $(p,0)$ $(p,q)$ $(0,q)$ . Those lattice points are divided by the diagonal $y = \frac{p}{q} \cdot x$ into $2$ halves. In this problem the number of lattice points on or below the diagonal and $x \ge 1$ is
$(p+1)(q+1)$ is the ... | E | 85 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positi... | The lower bound of $m$ is $\frac23 = \frac{20}{30}$ . Inside the rectangle with vertices $(0,0)$ $(30,0)$ $(30,20)$ $(0, 20)$ and diagonal $y = \frac23 x$ , there are $(30-1)(20-1) = 551$ lattice points inside, not including the edges. There are $9$ lattice points on diagonal $y = \frac23 x$ inside the rectangle, $551 ... | E | 85 |
2a070373c59e4bfe58a7696b2c94dfd6 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_2 | The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$ . What is the average of $a$ and $b$
$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$ | The arithmetic mean of the numbers $3, 5, 7, a,$ and $b$ is equal to $\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15$ . Solving for $a+b$ , we get $a+b=60$ . Dividing by $2$ to find the average of the two numbers $a$ and $b$ gives $\frac{60}{2}=\boxed{30}$ | C | 30 |
8ac11081c69023fb62de5acf84ec676b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_3 | Assuming $a\neq3$ $b\neq4$ , and $c\neq5$ , what is the value in simplest form of the following expression? \[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) ... | If $x\neq y,$ then $\frac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression: \[\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c... | A | 1 |
8ac11081c69023fb62de5acf84ec676b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_3 | Assuming $a\neq3$ $b\neq4$ , and $c\neq5$ , what is the value in simplest form of the following expression? \[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) ... | At $(a,b,c)=(4,5,6),$ the answer choices become
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } {-}\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}$
and the original expression becomes \[\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{1}.\] ~MRENTHUSIASM | A | 1 |
0abc4ab743b94eebaf4c89a5e03ba9ed | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_4 | A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $$0.50$ per mile, and her only expense is gasoline at $$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\te... | Since the driver travels $60$ miles per hour and each hour she uses $2$ gallons of gasoline, she spends $$4$ per hour on gas. If she gets $$0.50$ per mile, then she gets $$30$ per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is $\boxed{26}$ .
~mathsmiley | E | 26 |
0abc4ab743b94eebaf4c89a5e03ba9ed | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_4 | A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $$0.50$ per mile, and her only expense is gasoline at $$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\te... | The driver is driving for $2$ hours at $60$ miles per hour, she drives $120$ miles. Therefore, she uses $\frac{120}{30}=4$ gallons of gasoline. So, total she has $$0.50\cdot120-$2.00\cdot4=$60-$8=$52$ . So, her rate is $\frac{52}{2}=\boxed{26}$ .
~sosiaops | E | 26 |
b40e02d22d26944e2dfe4334c7a3569c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_5 | What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$ | Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
Case 1:
The equation yields $x^2-12x+34=2$ , which is equal to $(x-4)(x-8)=0$ . Therefore, the two values for the positive case is $4$ and $8$
Case 2:
Similarly, taking the nonpositive case for the value inside the... | C | 18 |
b40e02d22d26944e2dfe4334c7a3569c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_5 | What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$ | We have the equations $x^2-12x+32=0$ and $x^2-12x+36=0$
Notice that the second is a perfect square with a double root at $x=6$ , and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is $-(-12)$ or $12$ $12+6=\boxed{18}$ | C | 18 |
b40e02d22d26944e2dfe4334c7a3569c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_5 | What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$ | Completing the square gives \begin{align*} \left|(x-6)^2-2\right|&=2 \\ (x-6)^2-2&=\pm2. \hspace{15mm}(\bigstar) \end{align*} Note that the graph of $y=(x-6)^2-2$ is an upward parabola with the vertex $(6,-2)$ and the axis of symmetry $x=6;$ the graphs of $y=\pm2$ are horizontal lines.
We apply casework to $(\bigstar):... | C | 18 |
12029877d9f3f90d24f177f850e5b4de | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_6 | How many $4$ -digit positive integers (that is, integers between $1000$ and $9999$ , inclusive) having only even digits are divisible by $5?$
$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$ | The units digit, for all numbers divisible by 5, must be either $0$ or $5$ . However, since all digits are even, the units digit must be $0$ . The middle two digits can be 0, 2, 4, 6, or 8, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is one choice for the units digit, 5 choices for ... | B | 100 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\te... | Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$ , and the common sum is $\frac{50}{5}=\boxed{10}$ | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\te... | Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \boxed{10}$ as our answer.
~Baolan | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\te... | Taking the average of the first and last terms, $-10$ and $14$ , we have that the mean of the set is $2$ . There are 5 values in each row, column or diagonal, so the value of the common sum is $5\cdot2$ , or $\boxed{10}$ .
~Arctic_Bunny, edited by KINGLOGIC | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\te... | Let us consider the horizontal rows. Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ . Since the sum of the 25 numbers used is $-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\boxed{10}$ .
~cw357 | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\te... | The mean of the set of numbers is $(14-10) \div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)
One row of the square would be \[\square \square 2 \square \square\]
Adding the numbers would be
\[0, 1, 2, 3, 4\]
with a sum of $\boxed{10}$ | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\te... | If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\boxed{10}$ .
~MC413551 | null | 10 |
bc958ce2333164b4e62bdad7adc0b369 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_8 | What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$ | Looking at the numbers, you see that every set of $4$ has $3$ positive numbers and 1 negative number. Calculating the sum of the first couple sets gives us $2+10+18...+394$ . Clearly, this pattern is an arithmetic sequence. By using the formula we get $\frac{2+394}{2}\cdot 50=\boxed{9900}$ | B | 9900 |
bc958ce2333164b4e62bdad7adc0b369 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_8 | What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$ | Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of $50\cdot (-2)=-100$ . Summing the odd numbers is equivalent to summing the first 100 odd numbers, which is equal to $100^2=10000$ . Adding these two, we obtain the answer of $\boxed{9900}$ | B | 9900 |
bc958ce2333164b4e62bdad7adc0b369 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_8 | What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$ | We can break this entire sum down into $4$ integer bits, in which the sum is $2x$ , where $x$ is the first integer in this bit. We can find that the first sum of every sequence is $4x-3$ , which we plug in for the $50$ bits in the entire sequence is $1+2+3+\cdots+50=1275$ , so then we can plug it into the first term of... | B | 9900 |
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