https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Crystalflower&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T00:19:25ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_20&diff=1883742023 AMC 8 Problems/Problem 202023-01-30T05:08:56Z<p>Crystalflower: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Two integers are inserted into the list <math>3, 3, 8, 11, 28</math> to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?<br />
<br />
<math>\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61</math><br />
<br />
==Solution==<br />
To double the range we must find the current range, which is <math>28 - 3 = 25</math>, to then the double is <math>2(25) = 50</math> Since we dont want to change the median we need to get a value greater than <math>8</math> (as <math>8</math> would change the mode) for the smaller and <math>53</math> is fixed for the larger as anything less than <math>3</math> is not beneficial to the optimization. So taking our optimal values of <math>53</math> and <math>7</math> we have an answer of <math>53 + 7 = \boxed{\textbf{(D)}\ 60}</math>.<br />
<br />
~apex304, SohumUttamchandani, wuwang2002, TaeKim, CrystalFlower<br />
<br />
==Animated Video Solution==<br />
https://youtu.be/ItntB7vEafM<br />
<br />
~Star League (https://starleague.us)<br />
<br />
==Video Solution by OmegaLearn (Using Smart Sequence Analysis)==<br />
https://youtu.be/qNsgNa9Qq9M<br />
<br />
==Video Solution by Magic Square==<br />
https://youtu.be/-N46BeEKaCQ?t=3136<br />
<br />
==See Also== <br />
{{AMC8 box|year=2023|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_18&diff=1831092014 AMC 8 Problems/Problem 182022-11-26T21:16:33Z<p>Crystalflower: </p>
<hr />
<div>==Problem==<br />
Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?<br />
<br />
(A) all 4 are boys<br />
(B) all 4 are girls<br />
(C) 2 are girls and 2 are boys<br />
(D) 3 are of one gender and 1 is of the other gender<br />
(E) all of these outcomes are equally likely<br />
<br />
==Solution 1==<br />
We'll just start by breaking cases down. The probability of A occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. The probability of B occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>.<br />
<br />
The probability of C occurring is <math>\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}</math>, because we need to choose 2 of the 4 slots to be girls. <br />
<br />
For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is <math>\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}</math> because we need to choose 1 of the 4 slots to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is <math>\frac{1}{4} \cdot 2 = \frac{1}{2}</math>.<br />
<br />
<br />
So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}}.</math><br />
<br />
==Video Solution==<br />
https://youtu.be/3bF8BAvg0uY ~savannahsolver<br />
<br />
==See Also== <br />
{{AMC8 box|year=2014|num-b=17|num-a=19}} <br />
<br />
[[Category:Introductory Probability Problems]]<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_22&diff=1691712015 AMC 8 Problems/Problem 222022-01-03T04:29:42Z<p>Crystalflower: /* Video Solution */</p>
<hr />
<div>== Problem 22==<br />
<br />
On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br />
<br />
<math>\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080 </math><br />
<br />
==Solution ==<br />
As we read through this text, we find that the given information means that the number of students in the group has <math>12</math> factors, since each arrangement is a factor. The smallest integer with <math>12</math> factors is <math>2^2\cdot3\cdot5=\boxed{\textbf{(C) }60}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/E2om3xIvtYM<br />
<br />
~savannahsolver<br />
<br />
https://youtu.be/HISL2-N5NVg?t=5241<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_22&diff=1691702015 AMC 8 Problems/Problem 222022-01-03T04:29:29Z<p>Crystalflower: </p>
<hr />
<div>== Problem 22==<br />
<br />
On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br />
<br />
<math>\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080 </math><br />
<br />
==Solution ==<br />
As we read through this text, we find that the given information means that the number of students in the group has <math>12</math> factors, since each arrangement is a factor. The smallest integer with <math>12</math> factors is <math>2^2\cdot3\cdot5=\boxed{\textbf{(C) }60}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/E2om3xIvtYM<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_21&diff=1685892014 AMC 8 Problems/Problem 212021-12-27T23:38:05Z<p>Crystalflower: /* See Also */</p>
<hr />
<div>==Problem==<br />
The <math>7</math>-digit numbers <math>\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}</math> and <math>\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}</math> are each multiples of <math>3</math>. Which of the following could be the value of <math>C</math>?<br />
<br />
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8</math><br />
==Video Solution==<br />
https://youtu.be/6xNkyDgIhEE?t=2593<br />
<br />
==Solution 1==<br />
The sum of a number's digits<math>\mod{3}</math> is congruent to the number<math>\mod{3}</math>. <math>74A52B1 \mod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \mod{3}</math> is also congruent to 0. <math>7+4+5+2+1 \equiv 1 \pmod{3}</math>, so <math>A+B\equiv 2 \pmod{3}</math>. As we know, <math>326AB4C\equiv 0 \pmod{3}</math>, so <math>3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}</math>, and therefore <math>A+B+C\equiv 0 \pmod{3}</math>. We can substitute 2 for <math>A+B</math>, so <math>2+C\equiv 0 \pmod{3}</math>, and therefore <math>C\equiv 1\pmod{3}</math>. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is <math>\boxed{\textbf{(A) }1}</math>.<br />
<br />
==Solution 2==<br />
<br />
Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. <math>7 + 4 + 5 + 2 + 1 = 19</math>. To be a multiple of <math>3</math>, <math>A + B</math> has to be either <math>2</math> or <math>5</math> or <math>8</math>... and so on. We add up the numerical digits in the second number; <math>3 + 2 + 6 + 4 = 15</math>. We then add two of the selected values, <math>5</math> to <math>15</math>, to get <math>20</math>. We then see that C = <math>1, 4</math> or <math>7, 10</math>... and so on, otherwise the number will not be divisible by three. We then add <math>8</math> to <math>15</math>, to get <math>23</math>, which shows us that C = <math>1</math> or <math>4</math> or <math>7</math>... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be <math>1, 4,</math> and <math>7</math>. However, in the answer choices, there is no <math>7</math> or <math>4</math> or anything greater than <math>7</math>, but there is a <math>1</math>, so <math>\boxed{\textbf{(A) }1}</math> is our answer.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_21&diff=1685872014 AMC 8 Problems/Problem 212021-12-27T23:34:12Z<p>Crystalflower: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
The <math>7</math>-digit numbers <math>\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}</math> and <math>\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}</math> are each multiples of <math>3</math>. Which of the following could be the value of <math>C</math>?<br />
<br />
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8</math><br />
==Video Solution==<br />
https://youtu.be/6xNkyDgIhEE?t=2593<br />
<br />
==Solution 1==<br />
The sum of a number's digits<math>\mod{3}</math> is congruent to the number<math>\mod{3}</math>. <math>74A52B1 \mod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \mod{3}</math> is also congruent to 0. <math>7+4+5+2+1 \equiv 1 \pmod{3}</math>, so <math>A+B\equiv 2 \pmod{3}</math>. As we know, <math>326AB4C\equiv 0 \pmod{3}</math>, so <math>3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}</math>, and therefore <math>A+B+C\equiv 0 \pmod{3}</math>. We can substitute 2 for <math>A+B</math>, so <math>2+C\equiv 0 \pmod{3}</math>, and therefore <math>C\equiv 1\pmod{3}</math>. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is <math>\boxed{\textbf{(A) }1}</math>.<br />
<br />
==Solution 2==<br />
<br />
Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. <math>7 + 4 + 5 + 2 + 1 = 19</math>. To be a multiple of <math>3</math>, <math>A + B</math> has to be either <math>2</math> or <math>5</math> or <math>8</math>... and so on. We add up the numerical digits in the second number; <math>3 + 2 + 6 + 4 = 15</math>. We then add two of the selected values, <math>5</math> to <math>15</math>, to get <math>20</math>. We then see that C = <math>1, 4</math> or <math>7, 10</math>... and so on, otherwise the number will not be divisible by three. We then add <math>8</math> to <math>15</math>, to get <math>23</math>, which shows us that C = <math>1</math> or <math>4</math> or <math>7</math>... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be <math>1, 4,</math> and <math>7</math>. However, in the answer choices, there is no <math>7</math> or <math>4</math> or anything greater than <math>7</math>, but there is a <math>1</math>, so <math>\boxed{\textbf{(A) }1}</math> is our answer.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=20|num-a=22}}<br />
{{MAA Notice}}<br />
Another solution is using the divisbility rule of 3. Just add the digits of each number together.</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_18&diff=1685272014 AMC 8 Problems/Problem 182021-12-26T23:46:17Z<p>Crystalflower: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?<br />
<br />
<math> \textbf{(A) }\text{all 4 are boys}\\ \textbf{(B) }\text{all 4 are girls}\\ \textbf{(C) }\text{2 are girls and 2 are boys}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely} </math><br />
<br />
==Solution 1==<br />
We'll just start by breaking cases down. The probability of A occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. The probability of B occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>.<br />
<br />
The probability of C occurring is <math>\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}</math>, because we need to choose 2 of the 4 slots to be girls. <br />
<br />
For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is <math>\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}</math> because we need to choose 1 of the 4 slots to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is <math>\frac{1}{4} \cdot 2 = \frac{1}{2}</math>.<br />
<br />
<br />
So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}}.</math><br />
<br />
==Solution 2==<br />
We can also find out how many total cases there are for one solution. This will work, because before simplifying, the denominators of the fraction will be the same. Both <math>\text{A}</math> and <math>\text{B}</math> have <math>{4 \choose 0}=1</math> possibility. <math>\text{C}</math> will have <math>{4 \choose 2}=6</math> possibilities. <math>\text{D}</math> has <math>2\cdot{4 \choose 1}=8</math> possibilities (note that the problem did not say a specific gender.) Therefore, <math>\boxed{\left(\text{D}\right)\text{ 3 are of one gender and 1 is of the other gender}}</math> will have the greatest probability of occurring.<br />
<br />
==See Also== <br />
{{AMC8 box|year=2014|num-b=17|num-a=19}} <br />
<br />
[[Category:Introductory Probability Problems]]<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_11&diff=1685232014 AMC 8 Problems/Problem 112021-12-26T22:24:07Z<p>Crystalflower: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks?<br />
<br />
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad \textbf{(E) }10</math><br />
<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=rxrQLNxESW0<br />
<br />
==Solution 1==<br />
We can apply complementary counting and count the paths that DO go through the blocked intersection, which is <math>\dbinom{2}{1}\dbinom{3}{1}=6</math>. There are a total of <math>\dbinom{5}{2}=10</math> paths, so there are <math>10-6=4</math> paths possible. <math>\boxed{A}</math> is the correct answer.<br />
<br />
==Solution 2==<br />
We can make a diagram of the roads available to Jack.<br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(A--B--H--F--A);<br />
draw(D--C);<br />
draw(E--G);<br />
</asy><br />
Then, we can simply list the possible routes.<br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--H--F);<br />
draw(D--C,dotted);<br />
draw(E--G,dotted);<br />
draw(F--A--B,dotted);<br />
</asy><br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--C--E--G--F);<br />
draw(B--A--F,dotted);<br />
draw(D--E,dotted);<br />
draw(C--H--G,dotted);<br />
</asy><br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--C--D--F);<br />
draw(B--A--D,dotted);<br />
draw(E--G,dotted);<br />
draw(F--H--C,dotted);<br />
</asy><br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--A--F);<br />
draw(B--H--F,dotted);<br />
draw(D--C,dotted);<br />
draw(E--G,dotted);<br />
</asy><br />
There are 4 possible routes, so our answer is <math>\boxed{A}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_11&diff=1685222014 AMC 8 Problems/Problem 112021-12-26T22:23:05Z<p>Crystalflower: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks?<br />
<br />
<math>\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad \textbf{(E) }10</math><br />
<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=rxrQLNxESW0<br />
<br />
==Solution 1==<br />
We can apply complementary counting and count the paths that do go through the blocked intersection, which is <math>\dbinom{2}{1}\dbinom{3}{1}=6</math>. There are a total of <math>\dbinom{5}{2}=10</math> paths, so there are <math>10-6=4</math> paths possible. <math>\boxed{A}</math> is the correct answer.<br />
<br />
==Solution 2==<br />
We can make a diagram of the roads available to Jack.<br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(A--B--H--F--A);<br />
draw(D--C);<br />
draw(E--G);<br />
</asy><br />
Then, we can simply list the possible routes.<br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--H--F);<br />
draw(D--C,dotted);<br />
draw(E--G,dotted);<br />
draw(F--A--B,dotted);<br />
</asy><br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--C--E--G--F);<br />
draw(B--A--F,dotted);<br />
draw(D--E,dotted);<br />
draw(C--H--G,dotted);<br />
</asy><br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--C--D--F);<br />
draw(B--A--D,dotted);<br />
draw(E--G,dotted);<br />
draw(F--H--C,dotted);<br />
</asy><br />
<asy><br />
size(50);<br />
defaultpen(linewidth(0.8));<br />
pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0);<br />
draw(B--A--F);<br />
draw(B--H--F,dotted);<br />
draw(D--C,dotted);<br />
draw(E--G,dotted);<br />
</asy><br />
There are 4 possible routes, so our answer is <math>\boxed{A}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=File:The_new_aops_classroom.PNG&diff=163440File:The new aops classroom.PNG2021-10-13T00:11:32Z<p>Crystalflower: /* Summary */</p>
<hr />
<div>== Summary ==<br />
The new AoPS classroom! (Beta)</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Crystalflower&diff=140767User talk:Crystalflower2020-12-27T21:04:43Z<p>Crystalflower: </p>
<hr />
<div>Hello! welcome to my user talk lol</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Crystalflower&diff=140765User talk:Crystalflower2020-12-27T21:04:10Z<p>Crystalflower: Created page with ":D"</p>
<hr />
<div>:D</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=User:Crystalflower&diff=140764User:Crystalflower2020-12-27T21:03:36Z<p>Crystalflower: Created page with "hi"</p>
<hr />
<div>hi</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=User:Bsu1&diff=140762User:Bsu12020-12-27T21:00:22Z<p>Crystalflower: </p>
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<div>Hello!</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=Reaper&diff=140663Reaper2020-12-26T20:18:27Z<p>Crystalflower: /* FAQ */</p>
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<div>AoPS's famous button clicking game, Reaper.<br />
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To access any upcoming or current games, please hover your mouse over the FTW tab, and then click "Reaper". Alternatively, click [http://www.artofproblemsolving.com/reaper here].</div>Crystalflowerhttps://artofproblemsolving.com/wiki/index.php?title=Reaper&diff=140662Reaper2020-12-26T20:18:10Z<p>Crystalflower: </p>
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<div>AoPS's famous button clicking game, Reaper.<br />
<br />
To access any upcoming or current games, please hover your mouse over the FTW tab, and then click "Reaper". Alternatively, click [http://www.artofproblemsolving.com/reaper here].<br />
==FAQ==<br />
<i>Source:chaotic_iak</i><br /><br />
<b>What is Reaper?</b><br />
<p>Reaper is a game of strategy and time-management in AoPS. You can access it [http://www.artofproblemsolving.com/reaper/ here].</p><br />
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<b>How do I play?</b><br />
<p>Click to reap. The time shown on the timer will belong to you, and the timer will be reset. To win, meet the requirements, which is currently always "reach xyz hours/days first". Read the aforementioned sticky for more information.</p><br />
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<b>Why do I get much less time (0-10 seconds) than what the timer on Reaper shows?</b><br />
<p>It's called sniping and it's usual. Reaper is designed to update every 5-10 seconds, so in the meantime between a Reaper update and your reap, there might be another person reaped the time already (and Reaper hasn't shown this because it hasn't updated). Refresh your page before reaping to ensure Reaper has refreshed and you don't get sniped.<br />
<br>P.S. If you propose that Reaper should make its update rate higher, it will bring down the server.</p><br />
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<b>Why do I get much more time than what the timer on Reaper shows?</b><br />
<p>If you are lucky, you can get a multiplied reap. This is when the time you gain is multiplied by a certain number. A Double Reap is not that rare, but an Octuple Reap is!</p><br />
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<b>Where can I find who won games that aren't in the Hall of Fame?</b><br />
<p>Go to the [[Reaper Archives]].</p></div>Crystalflower