https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jkibbe&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T21:01:15ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_5&diff=2148562024 AMC 8 Problems/Problem 52024-02-08T13:35:34Z<p>Jkibbe: </p>
<hr />
<div>==Problem==<br />
Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of <math>6</math>. Which of the following integers cannot be the sum of the two numbers?<br />
<br />
<math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math><br />
<br />
==Solution 1==<br />
<br />
First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following: <br />
<br />
<math>\textbf{(A)}</math> is possible: <math>2\times 3</math> <br />
<br />
<math>\textbf{(C)}</math> is possible: <math>1\times 6</math> <br />
<br />
<math>\textbf{(D)}</math> is possible: <math>2\times 6</math> <br />
<br />
<math>\textbf{(E)}</math> is possible: <math>3\times 6</math><br />
<br />
The only integer that cannot be the sum is <math>\boxed{\textbf{(B)\ 6}}.</math><br />
<br />
-ILoveMath31415926535 & countmath1 & Nivaar<br />
<br />
==Video Solution 1 (easy to digest) by Power Solve==<br />
https://youtu.be/HE7JjZQ6xCk?si=2M33-oRy1ExY3_6l&t=239<br />
<br />
==Video Solution by Math-X (First fully understand the problem!!!)==<br />
https://youtu.be/BaE00H2SHQM?si=W8N4vwOx84KauOA6&t=1108<br />
<br />
~Math-X<br />
<br />
==Video Solution by NiuniuMaths (Easy to understand!)==<br />
https://www.youtube.com/watch?v=Ylw-kJkSpq8<br />
<br />
~NiuniuMaths<br />
<br />
==Video Solution 2 by SpreadTheMathLove==<br />
https://www.youtube.com/watch?v=L83DxusGkSY<br />
<br />
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==<br />
<br />
https://www.youtube.com/watch?v=51pqs80PUnY<br />
==Video Solution by Interstigation==<br />
https://youtu.be/ktzijuZtDas&t=296<br />
<br />
==See Also==<br />
{{AMC8 box|year=2024|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_21&diff=1915652023 AMC 8 Problems/Problem 212023-03-30T12:03:11Z<p>Jkibbe: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Alina writes the numbers <math>1, 2, \dots , 9</math> on separate cards, one number per card. She wishes to divide the cards into <math>3</math> groups of <math>3</math> cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?<br />
<br />
<math>\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math><br />
<br />
==Solution 1==<br />
<br />
First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. <math>1 + 2 \cdots + 9 = \frac{9(10)}{2} = 45</math>. Then, dividing by <math>3</math>, we have <math>\frac{45}{3} = 15</math>, so each group of <math>3</math> must have a sum of 15. To make the counting easier, we we will just see the possible groups 9 can be with. The possible groups 9 can be with with 2 distinct numbers are <math>(9, 2, 4)</math> and <math>(9, 1, 5)</math>. Going down each of these avenues, we will repeat the same process for <math>8</math> using the remaining elements in the list. Where there is only 1 set of elements getting the sum of <math>7</math>, <math>8</math> needs in both cases. After <math>8</math> is decided, the remaining 3 elements are forced in a group, yielding us an answer of <math>\boxed{\textbf{(C)}\ 2}</math> as our sets are <math>(9, 1, 5) (8, 3, 4) (7, 2, 6)</math> and <math>(9, 2, 4) (8, 1, 6) (7, 3 ,5)</math><br />
<br />
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat<br />
<br />
==Solution 2==<br />
The group with <math>5</math> must have the two other numbers adding up to <math>10</math>, since the sum of all the numbers is <math>(1 + 2 \cdots + 9) = \frac{9(10)}{2} = 45</math>. The sum of the numbers in each group must therefore be <math>\frac{45}{3}=15</math>. We can have <math>(1, 5, 9)</math>, <math>(2, 5, 8)</math>, <math>(3, 5, 7)</math>, or <math>(4, 5, 6)</math>. With the first group, we have <math>(2, 3, 4, 6, 7, 8)</math> left over. The only way to form a group of <math>3</math> numbers that add up to <math>15</math> is with <math>(3, 4, 8)</math> or <math>(2, 6, 7)</math>. One of the possible arrangements is therefore <math>(1, 5, 9) (3, 4, 8) (2, 6, 7)</math>. Then, with the second group, we have <math>(1, 3, 4, 6, 7, 9)</math> left over. With these numbers, there is no way to form a group of <math>3</math> numbers adding to <math>15</math>. Similarly, with the third group there is <math>(1, 2, 4, 6, 8, 9)</math> left over and we can make a group of <math>3</math> numbers adding to <math>15</math> with <math>(1, 6, 8)</math> or <math>(2, 4, 9)</math>. Another arrangement is <math>(3, 5, 7) (1, 6, 8) (2, 4, 9)</math>. Finally, the last group has <math>(1, 2, 3, 7, 8, 9)</math> left over. There is no way to make a group of <math>3</math> numbers adding to <math>15</math> with this, so the arrangements are <math>(1, 5, 9) (3, 4, 8) (2, 6, 7)</math> and <math>(3, 5, 7) (1, 6, 8) (2, 4, 9)</math>. There are <math>\boxed{\textbf{(C)}\ 2}</math> sets that can be formed. <br />
<br />
~Turtwig113<br />
<br />
==Video Solution 1 by OmegaLearn (Using Casework)==<br />
https://youtu.be/l1MfKj5MkWg<br />
<br />
==Animated Video Solution==<br />
https://youtu.be/_gpWj2lYers<br />
<br />
~Star League (https://starleague.us)<br />
<br />
==Video Solution by Magic Square==<br />
https://youtu.be/-N46BeEKaCQ?t=2853<br />
==Video Solution by Interstigation==<br />
https://youtu.be/1bA7fD7Lg54?t=2062<br />
<br />
==Video Solution by WhyMath==<br />
https://youtu.be/l9zexK9hiBo<br />
<br />
~savannahsolver<br />
<br />
==See Also== <br />
{{AMC8 box|year=2023|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_14&diff=927652017 AMC 8 Problems/Problem 142018-03-06T15:04:34Z<p>Jkibbe: /* Solution 2 */ two more 'he' -> 'she' -- both are female names as given in problem</p>
<hr />
<div>==Problem 14==<br />
<br />
Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only <math>80\%</math> of the problems she solved alone, but overall <math>88\%</math> of her answers were correct. Zoe had correct answers to <math>90\%</math> of the problems she solved alone. What was Zoe's overall percentage of correct answers?<br />
<br />
<math>\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98</math><br />
<br />
==Solution 1==<br />
<br />
Let the number of questions that they solved alone be <math>x</math>. Let the percentage of problems they correctly solve together be <math>a</math>%. <br />
As given, <cmath>\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}</cmath>. <br />
<br />
Hence, <math>a = 96</math>. <br />
<br />
Zoe got <math>\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}</math> problems right out of <math>2x</math>. Therefore, Zoe got <math>\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{\textbf{(C) } 93}</math> percent of the problems correct.<br />
<br />
==Solution 2==<br />
<br />
Assume the total amount of problems is <math>100</math> per half homework assignment, since we are dealing with percentages, and no values. Then, we know that Chloe got <math>80</math> problems correct by herself, and got <math>176</math> problems correct overall. We also know that Zoe had <math>90</math> problems she did alone correct. We can see that the total amount of correct problems Chloe had when Zoe and she did the homework together is <math>176-80=96</math>, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. Therefore Zoe has <math>96+90=186</math> problems out of <math>200</math> problems correct. This is <math>\boxed{\textbf{(C) } 93}</math> percent.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2017|num-b=13|num-a=15}}<br />
<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_14&diff=927642017 AMC 8 Problems/Problem 142018-03-06T15:00:29Z<p>Jkibbe: /* Solution 2 */ gender 'he' -> 'she'</p>
<hr />
<div>==Problem 14==<br />
<br />
Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only <math>80\%</math> of the problems she solved alone, but overall <math>88\%</math> of her answers were correct. Zoe had correct answers to <math>90\%</math> of the problems she solved alone. What was Zoe's overall percentage of correct answers?<br />
<br />
<math>\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98</math><br />
<br />
==Solution 1==<br />
<br />
Let the number of questions that they solved alone be <math>x</math>. Let the percentage of problems they correctly solve together be <math>a</math>%. <br />
As given, <cmath>\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}</cmath>. <br />
<br />
Hence, <math>a = 96</math>. <br />
<br />
Zoe got <math>\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}</math> problems right out of <math>2x</math>. Therefore, Zoe got <math>\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{\textbf{(C) } 93}</math> percent of the problems correct.<br />
<br />
==Solution 2==<br />
<br />
Assume the total amount of problems is <math>100</math> per half homework assignment, since we are dealing with percentages, and no values. Then, we know that Chloe got <math>80</math> problems correct by herself, and got <math>176</math> problems correct overall. We also know that Zoe had <math>90</math> problems she did alone correct. We can see that the total amount of correct problems Chloe had when Zoe and she did the homework together is <math>176-80=96</math>, which is the total amount of problems he got correct, subtracted by the number of correct problems he did alone. Therefore Zoe has <math>96+90=186</math> problems out of <math>200</math> problems correct. This is <math>\boxed{\textbf{(C) } 93}</math> percent.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2017|num-b=13|num-a=15}}<br />
<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems&diff=911692017 AMC 8 Problems2018-02-13T14:33:48Z<p>Jkibbe: /* Problem 9 */ extra info in previous version made problem too easy. updated to match actual question</p>
<hr />
<div>==Problem 1==<br />
<br />
Which of the following values is largest?<br />
<br />
<math>\textbf{(A) }2+0+1+7\qquad\textbf{(B) }2 \times 0 +1+7\qquad\textbf{(C) }2+0 \times 1 + 7\qquad\textbf{(D) }2+0+1 \times 7\qquad\textbf{(E) }2 \times 0 \times 1 \times 7</math><br />
<br />
[[2017 AMC 8 Problems/Problem 1|Solution<br />
]]<br />
<br />
==Problem 2==<br />
<br />
Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received 36 votes, then how many votes were cast all together?<br />
<asy><br />
draw((-1,0)--(0,0)--(0,1));<br />
draw((0,0)--(0.309, -0.951));<br />
filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray);<br />
filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray);<br />
draw(arc((0,0), (-1,0), (0.309, -0.951)));<br />
label("Colby", (-0.5, 0.5));<br />
label("25\%", (-0.5, 0.3));<br />
label("Alicia", (0.7, 0.2));<br />
label("45\%", (0.7, 0));<br />
label("Brenda", (-0.5, -0.4));<br />
label("30\%", (-0.5, -0.6));<br />
</asy><br />
<br />
<math>\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }100\qquad\textbf{(D) }106\qquad\textbf{(E) }120</math><br />
<br />
[[2017 AMC 8 Problems/Problem 2|Solution<br />
]]<br />
<br />
==Problem 3==<br />
<br />
What is the value of the expression <math>\sqrt{16\sqrt{8\sqrt{4}}}</math>?<br />
<br />
<math>\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16</math><br />
<br />
[[2017 AMC 8 Problems/Problem 3|Solution<br />
]]<br />
<br />
==Problem 4==<br />
<br />
When 0.000315 is multiplied by 7,928,564 the product is closest to which of the following?<br />
<br />
<math>\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000</math><br />
<br />
[[2017 AMC 8 Problems/Problem 4|Solution<br />
]]<br />
<br />
==Problem 5==<br />
<br />
What is the value of the expression <math>\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}</math>?<br />
<br />
<math>\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360</math><br />
<br />
[[2017 AMC 8 Problems/Problem 5|Solution<br />
]]<br />
<br />
==Problem 6==<br />
<br />
If the degree measures of the angles of a triangle are in the ratio <math>3:3:4</math>, what is the degree measure of the largest angle of the triangle?<br />
<br />
<math>\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90</math><br />
<br />
[[2017 AMC 8 Problems/Problem 6|Solution<br />
]]<br />
<br />
==Problem 7==<br />
<br />
Let <math>Z</math> be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of <math>Z</math>?<br />
<br />
<math>\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111</math><br />
<br />
[[2017 AMC 8 Problems/Problem 7|Solution<br />
]]<br />
<br />
==Problem 8==<br />
<br />
Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."<br />
<br />
(1) It is prime.<br />
<br />
(2) It is even.<br />
<br />
(3) It is divisible by 7.<br />
<br />
(4) One of its digits is 9.<br />
<br />
This information allows Malcolm to determine Isabella's house number. What is its units digit?<br />
<br />
<math>\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math><br />
<br />
[[2017 AMC 8 Problems/Problem 8|Solution<br />
]]<br />
<br />
==Problem 9==<br />
<br />
All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?<br />
<br />
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math><br />
<br />
[[2017 AMC 8 Problems/Problem 9|Solution<br />
]]<br />
<br />
==Problem 10==<br />
<br />
A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?<br />
<br />
<math>\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}</math><br />
<br />
[[2017 AMC 8 Problems/Problem 10|Solution<br />
]]<br />
<br />
==Problem 11==<br />
<br />
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?<br />
<br />
<math>\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369</math><br />
<br />
[[2017 AMC 8 Problems/Problem 11|Solution<br />
]]<br />
<br />
==Problem 12==<br />
<br />
The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?<br />
<br />
<math>\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124</math><br />
<br />
[[2017 AMC 8 Problems/Problem 12|Solution<br />
]]<br />
<br />
==Problem 13==<br />
<br />
Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4</math><br />
<br />
[[2017 AMC 8 Problems/Problem 13|Solution<br />
]]<br />
<br />
==Problem 14==<br />
<br />
Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only <math>80\%</math> of the problems she solved alone, but overall <math>88\%</math> of her answers were correct. Zoe had correct answers to <math>90\%</math> of the problems she solved alone. What was Zoe's overall percentage of correct answers?<br />
<br />
<math>\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98</math><br />
<br />
[[2017 AMC 8 Problems/Problem 14|Solution<br />
]]<br />
<br />
==Problem 15==<br />
<br />
In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.<br />
<asy><br />
fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray);<br />
fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray);<br />
label("$8$", (1, 0));<br />
label("$C$", (2, 0));<br />
label("$8$", (3, 0));<br />
label("$8$", (0, 1));<br />
label("$C$", (1, 1));<br />
label("$M$", (2, 1));<br />
label("$C$", (3, 1));<br />
label("$8$", (4, 1));<br />
label("$C$", (0, 2));<br />
label("$M$", (1, 2));<br />
label("$A$", (2, 2));<br />
label("$M$", (3, 2));<br />
label("$C$", (4, 2));<br />
label("$8$", (0, 3));<br />
label("$C$", (1, 3));<br />
label("$M$", (2, 3));<br />
label("$C$", (3, 3));<br />
label("$8$", (4, 3));<br />
label("$8$", (1, 4));<br />
label("$C$", (2, 4));<br />
label("$8$", (3, 4));</asy><br />
<br />
<math>\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36</math><br />
<br />
[[2017 AMC 8 Problems/Problem 15|Solution<br />
]]<br />
<br />
==Problem 16==<br />
<br />
In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>?<br />
<asy>draw((0,0)--(4,0)--(0,3)--(0,0));<br />
label("$A$", (0,0), SW);<br />
label("$B$", (4,0), ESE);<br />
label("$C$", (0, 3), N);<br />
label("$3$", (0, 1.5), W);<br />
label("$4$", (2, 0), S);<br />
label("$5$", (2, 1.5), NE);</asy><br />
<br />
<math>\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}</math><br />
<br />
[[2017 AMC 8 Problems/Problem 16|Solution<br />
]]<br />
<br />
==Problem 17==<br />
<br />
Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?<br />
<br />
<math>\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81</math><br />
<br />
[[2017 AMC 8 Problems/Problem 17|Solution<br />
]]<br />
<br />
==Problem 18==<br />
<br />
In the non-convex quadrilateral <math>ABCD</math> shown below, <math>\angle BCD</math> is a right angle, <math>AB=12</math>, <math>BC=4</math>, <math>CD=3</math>, and <math>AD=13</math>.<br />
<asy>draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0));<br />
label("$B$", (0, 0), SW);<br />
label("$A$", (12, 0), ESE);<br />
label("$C$", (2.4, 3.6), SE);<br />
label("$D$", (0, 5), N);</asy><br />
What is the area of quadrilateral <math>ABCD</math>?<br />
<br />
<math>\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36</math><br />
<br />
[[2017 AMC 8 Problems/Problem 18|Solution<br />
]]<br />
<br />
==Problem 19==<br />
<br />
For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ?<br />
<br />
<math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math><br />
<br />
[[2017 AMC 8 Problems/Problem 19|Solution<br />
]]<br />
<br />
==Problem 20==<br />
<br />
An integer between <math>1000</math> and <math>9999</math>, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?<br />
<br />
<math>\textbf{(A) }\frac{14}{75}\qquad\textbf{(B) }\frac{56}{225}\qquad\textbf{(C) }\frac{107}{400}\qquad\textbf{(D) }\frac{7}{25}\qquad\textbf{(E) }\frac{9}{25}</math><br />
<br />
[[2017 AMC 8 Problems/Problem 20|Solution<br />
]]<br />
<br />
==Problem 21==<br />
<br />
Suppose <math>a</math>, <math>b</math>, and <math>c</math> are nonzero real numbers, and <math>a+b+c=0</math>. What are the possible value(s) for <math>\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}</math>?<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1</math><br />
<br />
[[2017 AMC 8 Problems/Problem 21|Solution<br />
]]<br />
<br />
==Problem 22==<br />
<br />
In the right triangle <math>ABC</math>, <math>AC=12</math>, <math>BC=5</math>, and angle <math>C</math> is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br />
<asy><br />
draw((0,0)--(12,0)--(12,5)--(0,0));<br />
draw(arc((8.67,0),(12,0),(5.33,0)));<br />
label("$A$", (0,0), W);<br />
label("$C$", (12,0), E);<br />
label("$B$", (12,5), NE);<br />
label("$12$", (6, 0), S);<br />
label("$5$", (12, 2.5), E);</asy><br />
<br />
<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}</math><br />
<br />
[[2017 AMC 8 Problems/Problem 22|Solution<br />
]]<br />
<br />
==Problem 23==<br />
<br />
Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?<br />
<br />
<math>\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82</math><br />
<br />
[[2017 AMC 8 Problems/Problem 23|Solution<br />
]]<br />
<br />
==Problem 24==<br />
<br />
Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?<br />
<br />
<math>\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152</math><br />
<br />
[[2017 AMC 8 Problems/Problem 24|Solution<br />
]]<br />
<br />
==Problem 25==<br />
<br />
In the figure shown, <math>\overline{US}</math> and <math>\overline{UT}</math> are line segments each of length 2, and <math>m\angle TUS = 60^\circ</math>. Arcs <math>\overarc{TR}</math> and <math>\overarc{SR}</math> are each one-sixth of a circle with radius 2. What is the area of the region shown?<br />
<asy>draw((1,1.732)--(2,3.464)--(3,1.732));<br />
draw(arc((0,0),(2,0),(1,1.732)));<br />
draw(arc((4,0),(3,1.732),(2,0)));<br />
label("$U$", (2,3.464), N);<br />
label("$S$", (1,1.732), W);<br />
label("$T$", (3,1.732), E);<br />
label("$R$", (2,0), S);</asy><br />
<br />
<math>\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}</math><br />
<br />
[[2017 AMC 8 Problems/Problem 25|Solution<br />
]]<br />
<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_7&diff=900752017 AMC 8 Problems/Problem 72018-01-30T14:25:55Z<p>Jkibbe: added solution from https://artofproblemsolving.com/community/c5h1549286p9414164</p>
<hr />
<div>==Problem 7==<br />
<br />
Let <math>Z</math> be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of <math>Z</math>?<br />
<br />
<math>\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.</math> Clearly, <math>Z</math> is divisible by <math>\boxed{\textbf{(A)}\ 11}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can see that numbers like <math>247247</math> can be written as <math>ABCABC</math>. We can see that the alternating sum of digits is <math>C-B+A-C+B-A</math>, which is <math>0</math>. Because <math>0</math> is a multiple of <math>11</math>, any number <math>ABCABC</math> is a multiple of <math>11</math>, so the answer is <math>A</math><br />
<br />
-Baolan (hi MVMS)<br />
<br />
==Solution 3==<br />
<br />
The most important step is to realize that any number in the form <math>\overline{ABCABC} = \overline{ABC000}+\overline{ABC} = 1000\overline{ABC}+\overline{ABC} = 1001\overline{ABC}</math>. Thus every number in this form is divisible by <math>1001</math>, and the answer is <math>\textbf{(A) }11</math>, because it is the only choice that is a factor of <math>1001</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2017|num-b=6|num-a=8}}<br />
<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_24&diff=724642000 AMC 8 Problems/Problem 242015-10-14T11:25:58Z<p>Jkibbe: /* Solution */ typo - 'striaght'</p>
<hr />
<div>==Problem==<br />
<br />
If <math> \angle A = 20^\circ </math> and <math> \angle AFG =\angle AGF </math>, then <math> \angle B+\angle D = </math><br />
<br />
<asy><br />
pair A,B,C,D,EE,F,G;<br />
A = (0,0);<br />
B = (9,4);<br />
C = (21,0);<br />
D = (13,-12);<br />
EE = (4,-16);<br />
F = (13/2,-6);<br />
G = (8,0);<br />
draw(A--C--EE--B--D--cycle);<br />
label("$A$",A,W);<br />
label("$B$",B,N);<br />
label("$C$",C,E);<br />
label("$D$",D,SE);<br />
label("$E$",EE,SW);<br />
label("$F$",F,WSW);<br />
label("$G$",G,NW);</asy><br />
<br />
<math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math><br />
<br />
==Solution==<br />
<br />
As a strategy, think of how <math>\angle B + \angle D</math> would be determined, particularly without determining either of the angles individually, since it may not be possible to determine <math>\angle B</math> or <math>\angle D</math> alone. If you see <math>\triangle BFD</math>, the you can see that the problem is solved quickly after determining <math>\angle BFD</math>.<br />
<br />
But start with <math>\triangle AGF</math>, since that's where most of our information is. Looking at <math>\triangle AGF</math>, since <math>\angle F = \angle G</math>, and <math>\angle A = 20</math>, we can write:<br />
<br />
<math>\angle A + \angle G + \angle F = 180</math><br />
<br />
<math>20 + 2\angle F = 180</math><br />
<br />
<math>\angle AFG = 80</math><br />
<br />
By noting that <math>\angle AFG</math> and <math>\angle GFD</math> make a straight line, we know<br />
<br />
<math>\angle AFG + \angle GFD = 180</math><br />
<br />
<math>80 + \angle GFD = 180</math><br />
<br />
<math>\angle GFD = 100</math><br />
<br />
Ignoring all other parts of the figure and looking only at <math>\triangle BFD</math>, you see that <math>\angle B + \angle D + \angle F = 180</math>. But <math>\angle F</math> is the same as <math>\angle GFD</math>. Therefore:<br />
<br />
<math>\angle B + \angle D + \angle GFD = 180</math><br />
<math>\angle B + \angle D + 100 = 180</math><br />
<math>\angle B + \angle D = 80^\circ</math>, and the answer is thus <math>\boxed{D}</math><br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_10&diff=719232000 AMC 8 Problems/Problem 102015-09-03T14:53:26Z<p>Jkibbe: /* Problem */ typo</p>
<hr />
<div>==Problem==<br />
<br />
Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?<br />
<br />
<math>\text{(A)}\ 48 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55</math><br />
<br />
==Solution==<br />
<br />
Shea has grown <math>20\%</math>, so she was originally <math>\frac{60}{1.2}=50</math> inches tall which is a <math>60 - 50 = 10</math> inch increase. Ara also started off at <math>50</math> inches. Since Ara grew half as much as Shea, Ara grew <math>10 \div 2 = 5</math> inches. Therefore, Ara is now <math>50+5=55</math> inches tall which is choice <math>\boxed{E}.</math><br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jkibbehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_20&diff=435702011 AMC 8 Problems/Problem 202011-12-02T14:56:58Z<p>Jkibbe: misspelling of 'trapezoid'</p>
<hr />
<div>Quadrilateral <math>ABCD</math> is a trapezoid, <math>AD = 15</math>, <math>AB = 50</math>, <math>BC = 20</math>, and the altitude is <math>12</math>. What is the area of the trapezoid?<br />
<br />
<asy><br />
pair A,B,C,D;<br />
A=(3,20);<br />
B=(35,20);<br />
C=(47,0);<br />
D=(0,0);<br />
draw(A--B--C--D--cycle);<br />
dot((0,0));<br />
dot((3,20));<br />
dot((35,20));<br />
dot((47,0));<br />
label("A",A,N);<br />
label("B",B,N);<br />
label("C",C,S);<br />
label("D",D,S);<br />
draw((19,20)--(19,0));<br />
dot((19,20));<br />
dot((19,0));<br />
draw((19,3)--(22,3)--(22,0));<br />
label("12",(21,10),E);<br />
label("50",(19,22),N);<br />
label("15",(1,10),W);<br />
label("20",(41,12),E);</asy><br />
<br />
<math> \textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800 </math><br />
<br />
==Solution==<br />
<br />
<asy><br />
unitsize(1.5mm);<br />
defaultpen(linewidth(.9pt)+fontsize(10pt));<br />
dotfactor=3;<br />
<br />
pair A,B,C,D,X,Y;<br />
A=(9,12); B=(59,12); C=(75,0); D=(0,0); X=(9,0); Y=(59,0);<br />
draw(A--B--C--D--cycle);<br />
draw(A--X); draw(B--Y);<br />
<br />
pair[] ps={A,B,C,D,X,Y};<br />
dot(ps);<br />
<br />
label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);<br />
label("$X$",X,SE); label("$Y$",Y,S);<br />
label("$a$",D--X,S); label("$b$",Y--C,S);<br />
label("$15$",D--A,NW); label("$50$",B--A,N); label("$20$",B--C,NE); label("$12$",X--A,E); label("$12$",Y--B,W);<br />
</asy><br />
<br />
If you draw altitudes from <math>A</math> and <math>B</math> to <math>CD,</math> the trapezoid will be divided into two right triangles and a rectangle. You can find the values of <math>a</math> and <math>b</math> with the [[Pythagorean theorem]].<br />
<br />
<cmath>a=\sqrt{15^2-12^2}=\sqrt{81}=9</cmath><br />
<br />
<cmath>b=\sqrt{20^2-12^2}=\sqrt{256}=16</cmath><br />
<br />
<math>ABYX</math> is a rectangle so <math>XY=AB=50.</math><br />
<br />
<cmath>CD=a+XY+b=9+50+16=75</cmath><br />
<br />
The area of the trapezoid is<br />
<br />
<cmath>12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{\textbf{(D)}\ 750}</cmath><br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=19|num-a=21}}</div>Jkibbe