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a4ba286964e674a8d5c3afd3bb18197c | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$ | Notice that Alice has skipped the numbers $3n-1$ for $n=1,2,3,...,333$ .
Namely, \[3\cdot1-1,3\cdot2-1,3\cdot3-1,...,3\cdot333-1\] Thus the numbers that Barbara skips are \[3\cdot2-1,3\cdot5-1,3\cdot8-1,...\] or in a more general expression, $3(3n-1)-1$ for $n=1,2,3,...$ .
Namely, \[3\cdot(3\cdot1-1)-1,3\cdot(3\cdot2-1)-1,3\cdot(3\cdot3-1)-1,...\] Repeating the pattern until George, we have the first number he says, \[3(3(3(3(3(3\cdot1-1)-1)-1)-1)-1)-1= \boxed{365}\] In addition, note that the second number George says exceeds $1000$ | C | 365 |
a4ba286964e674a8d5c3afd3bb18197c | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$ | Similar to Solution 1 we find that the only number that George can say must leave a remainder of $2$ when divided by $3$ , and that it must also leave a remainder of $5$ when divided by $9$ . Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that our answer is $\boxed{365}$ | C | 365 |
a4ba286964e674a8d5c3afd3bb18197c | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$ | Every integer from 1 to 1000 can be written in the form $t_n+1$ in base 10, where $t_n$ is a trinary integer with no more than 7 significant digits. The important insight is that person $1\le{k}\le6$ will not say $n$ if and only if the $k$ th digit from the right of $t_n$ is 1. Therefore, the last 6 digits of $t_n$ must be 1, as the first 6 people never said the number. The only options for $t_n$ are thus $2111111$ $1111111$ , and $0111111$ . But, since George only said one number, the first two must have been too big for it to be $\le1000$ . Our answer is therefore $111111_3+1=\frac{3^6-1}{3-1}+1=364+1=\boxed{365}$ | C | 365 |
5c4905829c041883b52b8ff65a8532ca | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_25 | Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$ | There must be four rays emanating from $X$ that intersect the four corners of the square region. Depending on the location of $X$ , the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is $100$ -ray partitional (let this point be the bottom-left-most point).
We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining $96$ rays are divided among the other two triangular sectors, each sector with $48$ rays, thus dividing these two sectors into $49$ triangles of equal areas.
Let the distance from this corner point to the closest side be $a$ and the side of the square be $s$ . From this, we get the equation $\frac{a\times s}{2}=\frac{(s-a)\times s}{2}\times\frac1{49}$ . Solve for $a$ to get $a=\frac s{50}$ . Therefore, point $X$ is $\frac1{50}$ of the side length away from the two sides it is closest to. By moving $X$ $\frac s{50}$ to the right, we also move one ray from the right sector to the left sector, which determines another $100$ -ray partitional point. We can continue moving $X$ right and up to derive the set of points that are $100$ -ray partitional.
In the end, we get a square grid of points each $\frac s{50}$ apart from one another. Since this grid ranges from a distance of $\frac s{50}$ from one side to $\frac{49s}{50}$ from the same side, we have a $49\times49$ grid, a total of $2401$ $100$ -ray partitional points. To find the overlap from the $60$ -ray partitional, we must find the distance from the corner-most $60$ -ray partitional point to the sides closest to it. Since the $100$ -ray partitional points form a $49\times49$ grid, each point $\frac s{50}$ apart from each other, we can deduce that the $60$ -ray partitional points form a $29\times29$ grid, each point $\frac s{30}$ apart from each other. To find the overlap points, we must find the common divisors of $30$ and $50$ which are $1, 2, 5,$ and $10$ . Therefore, the overlapping points will form grids with points $s$ $\frac s{2}$ $\frac s{5}$ , and $\frac s{10}$ away from each other respectively. Since the grid with points $\frac s{10}$ away from each other includes the other points, we can disregard the other grids. The total overlapping set of points is a $9\times9$ grid, which has $81$ points. Subtract $81$ from $2401$ to get $2401-81=\boxed{2320}$ | C | 2320 |
5c4905829c041883b52b8ff65a8532ca | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_25 | Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$ | Position the square region $R$ so that the bottom-left corner of the square is at the origin. Then define $s$ to be the sidelength of $R$ and $X$ to be the point $(rs, qs)$ , where $0<r,q<1$
There must be four rays emanating from $X$ that intersect the four corners of $R$ . The areas of the four triangles formed by these rays are then $A_1=\frac{qs\times s}{2}$ $A_2=\frac{(s-rs)\times s}{2}$ $A_3=\frac{(s-qs)\times s}{2}$ , and $A_4=\frac{rs\times s}{2}$
If a point is $n$ -ray partitional, then there exist positive integers $a, b, c, d$ such that $a+b+c+d=n$ and $\frac{A_1}{a}=\frac{A_2}{b}=\frac{A_3}{c}=\frac{A_4}{d}$ . Substituting in our formulas for $A_1$ $A_2$ $A_3$ , and $A_4$ and canceling equal terms, we get \[\frac{q}{a}=\frac{1-r}{b}=\frac{1-q}{c}=\frac{r}{d}.\]
Taking $\frac{q}{a}=\frac{1-q}{c}$ and solving for $q$ , we get $q=\frac{a}{a+c}$ , and taking $\frac{1-r}{b}=\frac{r}{d}$ and solving for $r$ , we get $r=\frac{d}{b+d}$ . Finally, from $\frac{q}{a}=\frac{r}{d}$ , we have $qd=ar$ $\Leftrightarrow$ $\frac{ad}{a+c}=\frac{ad}{b+d}$ $\Leftrightarrow$ $a+c=b+d$
So for a point $X$ to be $100$ -ray partitional, $a+b+c+d=100$ , so $a+c=b+d=50$ $X$ must then be of the form $\left(\frac{d}{50}s, \frac{a}{50}s\right)$ . Since $X$ is in the interior of $R$ $a$ and $d$ can be any positive integer from $1$ to $49$ (with $b$ and $c$ just equaling $50-d$ and $50-a$ , respectively). Thus, there are $49\times 49=2401$ points that are $100$ -ray partitional.
However, the problem asks for points that are not only $100$ -ray partitional but also not $60$ -ray partitional. Points that are $60$ -ray partitional are of the form $\left(\frac{m}{30}s, \frac{n}{30}s\right)$ , where $m$ and $n$ are also positive integers. We count the number of points $\left(\frac{d}{50}s, \frac{a}{50}s\right)$ that can also be written in this form. For a given $d$ $\frac{d}{50}=\frac{m}{30}$ if and only if $m=\frac{3}{5}d$ , and likewise with $a$ and $n$ . We can then see that a point is both $100$ -ray partitional and $60$ -ray partitional if and only if $a$ and $d$ are both divisible by $5$ . There are $9$ integers between $1$ and $49$ that are divisible by $5$ , so out of our $2401$ points that are $100$ -ray partitional, $9\times 9=81$ are also $60$ -ray partitional.
Our answer then is just $2401-81=\boxed{2320}$ | C | 2320 |
5c4905829c041883b52b8ff65a8532ca | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_25 | Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$ | For the sake of simplicity, let $R$ be a $60 \times 60$ square and set the bottom-left point as the origin. Then, $R$ has vertices: \[(0,0), (60,0), (60,60), (0,60).\]
Now, let a point in the square have coordinates $(x, y).$
In order for the point to be $100-$ ray partitional, we must be able to make $100$ triangles with area $60^2/100 = 36.$ For it to not be $60$ -ray partitional, we cannot make $60$ triangles with area $60^2/60 = 60.$
When we draw such a triangle, it's base will always be on one of the sides of the square. If it is on the bottom side, then the height must be $y$ . So, the base of each triangle must be $\frac{72}{y}.$ So, there will be $\frac{60}{\frac{72}{y}} = \frac{60y}{72}$ total triangles on that side.
If the triangle is on the right side of the square, then the height has to be $60 - x.$ So, the base will be $\frac{72}{60-x}$ and there will be $\frac{60 (60-x)}{72}$ triangles.
If the triangle is on the top, the height will be $60-y$ , the base will be $\frac{72}{60-y}$ and there will be $\frac{60 (60-y)}{72}$ triangles along this side.
The triangles on the left will have height $x$ , base $\frac{72}{x}$ and $\frac{60x}{72}$ such triangles must exist.
Simplifying, we get $\frac{5y}{6}, \frac{5 (60-x)}{6}, \frac{5(60-y)}{6}, \frac{5x}{6}$ triangles on the respective sides of the square.
All of these numbers must be integers. Let $x = 60a$ and $y = 60b$ where $0 < a, b < 1.$
Then, the amounts become: \[50a, 50b, 50 - 50a, 50 - 50b.\]
As long as $50a$ and $50b$ are integers, $50 - 50a$ and $50 - 50b$ will also be integers.
For this to happen, $a$ and $b$ must be of the form $\frac{x}{50}$ where $1 \le x \le 49.$ So, we have a total of $49^2 = 2401$ points that are $100$ -ray partitional.
Now, we must calculate the number of $100$ - ray partitional points that are also $60$ -ray partitional.
Using a method similar to before, if a point is $60$ -ray partitional, then we must be able to make $30a, 30b, 30 - 30a, 30 -30b$ triangles on the different sides.
So, $30a$ and $30b$ must be integers. This means $a$ and $b$ have to be of the form $\frac{y}{30}$ where $0<y<30.$
If a point is both $100$ -ray partitional and $60$ -ray partitional, then it can be written as \[\frac{x}{50} = \frac{y}{30}\] . Note that whenever $x$ is divisible by $5$ , a $y$ will always exist to satisfy the above equation.
So, $x$ has to be in the range $(0, 50)$ and must be divisible by $5$ . There are $9$ possibilities, namely $5, 10, 15, 20, 25, 30, 35, 40, 45.$
The $x$ -coordinate of the point and the $y$ -coordinate of the point can each use any of these $9$ possibilities, giving $9^2 = 81$ numbers that are both $100$ -ray partitional and $60$ -ray partitional.
Overall, we are left with $2401 - 81 = \boxed{2320}$ solutions. | null | 2320 |
e1ff8c26f989ccfc068aa6e2b8eee099 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_2 | Josanna's test scores to date are $90, 80, 70, 60,$ and $85$ . Her goal is to raise here test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 82 \qquad\textbf{(C)}\ 85 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 95$ | The average of her current scores is $77$ . To raise it $3$ points, she needs an average of $80$ , and so after her $6$ tests, a sum of $480$ . Her current sum is $385$ , so she needs a $480 - 385 = \boxed{95}$ | E | 95 |
bfae4222fa1728dba2c3b7042a1907f2 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_3 | At a store, when a length or a width is reported as $x$ inches that means it is at least $x - 0.5$ inches and at most $x + 0.5$ inches. Suppose the dimensions of a rectangular tile are reported as $2$ inches by $3$ inches. In square inches, what is the minimum area for the rectangle?
$\textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75$ | The minimum dimensions of the rectangle are $1.5$ inches by $2.5$ inches. The minimum area is $1.5\times2.5=\boxed{3.75}$ square inches. | A | 3.75 |
7ba7073a064c5709cd1f7d4cb7b3b746 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_4 | LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B$ . How many dollars must LeRoy give to Bernardo so that they share the costs equally?
$\textbf{(A)}\ \frac{A + B}{2} \qquad\textbf{(B)}\ \dfrac{A - B}{2}\qquad\textbf{(C)}\ \dfrac{B - A}{2}\qquad\textbf{(D)}\ B - A \qquad\textbf{(E)}\ A + B$ | The difference in how much LeRoy and Bernardo paid is $B-A$ . To share the costs equally, LeRoy must give Bernardo half of the difference, which is $\boxed{2}$ | C | 2 |
7ba7073a064c5709cd1f7d4cb7b3b746 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_4 | LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B$ . How many dollars must LeRoy give to Bernardo so that they share the costs equally?
$\textbf{(A)}\ \frac{A + B}{2} \qquad\textbf{(B)}\ \dfrac{A - B}{2}\qquad\textbf{(C)}\ \dfrac{B - A}{2}\qquad\textbf{(D)}\ B - A \qquad\textbf{(E)}\ A + B$ | Since there are no restrictions on cost paid besides $A<B$ , we can use an example where $A = 40$ and $B = 50$ . Quickly, we realize the only way they could pay the same amount of money is if they both pay 45 dollars. This means LeRoy must give Bernardo $50 - 45 = 5$ . Looking at the answer choices we see only $\frac{50-40}{2} = 5$ works. $\boxed{2}$ | C | 2 |
7149376998549231ec7a716ba5adb28c | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_5 | In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161$ . What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad\textbf{(B)}\ 161 \qquad\textbf{(C)}\ 204 \qquad\textbf{(D)}\ 214 \qquad\textbf{(E)}\ 224$ | We have $161 = 7 \cdot 23.$ Since $a$ has two digits, the factors must be $23$ and $7,$ so $a = 32$ and $b = 7.$ Then, $ab = 7 \times 32 = \boxed{224}.$ | E | 224 |
db0e2e7b8e6baa99f114cb953a1a6242 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_6 | On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$ | Let $x$ represent the amount of candies Casper had at the beginning.
\begin{align*} \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ \frac{2}{3} x - 2 &= 18\\ \frac{2}{3} x &= 20\\ x &= \boxed{30} | A | 30 |
db0e2e7b8e6baa99f114cb953a1a6242 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_6 | On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$ | We work backwards. If he had 8 candies at the end, then before he gave candies to his sister he had 12 candies. This means that at the end of Halloween he had 18 candies, so before he gave candies to his brother he had 20 candies. Therefore, at the start he had $\boxed{30}$ | A | 30 |
db0e2e7b8e6baa99f114cb953a1a6242 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_6 | On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$ | A solve by algebra is more secure and safe (and usually faster), but you can also test the answer choices. We are lucky and option A works $\Longrightarrow \boxed{30}$ | A | 30 |
490c106f08a1e36ebfd7be780e48d870 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_7 | The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?
$\textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108$ | The sum of two angles in a triangle is $\frac{6}{5}$ of a right angle $\longrightarrow \frac{6}{5} \times 90 = 108$
If $x$ is the measure of the first angle, then the measure of the second angle is $x+30$ \[x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39\]
Now we know the measure of two angles are $39^{\circ}$ and $69^{\circ}$ . By the Triangle Sum Theorem, the sum of all angles in a triangle is $180^{\circ},$ so the final angle is $72^{\circ}$ . Therefore, the largest angle in the triangle is $\boxed{72}$ | B | 72 |
fe4480b7f3c67ca8e5426b3dbb353b42 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_10 | Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$ | The requested ratio is \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Using the formula for a geometric series, we have \[10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},\] which is very close to $\dfrac{10^{10}}{9},$ so the ratio is very close to $\boxed{9}.$ | B | 9 |
fe4480b7f3c67ca8e5426b3dbb353b42 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_10 | Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$ | The problem asks for the value of \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Written in base 10, we can find the value of $10^9 + 10^8 + \ldots + 10 + 1$ to be $1111111111.$ Long division gives us the answer to be $\boxed{9}.$ | B | 9 |
fe4480b7f3c67ca8e5426b3dbb353b42 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_10 | Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$ | Let $f(n)=\dfrac{10^n}{1+10+10^2+10^3+\cdots+10^{n-1}}$ . We are approximating $f(10)$ . Trying several small values of $n$ gives answers very close to $9$ , so our answer is $\boxed{9}\approx9.09.$ ~Technodoggo | B | 9 |
8d572f7070431611c02cc5c63e13e72c | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_11 | There are $52$ people in a room. what is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$ | Pretend you have $52$ people you want to place in $12$ boxes, because there are $12$ months in a year. By the Pigeonhole Principle , one box must have at least $\left\lceil \frac{52}{12} \right\rceil$ people $\longrightarrow \boxed{5}$ | D | 5 |
0fffe96650ab0fe344aaffc699abae7e | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_12 | Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?
$\textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}$ | Let $s$ be Keiko's speed in meters per second, $a$ be the length of the straight parts of the track, $b$ be the radius of the smaller circles, and $b+6$ be the radius of the larger circles. The length of the inner edge will be $2a+2b \pi$ and the length of the outer edge will be $2a+2\pi (b+6).$ Since it takes $36$ seconds longer for Keiko to walk on the outer edge,
\begin{align*} \frac{2a+2b \pi}{s} + 36 &= \frac{2a+2\pi (b+6)}{s}\\ 2a+2b\pi +36s &= 2a+2b\pi +12\pi\\ 36s&=12\pi\\ s&=\boxed{3} | A | 3 |
0fffe96650ab0fe344aaffc699abae7e | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_12 | Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?
$\textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}$ | It is basically the same as Solution 1 except you can completely disregard the straight edges of the track since it will take Keiko the same time to walk that length for both inner and outer circles. Instead, focus on the circular part. If the diameter of the smaller circle were $r,$ then the length of the smaller circle would be $r \pi$ and the length of the larger circle would be $(r+12) \pi.$ Since it still takes $36$ seconds longer,
\begin{align*} \frac{r \pi}{s} + 36 &= \frac{(r+12)\pi}{s}\\ r \pi + 36s &= r \pi + 12 \pi\\ s&=\boxed{3} | A | 3 |
31b72da5b27df85c7837a69f4cdb6100 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_13 | Two real numbers are selected independently at random from the interval $[-20, 10]$ . What is the probability that the product of those numbers is greater than zero?
$\textbf{(A)}\ \frac{1}{9} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{4}{9} \qquad\textbf{(D)}\ \frac{5}{9} \qquad\textbf{(E)}\ \frac{2}{3}$ | We will use complementary counting. The probability that the product is negative can be found by finding the probability that one number is positive and the other number is negative. The probability of a positive number being selected is $\frac13$ , and the probability of a negative number being selected is $\frac23$ . So, we see that the probability of the product being negative is \[2 \cdot \frac23 \cdot \frac13 = \frac49.\]
We multiply by $2$ because you can either pick the negative or positive number first.
Thus, the probability of the product being positive is \[1-\frac49 = \boxed{59}\] | D | 59 |
6a28f3a3c6fd71395cfb096cdae2e395 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_14 | A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$ | Let the sides of the rectangular parking lot be $a$ and $b$ . Then $a^2 + b^2 = 625$ and $ab = 168$ . Add the two equations together, then factor. \begin{align*} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align*} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \boxed{62}$ | C | 62 |
6a28f3a3c6fd71395cfb096cdae2e395 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_14 | A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$ | We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy $a^2+b^2=25^2$ . In other words, $(a,b,25)$ is a set of Pythagorean triples. Guessing and checking, we have $(7,24,25)$ as the triplet, as the area is $7 \cdot 24 = 168$ as requested. Therefore, the perimeter is $2(7+24)=\boxed{62}$ | C | 62 |
a2110e088fc6c82e1b34dea91accae12 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_17 | In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$ | We can let $\angle AEB$ be $4x$ and $\angle ABE$ be $5x$ because they are in the ratio $4 : 5$ . When an inscribed angle contains the diameter , the inscribed angle is a right angle . Therefore by triangle sum theorem, $4x+5x+90=180 \longrightarrow x=10$ and $\angle ABE = 50$
$\angle ABE = \angle BED$ because they are alternate interior angles and $\overline{AB} \parallel \overline{ED}$ . Opposite angles in a cyclic quadrilateral are supplementary , so $\angle BED + \angle BCD = 180$ . Use substitution to get $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{130}$ | C | 130 |
a2110e088fc6c82e1b34dea91accae12 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_17 | In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$ | Note $\angle ABE = \angle BED=50$ as before. The sum of the interior angles for quadrilateral $EBCD$ is $360$ . Denote the center of the circle as $P$ $\angle PDE = \angle PED = 50$ . Denote $\angle PDC = \angle PCD = x$ and $\angle PBC = \angle PCB = y$ . We wish to find $\angle BCD = x+y$ . Our equation is $(\angle PDE +\angle PED)+(\angle PDC+\angle PCD)+(\angle PBC + \angle PCB) = 360 \longrightarrow 2(50) + 2x +2y = 360$ . Our final equation becomes $2(x+y)+100 = 360$ . After subtracting $100$ and dividing by $2$ , our answer becomes $x+y=\boxed{130}$ | C | 130 |
a2110e088fc6c82e1b34dea91accae12 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_17 | In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$ | Note that $\overset{\Large\frown} {BE}$ intercepts $\angle BAE$ . Since, $\overset{\Large\frown} {BE}=180$ , thus $\angle BAE=90°$ (courtesy of the Inscribed Angles Theorem).
Since we know that $\angle BAE=90°$ , then $\angle AEB + \angle ABE = 90°$ , (courtesy of the Triangle Sum Theorem) and also $5\angle AEB = 4\angle ABE$ . By solving this variation, $\angle AEB = 40$ and $\angle ABE = 50$ . After that, due to the Alternate Interior Angles Theorem, $\angle ABE \cong \angle BED$ , which means $\angle BED = 50$
After doing some angle chasing, then these following facts should be true, $\overset{\Large\frown} {AB}=80$ $\overset{\Large\frown} {BD}=100$ $\overset{\Large\frown} {AE}=100$
Note that the arcs have to equal 360, so, $360=\overset{\Large\frown} {AB}+\overset{\Large\frown} {BD}+\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}$
$360=80+100+100+\overset{\Large\frown} {DE}$
$\overset{\Large\frown} {DE}=80$
Notice how $\overset{\Large\frown} {DB}$ intercepts $\angle BCD$ and that $\overset{\Large\frown} {DB}=\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}+\overset{\Large\frown} {AB}$
$\overset{\Large\frown} {DB}=80+100+80$
$\overset{\Large\frown} {DB}=260$
According to the Inscribed Angles Theorem, $2\angle BCD=\overset{\Large\frown} {DB}$ , therefore the answer is $\frac{260}{2}= \boxed{130}$ | C | 130 |
cf99ba14326157ec896f956013b6eb2f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | It is given that $\angle AMD \sim \angle CMD$ . Since $\angle AMD$ and $\angle CDM$ are alternate interior angles and $\overline{AB} \parallel \overline{DC}$ $\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM$ . Use the Base Angle Theorem to show $\overline{DC} \cong \overline{MC}$ . We know that $ABCD$ is a rectangle , so it follows that $\overline{MC} = 6$ . We notice that $\triangle BMC$ is a $30-60-90$ triangle, and $\angle BMC = 30^{\circ}$ . If we let $x$ be the measure of $\angle AMD,$ then \begin{align*} 2x + 30 &= 180\\ 2x &= 150\\ x &= \boxed{75} | E | 75 |
cf99ba14326157ec896f956013b6eb2f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | After finding $MC = 6,$ we can continue using trigonometry as follows.
We know that $\angle{BMC} = 180-2x$ and so $\sin (180-2x) = \frac{3}{6} = \frac{1}{2}$
It is obvious that $\sin (30) = \frac{1}{2}$ and so $180-2x=30.$
Solving, we have $x = \boxed{75}$ | null | 75 |
cf99ba14326157ec896f956013b6eb2f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | Let $\angle{DMC} = \angle{AMD} = \theta$ . If we let $AM = x$ , we have that $MD = \sqrt{x^2 + 9}$ , by the Pythagorean Theorem, and similarily, $MC = \sqrt{x^2 - 12x + 45}$ . Applying the law of cosine, we see that \[2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36\] and \[\tan (\theta) = \frac{3}{x}\] YAY!!! We have two equations for two variables... that are relatively difficult to deal with. Well, we'll try to solve it. First of all, note that $\sin (\theta) = \frac{3}{\sqrt{x^2+9}}$ , so solving for $\cos (\theta)$ in terms of $x$ , we get that $\cos (\theta) = \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9}$ . The equation now becomes
\[2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9} = 36\] Simplifying, we get
\[4x^4 - 48x^3 + 216x^2 - 432x + 324\]
Now, we apply the quartic formula to get
\[x = 6 \pm 3 \sqrt{3}\]
We can easily see that $x = 6 + 3 \sqrt{3}$ is an invalid solution. Thus, $x = 6 - 3 \sqrt{3}$
Finally, since $\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}$ $\theta = \frac{5 + 12n}{12} \pi$ , where $n$ is any integer. Converting to degrees, we have that $\theta = 75 + 180n$ . Since $0 < \theta < 90$ , we have that $\theta = \boxed{75}$ $\square$ | null | 75 |
cf99ba14326157ec896f956013b6eb2f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | We have $DC=CM=6$ . By the Pythagorean Theorem, $BM=\sqrt{6^2-3^2}=3\sqrt{3}$ , and thus $AM=6-3\sqrt{3}$ , We have $\tan(AMD)=\frac{6-3\sqrt{3}}{3}=2+\sqrt{3}$ , or $\angle AMD=\boxed{75}$ ~awsomek | null | 75 |
cf99ba14326157ec896f956013b6eb2f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | Let $\angle AMD=\angle DMC=\theta$ . Thus, $\angle BMC=180-2\theta\implies\angle MCB=2\theta-90$ . Since all angles should be positive, $\theta>45^\circ$ , narrowing the options to D and E. Trying $60^\circ$ (option D), $\Delta AMD$ is a 30-60-90 triangle. $AD=3$ , so it follows that $AM=\sqrt3$ .
Since $\angle BMC=180-2\theta$ $\angle BMC=60^\circ$ , too. However, that would imply that $\Delta MBC$ is also a $30-60-90$ triangle, which would, in turn, imply that $MB=3\sqrt3$ , since $BC=3$ . We know that $AM+MB=AB$ and $AB=6,$ but we know that $AM=\sqrt3$ and $MB=3\sqrt3$ $AM+MB$ is clearly not $6$ , so this is not possible.
Thus, the answer must be $\boxed{75}$ .
~ Technodoggo | E | 75 |
d8df0a9f6cb5300f8404b0c9c3b961b9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_19 | What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$ | First, square both sides, and isolate the absolute value. \begin{align*} 5|x|+8&=x^2-16\\ 5|x|&=x^2-24\\ |x|&=\frac{x^2-24}{5}. \\ \end{align*} Solve for the absolute value and factor.
Case 1: $x=\frac{x^2-24}{5}$
Multiplying both sides by $5$ gives us \[5x=x^2-24.\] Rearranging and factoring, we have \begin{align*} x^2-5x-24 &=0, \\ (x-8)(x+3) &= 0.\\ \end{align*}
Case 2: $x=\frac{-x^2+24}{5}$
As above, we multiply both sides by $5$ to find \[5x=-x^2+24.\] Rearranging and factoring gives us \begin{align*} x^2+5x-24 &=0, \\ (x+8)(x-3) &= 0. \\ \end{align*}
Combining these cases, we have $x= -8, -3, 3, 8$ . Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for $x$ back into the original equation to ensure it works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways.
Trying $|x|=|3|$ , we have \begin{align*} \sqrt{5|3|+8}&=\sqrt{3^2-16}, \\ \sqrt{15+8}&=\sqrt{9-16}, \\ \sqrt{23} &\not= \sqrt{-7}.\\ \end{align*} Therefore, $x = 3$ and $x= -3$ are extraneous.
Checking $|x|=|8|$ , we have \begin{align*} \sqrt{5|8|+8}&=\sqrt{8^2-16}, \\ \sqrt{40+8}&=\sqrt{64-16}, \\ \sqrt{48}&=\sqrt{48}.\\ \end{align*}
The roots of our original equation are $-8$ and $8$ and product is $-8 \times 8 = \boxed{64}$ | A | 64 |
d8df0a9f6cb5300f8404b0c9c3b961b9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_19 | What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$ | Square both sides, to get $5|x| + 8 = x^2-16$ . Rearrange to get $x^2 - 5|x| - 24 = 0$ . Seeing that $x^2 = |x|^2$ , substitute to get $|x|^2 - 5|x| - 24 = 0$ . We see that this is a quadratic in $|x|$ . Factoring, we get $(|x|-8)(|x|+3) = 0$ , so $|x| = \{8,-3\}$ . Since the radicand of the equation can't be negative, the sole solution is $|x| = 8$ . Therefore, $x$ can be $8$ or $-8$ . The product is then $-8 \times 8 = \boxed{64}$ | A | 64 |
d8df0a9f6cb5300f8404b0c9c3b961b9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_19 | What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$ | First we note that $x \in (-\infty,-4] \cup [4,\infty]$ . This will help us later with finding extraneous solutions.
Next, we have two cases: \[\text{IF } x\leq 4:\] \[\text{ } \sqrt{-5x+8} = \sqrt{x^2-16} \implies x^2+5x-24=0 \rightarrow x = -8,3\] .
We note that $3$ is not in the range of possible $x$ 's and thus is not a solution.
\[\text{IF } x\geq 4:\] \[\text{ } \sqrt{5x+8} = \sqrt{x^2-16} \implies x^2-5x-24=0 \rightarrow x = -3,8\] .
We again not that $-3$ is an extraneous solution.
Thus, we have the two solutions $-8$ and $8$ . Therefore product is $-8 \cdot 8 = \boxed{64}$ | A | 64 |
d0e8308d0ecbb1c3045995f0ca7f7b98 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | The largest difference, $9,$ must be between $w$ and $z.$
The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two differences can't be right next to each other because they would make a difference of $8,$ which isn't given as a possibility in the problem. This means $1$ must be the difference between $y$ and $x.$ We can express the possible configurations as the lines.
If we look at the first number line, you can express $x$ as $w-5,$ $y$ as $w-6,$ and $z$ as $w-9.$ Since the sum of all these integers equal $44$ \begin{align*} w+w-5+w-6+w-9&=44\\ 4w&=64\\ w&=16 \end{align*} You can do something similar to this with the second number line to find the other possible value of $w.$ \begin{align*} w+w-3+w-4+w-9&=44\\ 4w&=60\\ w&=15 \end{align*} The sum of the possible values of $w$ is $16+15 = \boxed{31}$ | B | 31 |
d0e8308d0ecbb1c3045995f0ca7f7b98 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | First, like Solution 1, we know that $w-z=9 \ \text{(1)}$ , because no two numbers could have a larger difference. Next, we find the sum of all the differences; since $w$ is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes $3w$ . Continuing in this way, we find that \[3w+x-y-3z=28 \ \text{(2)}\] . Now, we can subtract $3w-3z=27$ from (2) to get $x-y=1 \ \text{(3)}$ . Also, adding (2) with $w+x+y+z=44$ gives $4w+2x-2z=72$ , or $2w+x-z=36$ . Subtracting (1) from this gives $w+x=27$ . Since we know $w-z$ and $x-y$ , we find that \[(w-z)+(x-y)=(w-y)+(x-z)=9+1=10\] . This means that $w-y$ and $x-z$ must be 4 and 6, in some order. If $w-y=6$ , then subtracting this from (3) gives $(w-y)-(x-y)=6-1=5$ , so $w-x=5$ . This means that $(w-x)+(w+x)=2w=27+5=32$ , so $w=16$ . Similarly, $w$ can also equal $15$
Now if you are in a rush, you most likely would have answered $16+15=\boxed{31}$ . But we do have to check if these work. In fact, they do, giving solutions $(w,x,y,z)=(16, 11, 10, 7)$ and $(w,x,y,z)=(15, 12, 11, 6)$ | B | 31 |
d0e8308d0ecbb1c3045995f0ca7f7b98 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | Let $w - x = a$ $w - y = b$ $w - z = c$ . As above, we know that $c = 9$ . Thus, $a < b < c$ .
So, we have $w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44$ . This means $a + b + 9$ is a multiple of $4$ . Testing values of $a$ and $b$ , we find $(a, b, c) = (1, 6, 9), (3, 4, 9),$ and $(5, 6, 9)$ all satisfy this relation. The corresponding $(w, x, y, z)$ sets are $(15, 14, 9, 6), (15, 12, 11, 6),$ and $(16, 11, 10, 7)$ . The first set does not satisfy the given conditions, but the other two do. Thus, $w = 15$ and $w = 16$ are both possible solutions so the answer is $16+15=\boxed{31}$ | B | 31 |
d0e8308d0ecbb1c3045995f0ca7f7b98 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | From the problem, we know that $w+x+y+z=44$ . Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are $(w-x)$ $(w-y)$ $(w-z)$ $(x-y)$ $(x-z)$ $(y-z)$ , the sum of $(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)$ is equal to the sum of 1, 3, 4, 5, 6, 9, so $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ . Simplifying, we get $3w-3z+x-y=28$ . Adding $w+x+y+z=44$ and $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ , we get $4w+2x-2z=72$ , and simplifying we get $2w+x-z=36$ . Since $x-z$ is one of our positive differences, we can start guessing values for $w$ , and if the equation simplifies to one of our numerical positive differences, that value of $w$ should work. We can start at $w=18$ and keep going down, because our sum has to be positive. For $w=18$ $x-z=0$ , which is not one of our sums. For $w=17$ $x-z=2$ ,which is not one of our sums. For $w=16$ $x-z=4$ , which is one of our sums, so 16 works. For $w=15$ $x-z=6$ , which is one of our sums, so 15 works. For $w=14$ $x-z=8$ , which is not one of our sums. If we keep going, $x-z$ will soon exceed 10 and exceed all our sums, so any value below $w=14$ will not work. Therefore, our only solutions for $w$ are 15 and 16, which means our sum is $\boxed{31}$ . You can check that 15 and 16 work by forming a string of 4 numbers as shown above. | B | 31 |
d0e8308d0ecbb1c3045995f0ca7f7b98 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | Because we know that $w>x>y>z$ and that the positive differences are $1, 3, 4, 5, 6, 9$ , we can immediately come to the conclusion that $w-z= 9$ (because w is the largest integer and z is the smallest integer, so their difference must be the greatest). With this we have $3$ equations, we have that $x+y+z+w=44$ (from the problem), $w-z=9$ and because we can add up all the possible possible differences (as shown in the previous solutions), we get that $3w+x-y-3z=28$ . With these equations, we eventually manipulate these equations by doing the first equation minus 3 times the second to get $x-y=1$ . We can also add the first and third equation and subtract the second equation to get $w+x=27$ thus we know that $w-x$ can be $3, 4, 5,$ and $6$ (note: 1 and 9 are not possible because we know that $x-y=1$ and $w-z=9$ and the remaining differences can only be taken by 1 pair, so $w-x$ cannot be equal to 1 or 9). However, in order to get an integer value for x and z, we find that $w-z$ can only be equal to 3 and 5. Thus, by solving these, we see that $w= 15$ and $w=16$ $15+16=\boxed{31}$ | B | 31 |
7ceea100a04d3a43b041f5418c9a38c9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$
To compute this, we use a clever application of the binomial theorem
$\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}$
In all of the other terms, the power of $10$ is greater than $2$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:
$\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}$
Therefore, the hundreds digit is $\boxed{6}.$ | D | 6 |
7ceea100a04d3a43b041f5418c9a38c9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem , it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$
In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have \[2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.\]
In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have $\begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}$
After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{6}.$ | D | 6 |
7ceea100a04d3a43b041f5418c9a38c9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | Notice that the hundreds digit of $2011^{2011}$ won't be affected by $2000$ . Essentially we could solve the problem by finding the hundreds digit of $11^{2011}$ . Powers of $11$ are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of $11$ can be found by reading rows of the triangle and adding extra numbers up. [add source] For example, the sixth row of the triangle is $1, 5, 10, 10, 5,$ and $1$ . Adding all numbers from right to left, we get $161051$ , which is also $11^5$ . In other words, each number is $10^n$ steps from the right side of the row. The hundreds digit is $0$ . We can do the same for $11^{2011}$ , but we only need to find the $3$ digits from the right. Observing, every $3$ number from the right is $1 + 2 + 3... + n$ . So to find the third number from the right on the row of $11^{2011}$ \[f(11^n) = 1 + 2 + 3... + (n-1),\] or $\frac{(2010 \cdot 2011)}{2}$ , or $2021055$ . The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously $2011$ . We must carry the $1$ in $2011$ 's tens digit to the $5$ in $2021055$ 's unit digit to get $\boxed{6}$ . The one at the very end of the row doesn't affect anything, so we can leave it alone. | D | 6 |
7ceea100a04d3a43b041f5418c9a38c9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | Since we are only looking at the last 3 digits and $2011^2$ has the same last 3 digits as $11^2$ , we can find $11^{2011}$ instead.
After this, we can repeatedly multiply the last 3 digits by 11 and take the last 3 digits of that product. We discover that $11^{51}$ 's last 2 digits are -11, the same as $11^1$
From this information, we can figure out $11^{11}$ and $11^{61}$ end in 611. Adding various multiples of 50 to the exponent gives us the fact that $11^{2011}$ 's last digits are 611. We get $\boxed{6}$ .
-ThisUsernameIsTaken | D | 6 |
7ceea100a04d3a43b041f5418c9a38c9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | We know that the hundreds digit of $2011^{2011}$ is just the hundreds digit of $11^{2011} \equiv 011 \pmod{1000}$ . If we actually take a look at the powers of $11 \pmod{1000}$ , we notice a pattern:
\[11^{1} \equiv 011 \pmod{1000}\] \[11^{2} \equiv 121 \pmod{1000}\] \[11^{3} \equiv 331 \pmod{1000}\] \[11^{4} \equiv 641 \pmod{1000}\] \[11^{5} \equiv 051 \pmod{1000}\] \[11^{6} \equiv 561 \pmod{1000}\] Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by $11$ out (if you generally multiply a few numbers by $11$ , you can see why this pattern holds).
\[11^{7} \equiv 171 \pmod{1000}\] \[11^{8} \equiv 881 \pmod{1000}\] \[11^{9} \equiv 691 \pmod{1000}\] \[11^{10} \equiv 601 \pmod{1000}\] \[11^{11} \equiv 611 \pmod{1000}\] Since $11^{11}$ ends in _ $11$ , the pattern "cycles" every $10$ times. $\frac{2011}{10}=201$ remainder $1$ , meaning that $2011^{2011} \equiv 11^{2011} \equiv 611 \pmod{1000}$ , giving us a hundreds digit of $\boxed{6}$ | D | 6 |
7ceea100a04d3a43b041f5418c9a38c9 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_23 | What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | We know that the hundreds digit of $2011^{11}$ is just the hundreds digit of $11^{11}$ (mod 1000). If we actually take a look at the powers of 11, we notice a pattern:
\[11^{1} = 11\]
\[11^{2} = 121\]
\[11^{3} = 1331\]
\[11^{4} = 14641\]
We notice that the hundreds digit just increases by 1 starting from 1(the hundreds digit of $11^{1}$ ). All we have to do now is add all the numbers from 1 to 11 (since the hundreds digit increases by 1 with every power of 11 that comes after) to get the hundreds digit of $11^{11}$
$1 + 2 + 3 + 4.... + 11 = 66$ . Thus the answer is $\boxed{6}$ | D | 6 |
91891e3f8ebfc93f63f6565fe87f7be7 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_1 | Mary's top book shelf holds five books with the following widths, in centimeters: $6$ $\dfrac{1}{2}$ $1$ $2.5$ , and $10$ . What is the average book width, in centimeters?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | To find the average, we add up the widths $6$ $\dfrac{1}{2}$ $1$ $2.5$ , and $10$ , to get a total sum of $20$ . Since there are $5$ books, the average book width is $\frac{20}{5}=4$ The answer is $\boxed{4}$ | D | 4 |
ad81507622c495aac2e00ff97b8ae94d | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_3 | Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 18 \qquad \mathrm{(D)}\ 25 \qquad \mathrm{(E)}\ 29$ | Let $x$ be the number of marbles Tyrone gave to Eric. Then, $97-x = 2\cdot(11+x)$ . Solving for $x$ yields $75=3x$ and $x = 25$ . The answer is $\boxed{25}$ | D | 25 |
ad81507622c495aac2e00ff97b8ae94d | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_3 | Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 18 \qquad \mathrm{(D)}\ 25 \qquad \mathrm{(E)}\ 29$ | Since the number of balls Tyrone and Eric have a specific ratio when Tyrone gives some of his balls to Eric, we can divide the total amount of balls by their respective ratios. Altogether, they have $97+11$ balls, or $108$ balls, and it does not change when Tyrone gives some of his to Eric, since none are lost in the process. After Tyrone gives some of his balls away, the ratio between the balls is $2:1$ , and added together is $108$ . The two numbers add up to $3$ and we divide $108$ by $3$ , and the outcome is $36$ . This is the number of balls Eric has, and so doubling that results in the number of balls Tyrone has, which is $72$ $97-72$ is $25$ , and $36-11$ is $25$ , thus proving our statement true, and the answer is $\boxed{25}$ | D | 25 |
80de1f886a8739420ba829bdaaee71cf | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_4 | A book that is to be recorded onto compact discs takes $412$ minutes to read aloud. Each disc can hold up to $56$ minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?
$\mathrm{(A)}\ 50.2 \qquad \mathrm{(B)}\ 51.5 \qquad \mathrm{(C)}\ 52.4 \qquad \mathrm{(D)}\ 53.8 \qquad \mathrm{(E)}\ 55.2$ | Assuming that there are fractions of compact discs, it would take $412/56 ~= 7.357$ CDs to have equal reading time. However, since the number of discs must be a whole number, there are at least 8 CDs, in which case there would be $412/8 = 51.5$ minutes of reading on each of the 8 discs. The answer is $\boxed{51.5}$ | B | 51.5 |
80de1f886a8739420ba829bdaaee71cf | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_4 | A book that is to be recorded onto compact discs takes $412$ minutes to read aloud. Each disc can hold up to $56$ minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?
$\mathrm{(A)}\ 50.2 \qquad \mathrm{(B)}\ 51.5 \qquad \mathrm{(C)}\ 52.4 \qquad \mathrm{(D)}\ 53.8 \qquad \mathrm{(E)}\ 55.2$ | We look at the options, and see which one is divisible by 412. We try A, and see that you don't get a whole number. Next, we try B. We see that the B is divisible by 412 and find that $\boxed{51.5}$ is our answer. | B | 51.5 |
4d039e94ae1160fda55671bb60c31081 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_5 | The area of a circle whose circumference is $24\pi$ is $k\pi$ . What is the value of $k$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 36 \qquad \mathrm{(E)}\ 144$ | If the circumference of a circle is $24\pi$ , the radius would be $12$ . Since the area of a circle is $\pi r^2$ , the area is $144\pi$ . The answer is $\boxed{144}$ | E | 144 |
4d039e94ae1160fda55671bb60c31081 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_5 | The area of a circle whose circumference is $24\pi$ is $k\pi$ . What is the value of $k$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 36 \qquad \mathrm{(E)}\ 144$ | By definition, $\pi$ is the ratio of the circumference to the diameter. Since the circumference is $24\pi$ , the diameter must be $24$ and the radius is $12$ . Therefore, by the area of circle formula $A=\pi r^{2}$ the area is $12^{2}\pi=144\pi$ and $k=144 \Longrightarrow \boxed{144}$ | E | 144 |
fa25321dd2bda305ea3f82106e7ddd87 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_7 | Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ \sqrt{2} \qquad \mathrm{(C)}\ \sqrt{3} \qquad \mathrm{(D)}\ 2 \qquad \mathrm{(E)}\ 2\sqrt{2}$ | Crystal runs north one mile, then her next two moves can be broken up into four individual moves: for her northeast section, it forms a $45-45-90$ triangle whose legs are each $\frac{\sqrt{2}}{2}$ . For her southeast section, it is also a $45-45-90$ triangle whose legs are each $\frac{\sqrt{2}}{2}$ . Notice that the two of the legs cancel each other out; she moves north $\frac{\sqrt{2}}{2}$ units and also south $\frac{\sqrt{2}}{2}$ units. So her net movement these two moves is $\sqrt{2}$ to the right. Finally, after the third move, she is at the corner of a right triangle with legs $1$ and $\sqrt{2}$ . Using the Pythagorean theorem, $d^{2}=1^{2}+\left(\sqrt{2}\right)^{2}=1+2=3$ and $d=\sqrt{3} \Longrightarrow \boxed{3}$ | C | 3 |
7f1e3219e7a7ae04f173571ce1c45c9d | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_8 | Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$ | Tony works $2$ hours a day and is paid $0.50$ dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is $12$ years old, he gets $12$ dollars a day. We also know that he worked $50$ days and earned $630$ dollars. If he was $12$ years old at the beginning of his working period, he would have earned $12 * 50 = 600$ dollars. If he was $13$ years old at the beginning of his working period, he would have earned $13 * 50 = 650$ dollars. Because he earned $630$ dollars, we know that he was $13$ for some period of time, but not the whole time, because then the money earned would be greater than or equal to $650$ . This is why he was $12$ when he began, but turned $13$ sometime during the six month period and earned $630$ dollars in total. So the answer is $13$ .The answer is $\boxed{13}$ . We could find out for how long he was $12$ and $13$ $12 \cdot x + 13 \cdot (50-x) = 630$ . Then $x$ is $20$ and we know that he was $12$ for $20$ days, and $13$ for $30$ days. Thus, the answer is $13$ | D | 13 |
7f1e3219e7a7ae04f173571ce1c45c9d | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_8 | Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$ | Let $x$ equal Tony's age at the end of the period. We know that his age changed during the time period (since $630$ does not evenly divide $50$ ). Thus, his age at the beginning of the time period is $x - 1$
Let $d$ be the number of days Tony worked while his age was $x$ . We know that his earnings every day equal his age (since $2 \cdot 0.50 = 1$ ). Thus, \[x \cdot d + (x - 1)(50 - d) = 630\] \[x\cdot d + 50x - x\cdot d - 50 + d = 630\] \[50x + d = 680\] \[x = \dfrac{680 - d}{50}\] Since $0 < d <50$ $d = 30$ . Then we know that $50x = 650$ and $x = \boxed{13}$ | D | 13 |
7f1e3219e7a7ae04f173571ce1c45c9d | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_8 | Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$ | Since Tony worked for $50$ days, he has worked for $100$ hours. Let $k$ be his hourly wage. Then, we know that $100k = 630$
Dividing both sides by $50$ , we get that $2k$ (Daily wage, which is equal to his age) $\approx$ $12.6$ . Since we now know that his age was either 12 or 13 during the 6 month span, and we are asked to find his age at the end of this time, the answer is $\boxed{13}$ | D | 13 |
37dc2f3c18e482162b685ea7d956527b | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_9 | $\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$ | $x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$
It follows that $x$ is $969$ , so the sum of the digits is $\boxed{24}$ | E | 24 |
37dc2f3c18e482162b685ea7d956527b | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_9 | $\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$ | For $x+32$ to be a four-digit number, $x$ is in between $968$ and $999$ . The palindromes in this range are $969$ $979$ $989$ , and $999$ , so the sum of the digits of $x$ can be $24$ $25$ $26$ , or $27$ . Only $\boxed{24}$ is an option, and upon checking, $x+32=1001$ is indeed a palindrome. | E | 24 |
37dc2f3c18e482162b685ea7d956527b | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_9 | $\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$ | Since we know $x+32$ to be $1 a a 1$ and the only palindrome that works is $0 = a$ , that means $x+32 = 1001$ , and so $x = 1001 - 32 = 969$ . So, $9$ $6$ $9$ $\boxed{24}$ .
~songmath20 | E | 24 |
95021b59c88252e38119e08ef740b11e | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_10 | Marvin had a birthday on Tuesday, May 27 in the leap year $2008$ . In what year will his birthday next fall on a Saturday?
$\mathrm{(A)}\ 2011 \qquad \mathrm{(B)}\ 2012 \qquad \mathrm{(C)}\ 2013 \qquad \mathrm{(D)}\ 2015 \qquad \mathrm{(E)}\ 2017$ | $\boxed{2017}$ $2017$ | E | 2017 |
ab5dde89a6f93af911f7333508ae3893 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_11 | The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$ . What is $b - a$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$ | Since we are given the range of the solutions, we must re-write the inequalities so that we have $x$ in terms of $a$ and $b$
$a\le 2x+3\le b$
Subtract $3$ from all of the quantities:
$a-3\le 2x\le b-3$
Divide all of the quantities by $2$
$\frac{a-3}{2}\le x\le \frac{b-3}{2}$
Since we have the range of the solutions, we can make it equal to $10$
$\frac{b-3}{2}-\frac{a-3}{2} = 10$
Multiply both sides by 2.
$(b-3) - (a-3) = 20$
Re-write without using parentheses.
$b-3-a+3 = 20$
Simplify.
$b-a = 20$
We need to find $b - a$ for the problem, so the answer is $\boxed{20}$ | D | 20 |
ab5dde89a6f93af911f7333508ae3893 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_11 | The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$ . What is $b - a$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$ | Without loss of generality , let the interval of solutions be $[0, 10]$ (or any real values $[p, 10+p]$ ). Then, substitute $0$ and $10$ to $x$ . This gives $b=23$ and $a=3$ . So, the answer is $23-3=\boxed{20}$ .
~ bearjere | D | 20 |
544b11f701e934157fde5fc265ad689f | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_12 | Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?
$\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4$ | The water tower holds $\frac{100000}{0.1} = 1000000$ times more water than Logan's miniature. The volume of a sphere is: $V=\dfrac{4}{3}\pi r^3$ . Since we are comparing the heights (m), we should compare the radii (m) to find the ratio. Since, the radius is cubed, Logan should make his tower $\sqrt[3]{1000000} = 100$ times shorter than the actual tower. This is $\frac{40}{100} = \boxed{0.4}$ | C | 0.4 |
62f620a2b673157b9914579e1ff36b0f | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_14 | Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ$ | Let $\angle BAE = \angle ACD = x$
\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\ \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ \angle EAC &= 60^\circ - x\\ \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}
Since $\frac{AC}{AB} = \frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\circ$ , triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90}$ | C | 90 |
d499d219290cb8497318bb6e687b2dc9 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_15 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "LeRoy is a frog."
LeRoy: "Chris is a frog."
Mike: "Of the four of us, at least two are toads."
How many of these amphibians are frogs?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
As Mike is a frog, his statement is false, hence there is at most one toad.
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
Hence we must have one toad and $\boxed{3}$ frogs. | D | 3 |
d499d219290cb8497318bb6e687b2dc9 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_15 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "LeRoy is a frog."
LeRoy: "Chris is a frog."
Mike: "Of the four of us, at least two are toads."
How many of these amphibians are frogs?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.
Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means that none or one of the amphibians is a toad (the opposite of the statement $n\geq2$ is $n<2$ , or $n=0, 1$ ). However, this is absurd because we assumed Brian is a toad, and we know one of Chris and LeRoy is a toad. So our assumption leads to a contradiction.
Hence Brian must be a frog, and he and Mike are the same species. Mike is also a frog. One of Chris and LeRoy is a frog. There are $3$ frogs in total $\Longrightarrow \boxed{3}$ | D | 3 |
70844fe9186d7276755bc4fe057832db | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_16 | Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$ | By the Angle Bisector Theorem , we know that $\frac{AB}{BC} = \frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = 16$ ), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$ | B | 33 |
70844fe9186d7276755bc4fe057832db | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_16 | Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$ | We find that $\frac{AB}{BC}=\frac{3}{8}$ by the Angle Bisector Theorem so we let the lengths be $3n$ and $8n$ , respectively where $n$ is a positive integer. Also since $AD=3$ and $BC=8$ , we notice that the perimeter of the triangle is the sum of these, namely $3n+8n+3+8=11n+11.$ This can be factored into $11(n+1)$ and so the sum must be a multiple of $11$ . The only answer choice which is a multiple of $11$ is $\boxed{33}$ . ~mathboy282 | B | 33 |
bf75959f7edfd0a7d9ed3f327a8ef11f | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_17 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$ | Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$ . The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$ | A | 7 |
bf75959f7edfd0a7d9ed3f327a8ef11f | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_17 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$ | We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as
the central $2\times 2\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.
Hence the total volume of the cuts is $3(2 \times 2 \times 3) - 2(2\times 2\times 2) = 36 - 16 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$ | A | 7 |
bf75959f7edfd0a7d9ed3f327a8ef11f | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_17 | A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$ | We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.
Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box.
$12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \boxed{7}$ | A | 7 |
537cc91911af1b37ae36ff42709011c6 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_19 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$ | It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$ . Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.
If we extend $BC$ $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$ $BC$ and $DE$ meet at $Y$ , and $DE$ and $FA$ meet at $Z$ , we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$ $CDY$ and $EFZ$ , of side length $1$ . The area of $ABCDEF$ is therefore
$\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$
Based on the initial conditions,
\[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\]
Simplifying this gives us $r^2-6r+1 = 0$ . By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{6}$ | E | 6 |
c7657d0f0e4de09d34fafaf32d6399b0 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21 | The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$
$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$ | By Vieta's Formulas , we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$ . Again Vieta's Formulas tell us that $2010$ is the product of the three integer roots. Also, $2010$ factors into $2\cdot3\cdot5\cdot67$ . But, since the polynomial has only three roots, two of the four prime factors must be multiplied so that we are left with three roots. To minimize $a$ $2$ and $3$ should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is $\boxed{78}$ .
~JimPickens | A | 78 |
c7657d0f0e4de09d34fafaf32d6399b0 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21 | The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$
$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$ | We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$
$(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$
We do not care about $+bx$ in this case, because we are only looking for $a$ . We know that the constant term is $-2010=-(2\cdot 3\cdot 5\cdot 67)$ We are trying to minimize a, such that we have $-ax^2$ Since we have three positive solutions, we have $(x-a)(x-b)(x-c)$ as our factors. We have to combine two of the factors of $2\cdot 3\cdot 5\cdot 67$ , and then sum up the $3$ resulting factors. Since we are minimizing, we choose $2$ and $3$ to combine together. We get $(x-6)(x-5)(x-67)$ which gives us a coefficient of $x^2$ of $-6-5-67=-78$ Therefore $-a=-78$ or $a=\boxed{78}$ | A | 78 |
c7657d0f0e4de09d34fafaf32d6399b0 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21 | The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$
$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$ | We want the polynomial $x^3-ax^2+bx-2010$ to have POSITIVE integer roots. That means we want to factor it in to the form $(x-a)(x-b)(x-c).$ We therefore want the prime factorization for $2010$ . The prime factorization of $2010$ is $2 \cdot 3 \cdot 5 \cdot 67$ . We want the smallest difference of the $3$ roots since by Vieta's formulas $a$ is the sum of the $3$ roots.
We proceed to factorize it in to $(x-5)(x-6)(x-67)$ . Therefore, our answer is $5+6+67$ $\boxed{78}$ | A | 78 |
7e184165b0cc885b1c6b37c372f59a5c | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_22 | Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140$ | To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only $1$ way to connect the points such that a triangle is formed in the circle's interior (this is because we want no two chords to be nonintersecting ~Williamgolly). Therefore, the answer is ${{8}\choose{6}}$ , which is equivalent to $\boxed{28}$ | A | 28 |
7e184165b0cc885b1c6b37c372f59a5c | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_22 | Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140$ | To make a triangle, where the $3$ points are arranged on a circle, you just need to choose $3$ points because no $3$ points are arranged in a straight line on a circle, meaning that to count the number of triangles we get $\binom{8}{3}=56$ . However, this is incorrect because of the restriction that no three chords intersect in a single point in the circle meaning that the answer is most likely less than $56$ , and the only answer choice that satisfies this condition is $\boxed{28}$ ~Batmanstark | A | 28 |
f7d8008a0b825189a3347fd406ed19ac | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_23 | Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$ , the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) < \frac{1}{2010}$
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005$ | The probability of drawing a white marble from box $k$ is $\frac{k}{k + 1}$ , and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$
To stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \ldots, n-1,$ and draw a red marble from box $n.$ Thus, \[P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.\]
So, we must have $\frac{1}{n(n + 1)} < \frac{1}{2010}$ or $n(n+1) > 2010.$
Since $n(n+1)$ increases as $n$ increases, we can simply test values of $n$ ; after some trial and error, we get that the minimum value of $n$ is $\boxed{45}$ , since $45(46) = 2070$ but $44(45) = 1980.$ | A | 45 |
4aaba0c188abd71dbae9398b50dd91c9 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_25 | Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$ , then his sequence contains $5$ numbers:
\[\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}\]
Let $N$ be the smallest number for which Jim’s sequence has $8$ numbers. What is the units digit of $N$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 3 \qquad \mathrm{(C)}\ 5 \qquad \mathrm{(D)}\ 7 \qquad \mathrm{(E)}\ 9$ | We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ and $x>1^2$ , which is $x=2$
We repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^2=2$ for the row after that. However, at the fourth row, we see that solving $x-1^2=3$ yields $x=4$ , in which case it would be incorrect since $1^2=1$ is not the greatest perfect square less than or equal to $x$ . So we make it a $2^2$ and solve $x-2^2=3$ . We continue on using this same method where we increase the perfect square until $x$ can be made bigger than it. When we repeat this until we have $8$ rows, we get:
\[\begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array}\]
Hence the solution is the last digit of $7223$ , which is $\boxed{3}$ | B | 3 |
4aaba0c188abd71dbae9398b50dd91c9 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_25 | Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$ , then his sequence contains $5$ numbers:
\[\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}\]
Let $N$ be the smallest number for which Jim’s sequence has $8$ numbers. What is the units digit of $N$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 3 \qquad \mathrm{(C)}\ 5 \qquad \mathrm{(D)}\ 7 \qquad \mathrm{(E)}\ 9$ | Notice that to get the previous term, we must add the smallest square number, (let's call it $n^2$ ) such that the sum is less than $(n+1)^2$ . Otherwise, instead of subtracting $n^2$ from the previous term, we're subtracting a greater square number.
Remember that $(x+1)^2 = x^2 + 2x + 1$ . Recall that to find the previous term, we must add a square number such that it is less than the next square number. $a + n^2 < (n+1)^2$ . For this to be true, $a < 2n + 1$ . What that means is that given a term $a$ , we can find the previous term by adding $n^2$ where $n > \frac {a-1}{2}$
For example, to find the term that precedes $167$ , we know that $n>166/2 = 83$ . Therefore, $n=84$ and the previous term is $167 + 84^2 = 7223$ . The last digit of $7223$ is $3 \Rightarrow \boxed{3}$ | B | 3 |
c9151da0263d54d25c1cab3d935e6f97 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_1 | What is $100(100-3)-(100\cdot100-3)$
$\textbf{(A)}\ -20,000 \qquad \textbf{(B)}\ -10,000 \qquad \textbf{(C)}\ -297 \qquad \textbf{(D)}\ -6 \qquad \textbf{(E)}\ 0$ | $100(100-3)-(100\cdot{100}-3)=10000-300-10000+3=-300+3=\boxed{297}$ | C | 297 |
e17a76d9e3addb02c533988521ec62cc | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_2 | Makarla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35$ | The total number of minutes in her $9$ -hour work day is $9 \times 60 = 540.$ The total amount of time spend in meetings in minutes is $45 + 45 \times 2 = 135.$ The answer is then $\frac{135}{540}$ $= \boxed{25}$ | C | 25 |
586ca3fd2bf91c32e52a148d14e15dea | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_3 | A drawer contains red, green, blue, and white socks with at least 2 of each color. What is
the minimum number of socks that must be pulled from the drawer to guarantee a matching
pair?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | After you draw $4$ socks, you can have one of each color, so (according to the pigeonhole principle ), if you pull $\boxed{5}$ then you will be guaranteed a matching pair. | C | 5 |
a393ab9c950e14716af0139191e7d3bc | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_4 | For a real number $x$ , define $\heartsuit(x)$ to be the average of $x$ and $x^2$ . What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$ | The average of two numbers, $a$ and $b$ , is defined as $\frac{a+b}{2}$ . Thus the average of $x$ and $x^2$ would be $\frac{x(x+1)}{2}$ . With that said, we need to find the sum when we plug, $1$ $2$ and $3$ into that equation. So:
\[\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{10}\] | C | 10 |
f9551f20dbd809f632208fee1a0ec666 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_5 | A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | $31 \equiv 3 \pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\boxed{3}$ | B | 3 |
c7d503bb6a683b6cbb1632e1c4c03fa6 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_6 | A circle is centered at $O$ $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$ . What is the degree measure of $\angle CAB$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$ | Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $\triangle COA$ is an isosceles triangle. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle COA = 180 - 50 = 130^{\circ}$ . Since $\triangle COA$ is isosceles, then $\angle OCA \cong \angle OAC$ . They also sum to $50^{\circ}$ , so each angle is $\boxed{25}$ | B | 25 |
c7d503bb6a683b6cbb1632e1c4c03fa6 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_6 | A circle is centered at $O$ $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$ . What is the degree measure of $\angle CAB$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$ | Note that $\angle AOC = 180^\circ - 50^\circ = 130^\circ$ . Because triangle $AOC$ is isosceles, $\angle CAB = (180^\circ - 130^\circ)/2 = \boxed{25}$ | null | 25 |
7383630f07a7fa8e6674537dda78b6e3 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_7 | A triangle has side lengths $10$ $10$ , and $12$ . A rectangle has width $4$ and area equal to the
area of the triangle. What is the perimeter of this rectangle?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$ | The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - \left( \dfrac{12}{2} \right)^2} = \sqrt{64} = 8$
Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$
We have that the area of the rectangle is the same as the area of the triangle, namely $48$ . We also have the width of the rectangle: $4$
The length of the rectangle therefore is: $l = \dfrac{48}{4} = 12$
The perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$
The answer is:
$\boxed{32}$ | D | 32 |
7383630f07a7fa8e6674537dda78b6e3 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_7 | A triangle has side lengths $10$ $10$ , and $12$ . A rectangle has width $4$ and area equal to the
area of the triangle. What is the perimeter of this rectangle?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$ | An alternative way to find the area of the triangle is by using Heron's formula, $A=\sqrt{(s)(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is $(10+10+12)/2 = 16$ . Thus the area equals $\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48.$ We know that the width of the rectangle is $4$ , so $48/4 = 12$ , which is the length. The perimeter of the rectangle is $2(4+12) =$ $\boxed{32}$ | D | 32 |
7383630f07a7fa8e6674537dda78b6e3 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_7 | A triangle has side lengths $10$ $10$ , and $12$ . A rectangle has width $4$ and area equal to the
area of the triangle. What is the perimeter of this rectangle?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$ | Note that a triangle with side lengths $10,10,12$ is essentially $2$ “6,8,10” right triangles stuck together. Hence, the height is $8$ , and our area is $48$
So, the length of the rectangle is $\frac{48}{4}=12$ , and our perimeter $P=2(4+12)=\boxed{32}$ | D | 32 |
0b227e936ba70920d9f6d3536ee459f1 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_11 | A shopper plans to purchase an item that has a listed price greater than $\textdollar 100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\textdollar 30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\textdollar 100$ Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is $y - x$
$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ 80 \qquad \textbf{(E)}\ 100$ | Let the listed price be $(100 + p)$ , where $p > 0$
Coupon A saves us: $0.15(100+p) = (0.15p + 15)$
Coupon B saves us: $30$
Coupon C saves us: $0.25p$
Now, the condition is that A has to be greater than or equal to either B or C which gives us the following inequalities:
$A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100$
$A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150$
We see here that the greatest possible value for $p$ is $150$ , thus $y = 100 + 150 = 250$ and the smallest value for $p$ is $100$ so $x = 100 + 100 = 200$
The difference between $y$ and $x$ is $y - x = 250 - 200 = \boxed{50}$ | A | 50 |
e36787062547c4f081a8fc4b01a66c45 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_12 | At the beginning of the school year, $50\%$ of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of $x$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$ | Clearly, the minimum possible value would be $70 - 50 = 20\%$ . The maximum possible value would be $30 + 50 = 80\%$ . The difference is $80 - 20 = \boxed{60}$ | D | 60 |
f638769b7e82cb168547f3552072b416 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_13 | What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$
$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124$ | We evaluate this in cases:
Case 1 $x<30$
When $x<30$ we are going to have $60-2x>0$ . When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$ . Therefore we have $x=|2x-(60-2x)|$ $x=|2x-60+2x|\implies x=|4x-60|$
Subcase 1 $30>x>15$
When $30>x>15$ we are going to have $4x-60>0$ . When this happens, we can express $|4x-60|$ as $4x-60$ .
Therefore we get $x=4x-60\implies -3x=-60\implies x=20$ . We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$ . Therefore $20$ is one possible solution.
Subcase 2 $x<15$
When $x<15$ we are going to have $4x-60<0$ , therefore $|4x-60|$ can be expressed in the form $60-4x$ .
We have the equation $x=60-4x\implies 5x=60\implies x=12$ . Since $12$ is less than $15$ $12$ is another possible solution. $x=|2x-|60-2x||$
Case 2 $x>30$
When $x>30$ $60-2x<0$ . When $x<0$ we can express this in the form $-x$ . Therefore we have $-(60-2x)=2x-60$ . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have $x=|2x-(2x-60)|$
$x=|2x-2x+60|$
$x=|60|$
$x=60$
We have now evaluated all the cases, and found the solution to be $\{60,12,20\}$ which have a sum of $\boxed{92}$ | C | 92 |
f638769b7e82cb168547f3552072b416 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_13 | What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$
$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124$ | From the equation $x = \left|2x-|60-2x|\right|$ , we have $x = 2x-|60-2x|$ , or $-x = 2x-|60-2x|$ . Therefore, $x=|60-2x|$ , or $3x=|60-2x|$ . From here we have four possible cases:
1. $x=60-2x$ ; this simplifies to $3x=60$ , so $x=20$
2. $-x=60-2x$ ; this simplifies to $x=60$
3. $3x=60-2x$ ; this simplifies to $5x=60$ , so $x=12$
4. $-3x=60-2x$ ; this simplifies to $-x=60$ , so $x=-60$ . However, this solution is extraneous because the absolute value of $2x-|60-2x|$ cannot be negative.
The sum of all of the solutions of $x$ is $20+60+12=\boxed{92}$ | C | 92 |
17a7470535b19cb604b2590a93759a02 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_15 | On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99$ . What is the maximum number of questions that Jesse could have answered correctly?
$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 33$ | Let $a$ be the amount of questions Jesse answered correctly, $b$ be the amount of questions Jesse left blank, and $c$ be the amount of questions Jesse answered incorrectly. Since there were $50$ questions on the contest, $a+b+c=50$ . Since his total score was $99$ $4a-c=99$ . Also, $a+c\leq50 \Rightarrow c\leq50-a$ . We can substitute this inequality into the previous equation to obtain another inequality: $4a-(50-a)\leq99 \Rightarrow 5a\leq149 \Rightarrow a\leq \frac{149}5=29.8$ . Since $a$ is an integer, the maximum value for $a$ is $\boxed{29}$ | C | 29 |
bc2978cf930fbb96572a9dbe1faa7197 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_17 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. How many schools are in the city?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$ | There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $23 \Rightarrow \boxed{23}$ | B | 23 |
bc2978cf930fbb96572a9dbe1faa7197 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_17 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. How many schools are in the city?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$ | Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$
Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$
Also, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \implies a \ge 33$
Putting it all together: $33 \le a < 37$ and $a \equiv 2 \pmod{3}$ , so clearly $a = 35$ , and the number of schools as we got before is $(2a-1)/3 = 69/3 = \boxed{23}$ | null | 23 |
98d9ebbec45aa124e34b57714050b9bf | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_19 | A circle with center $O$ has area $156\pi$ . Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$ , and point $O$ is outside $\triangle ABC$ . What is the side length of $\triangle ABC$
$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$ | The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$
Because $OA=4\sqrt{3} < \sqrt{156}, A$ must be in the interior of circle $O.$
Let $s$ be the unknown value, the sidelength of the triangle, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle ABC$ is equilateral, $BX=\frac{s}{2}$ and $AX=\frac{s\sqrt{3}}{2}.$ We are given $AO=4\sqrt{3}.$ Use the Pythagorean Theorem and solve for $s.$
\begin{align*} (\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\ 156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\ 0 &= s^2 + 12s - 108\\ 0 &= (s-6)(s+18)\\ s &= \boxed{6} | B | 6 |
98d9ebbec45aa124e34b57714050b9bf | https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_19 | A circle with center $O$ has area $156\pi$ . Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$ , and point $O$ is outside $\triangle ABC$ . What is the side length of $\triangle ABC$
$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$ | We can use the same diagram as Solution 1 and label the side length of $\triangle ABC$ as $s$ . Using congruent triangles, namely the two triangles $\triangle BOA$ and $\triangle COA$ , we get that $\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150$ . From this, we can use the Law of Cosines , to get \[s^2 + (4 \sqrt{3})^2 - 2 \times s \times 4 \sqrt{3} \times - \frac{\sqrt{3}}{2} = (2 \sqrt{39})^2\] Simplifying, we get \[s^2 + 12s + 48 = 156 \implies s^2 + 12s - 108 = 0\] We can factor this to get \[(x-6)(x+18)\] Lengths must be non-negative, so the answer is $\boxed{6}$ ~bryan gao | B | 6 |
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