https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Technetiumdolomite&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T21:53:16ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_17&diff=1308751998 AJHSME Problems/Problem 172020-08-06T23:57:51Z<p>Technetiumdolomite: /* Solution 2 */</p>
<hr />
<div>==Don't Crowd the Isles==<br />
<br />
Problems 15, 16, and 17 all refer to the following:<br />
<br />
<center><br />
In the very center of the Irenic Sea lie the beautiful Nisos Isles. In 1998 the number of people on these islands is only 200, but the population triples every 25 years. Queen Irene has decreed that there must be at least 1.5 square miles for every person living in the Isles. The total area of the Nisos Isles is 24,900 square miles.<br />
</center><br />
<br />
===Problem 17===<br />
<br />
In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?<br />
<br />
<math>\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}</math><br />
<br />
==Solution 1==<br />
<br />
We can divide the total area by how much will be occupied per person:<br />
<br />
<math>\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}</math> can stay on the island at its maximum capacity.<br />
<br />
We can divide this by the current population the island in the year <math>1998</math> to see by what factor the population increases:<br />
<br />
<math>\frac{16600}{200}=83</math>-fold increase in population.<br />
<br />
Thus, the population increases by a factor <math>83</math>. This is very close to <math>3 \times 3 \times 3 \times 3 = 81</math>, and so there are about <math>4</math> triplings of the island's population.<br />
<br />
It takes <math>4\times25=100=\boxed{C}</math> years to triple the island's population four times in succession.<br />
<br />
==Solution 2==<br />
<br />
We can continue the pattern, and because the pattern increases numbers rapidly, it won't be hard. <math>\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}</math> can live on the island at its maximum capacity. <br />
<br />
<math>1998: 200 \text{ people}</math><br />
<br><br />
<math>2023: 600 \text{ people}</math><br />
<br><br />
<math>2048: 1800 \text{ people}</math><br />
<br><br />
<math>2073: 5400 \text{ people}</math><br />
<br><br />
<math>2098: 16200 \text{ people}</math><br />
<br><br />
After the year <math>2098</math>, it will not be possible for the next increase to occur, because when <math>16200</math> is tripled, it is way more than the maximum capacity.<br />
<br><br />
Thus, the answer is <math>2098-1998 = 100</math>, or <math>\boxed{C}</math><br />
<br />
== See also ==<br />
{{AJHSME box|year=1998|num-b=16|num-a=18}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_9&diff=1308731998 AJHSME Problems/Problem 92020-08-06T23:49:22Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
For a sale, a store owner reduces the price of a <math>\$10</math> scarf by <math>20\% </math>. Later the price is lowered again, this time by one-half the reduced price. The price is now<br />
<br />
<math>\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}</math><br />
<br />
==Solution==<br />
===Solution 1===<br />
<br />
<math>100\%-20\%=80\%</math><br />
<br />
<math>10\times80\%=10\times0.8</math><br />
<br />
<math>10\times0.8=8</math><br />
<br />
<math>\frac{8}{2}=4=\boxed{C}</math><br />
<br />
===Solution 2===<br />
<br />
The first discount has percentage 20, which is then discounted again for half of the already discounted price.<br />
<br />
<math>100-20=80</math><br />
<br />
<math>\frac{80}{2}=40</math><br />
<br />
<math>40\%\times10=10\times0.4=4=\boxed{C}</math><br />
<br />
== See also ==<br />
{{AJHSME box|year=1998|num-b=8|num-a=10}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_12&diff=1275482003 AMC 8 Problems/Problem 122020-07-05T23:20:55Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>==Problem==<br />
When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces that can be seen is divisible by <math>6</math>?<br />
<br />
<math> \textbf{(A)}\ 1/3\qquad\textbf{(B)}\ 1/2\qquad\textbf{(C)}\ 2/3\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 1 </math><br />
<br />
==Solution==<br />
All the possibilities where <math>6</math> is on any of the five sides is always divisible by six, and <math>1 \times 2 \times 3 \times 4 \times 5</math> is divisible by <math>6</math> since <math>2 \times 3 = 6</math>. So, the answer is <math>\boxed{\textbf{(E)}\ 1}</math> because the outcome is always divisible by <math>6</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2003|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_9&diff=1222982001 AMC 8 Problems/Problem 92020-05-11T23:57:48Z<p>Technetiumdolomite: </p>
<hr />
<div><br />
<br />
==Problem==<br />
To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.<br />
<br />
<asy><br />
for (int a = 0; a < 7; ++a)<br />
{<br />
for (int b = 0; b < 8; ++b)<br />
{<br />
dot((a,b));<br />
}<br />
}<br />
<br />
draw((3,0)--(0,5)--(3,7)--(6,5)--cycle);<br />
</asy><br />
The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?<br />
<br />
<math>\text{(A)}\ 63 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 180 \qquad \text{(D)}\ 189 \qquad \text{(E)}\ 264</math><br />
<br />
==Solution==<br />
<br />
<br />
The large grid has dimensions three times that of the small grid, so its dimensions are <math> 3(6)\times3(7) </math>, or <math> 18\times21 </math>, so the area is <math> (18)(21)=378 </math>. The area of the kite is half the product of its diagonals, and the diagonals are the dimensions of the rectangle, so the area of the kite is <math> \frac{(18)(21)}{2}=189 </math>. Thus, the area of the remaining gold is <math> 378-189=189, \boxed{\text{D}} </math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2001|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_8&diff=1222972001 AMC 8 Problems/Problem 82020-05-11T23:53:33Z<p>Technetiumdolomite: </p>
<hr />
<div>==Problem==<br />
To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.<br />
<br />
<asy><br />
for (int a = 0; a < 7; ++a)<br />
{<br />
for (int b = 0; b < 8; ++b)<br />
{<br />
dot((a,b));<br />
}<br />
}<br />
<br />
draw((3,0)--(0,5)--(3,7)--(6,5)--cycle);<br />
</asy><br />
<br />
Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need?<br />
<br />
<math>\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 38 \qquad \text{(E)}\ 39</math><br />
<br />
==Solution==<br />
<br />
Each diagonal of the large kite is <math> 3 </math> times the length of the corresponding diagonal of the short kite since it was made with a grid <math> 3 </math> times as long in each direction. The diagonals of the small kite are <math> 6 </math> and <math> 7 </math>, so the diagonals of the large kite are <math> 18 </math> and <math> 21 </math>, and the amount of bracing Genevieve needs is the sum of these lengths, which is <math> 39, \boxed{\text{E}} </math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2001|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_7&diff=1222962001 AMC 8 Problems/Problem 72020-05-11T23:50:20Z<p>Technetiumdolomite: </p>
<hr />
<div>==Problem==<br />
To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.<br />
<asy><br />
for (int a = 0; a < 7; ++a)<br />
{<br />
for (int b = 0; b < 8; ++b)<br />
{<br />
dot((a,b));<br />
}<br />
}<br />
<br />
draw((3,0)--(0,5)--(3,7)--(6,5)--cycle);<br />
</asy><br />
<br />
What is the number of square inches in the area of the small kite?<br />
<br />
<math>\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25</math><br />
==Solution 1==<br />
The area of a kite is half the product of its diagonals. The diagonals have lengths of <math> 6 </math> and <math> 7 </math>, so the area is <math> \frac{(6)(7)}{2}=21, \boxed{\text{A}} </math>.<br />
<br />
==Solution 2==<br />
Drawing in the diagonals of the kite will form four right triangles on the "inside" part of the grid. Drawing in the border of the 7 by 6 grid will form four right triangles on the "outside" part of the grid. Since each right triangle on the inside can be paired with a congruent right triangle that is on the outside, the area of the kite is half the total area of the grid, or <math> \frac{(6)(7)}{2}=21, \boxed{\text{A}} </math>.<br />
<br />
==Solution 3==<br />
Pick's Theorem states: <cmath>\frac{\text{number of boundary points}}{2}+\text{number of interior points}-1</cmath> as the area of a figure on a grid. Counting, we see there are <math>4</math> boundary points and <math>20</math> interior points. Therefore, we have <cmath>\frac{4}{2}+20-1\implies 20+1\implies 21.</cmath> Hence, the answer is <math>\boxed{\text{(A)}}</math> <math>\hspace{0.2cm} \square</math><br />
==See Also==<br />
{{AMC8 box|year=2001|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_11&diff=1221222002 AMC 8 Problems/Problem 112020-05-07T03:26:50Z<p>Technetiumdolomite: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?<br />
<br />
<asy><br />
path p=origin--(1,0)--(1,1)--(0,1)--cycle;<br />
draw(p);<br />
draw(shift(3,0)*p);<br />
draw(shift(3,1)*p);<br />
draw(shift(4,0)*p);<br />
draw(shift(4,1)*p);<br />
draw(shift(7,0)*p);<br />
draw(shift(7,1)*p);<br />
draw(shift(7,2)*p);<br />
draw(shift(8,0)*p);<br />
draw(shift(8,1)*p);<br />
draw(shift(8,2)*p);<br />
draw(shift(9,0)*p);<br />
draw(shift(9,1)*p);<br />
draw(shift(9,2)*p);</asy><br />
<br />
<math>\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
The first square has a sidelength of <math>1</math>, the second square <math>2</math>, and so on. The seventh square has <math>7</math> and is made of <math>7^2=49</math> unit tiles. The sixth square has <math>6</math> and is made of <math>6^2=36</math> unit tiles. The seventh square has <math>49-36=\boxed{\text{(C)}\ 13}</math> more tiles than the sixth square.<br />
<br />
===Solution 2===<br />
The edge of each square is one tile length longer than the edge of the previous square, which means that each square has <math>2*</math> edge length <math>- 1</math> more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh square has <math>2*7-1=13</math> more tiles than the sixth square, which is <math>\boxed{\text{(C)}\ 13}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_11&diff=1221212002 AMC 8 Problems/Problem 112020-05-07T03:25:04Z<p>Technetiumdolomite: </p>
<hr />
<div>==Problem==<br />
<br />
A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?<br />
<br />
<asy><br />
path p=origin--(1,0)--(1,1)--(0,1)--cycle;<br />
draw(p);<br />
draw(shift(3,0)*p);<br />
draw(shift(3,1)*p);<br />
draw(shift(4,0)*p);<br />
draw(shift(4,1)*p);<br />
draw(shift(7,0)*p);<br />
draw(shift(7,1)*p);<br />
draw(shift(7,2)*p);<br />
draw(shift(8,0)*p);<br />
draw(shift(8,1)*p);<br />
draw(shift(8,2)*p);<br />
draw(shift(9,0)*p);<br />
draw(shift(9,1)*p);<br />
draw(shift(9,2)*p);</asy><br />
<br />
<math>\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
The first square has a sidelength of <math>1</math>, the second square <math>2</math>, and so on. The seventh square has <math>7</math> and is made of <math>7^2=49</math> unit tiles. The sixth square has <math>6</math> and is made of <math>6^2=36</math> unit tiles. The seventh square has <math>49-36=\boxed{\text{(C)}\ 13}</math> more tiles than the sixth square.<br />
<br />
===Solution 2===<br />
The edge of each square is one tile length longer than the edge of the previous square, which means that each square has <math>2*</math> edge length <math>- 1</math> more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh square has <math>2^27-1=13</math> more tiles than the sixth square, which is <math>\boxed{\text{(C)}\ 13}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_9&diff=1221202002 AMC 8 Problems/Problem 92020-05-07T03:21:29Z<p>Technetiumdolomite: </p>
<hr />
<div>==Juan's Old Stamping Grounds==<br />
<br />
Problems 8,9 and 10 use the data found in the accompanying paragraph and table:<br />
<br />
<center><br />
Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and<br />
France, 6 cents each, Peru 4 cents each, and Spain 5 cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.)<br />
</center><br />
<br />
<asy><br />
/* AMC8 2002 #8, 9, 10 Problem */<br />
size(3inch, 1.5inch);<br />
for ( int y = 0; y &lt;= 5; ++y )<br />
{<br />
draw((0,y)--(18,y));<br />
}<br />
draw((0,0)--(0,5));<br />
draw((6,0)--(6,5));<br />
draw((9,0)--(9,5));<br />
draw((12,0)--(12,5));<br />
draw((15,0)--(15,5));<br />
draw((18,0)--(18,5));<br />
draw(scale(0.8)*"50s", (7.5,4.5));<br />
draw(scale(0.8)*"4", (7.5,3.5));<br />
draw(scale(0.8)*"8", (7.5,2.5));<br />
draw(scale(0.8)*"6", (7.5,1.5));<br />
draw(scale(0.8)*"3", (7.5,0.5));<br />
draw(scale(0.8)*"60s", (10.5,4.5));<br />
draw(scale(0.8)*"7", (10.5,3.5));<br />
draw(scale(0.8)*"4", (10.5,2.5));<br />
draw(scale(0.8)*"4", (10.5,1.5));<br />
draw(scale(0.8)*"9", (10.5,0.5));<br />
draw(scale(0.8)*"70s", (13.5,4.5));<br />
draw(scale(0.8)*"12", (13.5,3.5));<br />
draw(scale(0.8)*"12", (13.5,2.5));<br />
draw(scale(0.8)*"6", (13.5,1.5));<br />
draw(scale(0.8)*"13", (13.5,0.5));<br />
draw(scale(0.8)*"80s", (16.5,4.5));<br />
draw(scale(0.8)*"8", (16.5,3.5));<br />
draw(scale(0.8)*"15", (16.5,2.5));<br />
draw(scale(0.8)*"10", (16.5,1.5));<br />
draw(scale(0.8)*"9", (16.5,0.5));<br />
label(scale(0.8)*"Country", (3,4.5));<br />
label(scale(0.8)*"Brazil", (3,3.5));<br />
label(scale(0.8)*"France", (3,2.5));<br />
label(scale(0.8)*"Peru", (3,1.5));<br />
label(scale(0.8)*"Spain", (3,0.5));<br />
label(scale(0.9)*"Juan's Stamp Collection", (9,0), S);<br />
label(scale(0.9)*"Number of Stamps by Decade", (9,5), N);</asy><br />
<br />
==Problem==<br />
In dollars and cents, how much did his South American stampes issued before the ’70s cost him?<br />
<br />
<math>\text{(A)}\ \$0.40 \qquad \text{(B)}\ \$1.06 \qquad \text{(C)}\ \$1.80 \qquad \text{(D)}\ \$2.38 \qquad \text{(E)}\ \$2.64</math><br />
<br />
==Solution==<br />
Brazil 50s and 60s total 11 stamps with each 6 cents, Peru 50s and 60s total 10 stamps with each 4 cents. So total <math>11*0.06+10*0.04 = \boxed{\text{(B)}\ \$1.06}</math> <br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_9&diff=1221192002 AMC 8 Problems/Problem 92020-05-07T03:20:19Z<p>Technetiumdolomite: </p>
<hr />
<div>==Juan's Old Stamping Grounds==<br />
<br />
Problems 8,9 and 10 use the data found in the accompanying paragraph and table:<br />
<br />
<center><br />
Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and<br />
France, 6 cents each, Peru 4 cents each, and Spain 5 cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.)<br />
</center><br />
<br />
<asy><br />
/* AMC8 2002 #8, 9, 10 Problem */<br />
size(3inch, 1.5inch);<br />
for ( int y = 0; y &lt;= 5; ++y )<br />
{<br />
draw((0,y)--(18,y));<br />
}<br />
draw((0,0)--(0,5));<br />
draw((6,0)--(6,5));<br />
draw((9,0)--(9,5));<br />
draw((12,0)--(12,5));<br />
draw((15,0)--(15,5));<br />
draw((18,0)--(18,5));<br />
draw(scale(0.8)*"50s", (7.5,4.5));<br />
draw(scale(0.8)*"4", (7.5,3.5));<br />
draw(scale(0.8)*"8", (7.5,2.5));<br />
draw(scale(0.8)*"6", (7.5,1.5));<br />
draw(scale(0.8)*"3", (7.5,0.5));<br />
draw(scale(0.8)*"60s", (10.5,4.5));<br />
draw(scale(0.8)*"7", (10.5,3.5));<br />
draw(scale(0.8)*"4", (10.5,2.5));<br />
draw(scale(0.8)*"4", (10.5,1.5));<br />
draw(scale(0.8)*"9", (10.5,0.5));<br />
draw(scale(0.8)*"70s", (13.5,4.5));<br />
draw(scale(0.8)*"12", (13.5,3.5));<br />
draw(scale(0.8)*"12", (13.5,2.5));<br />
draw(scale(0.8)*"6", (13.5,1.5));<br />
draw(scale(0.8)*"13", (13.5,0.5));<br />
draw(scale(0.8)*"80s", (16.5,4.5));<br />
draw(scale(0.8)*"8", (16.5,3.5));<br />
draw(scale(0.8)*"15", (16.5,2.5));<br />
draw(scale(0.8)*"10", (16.5,1.5));<br />
draw(scale(0.8)*"9", (16.5,0.5));<br />
label(scale(0.8)*"Country", (3,4.5));<br />
label(scale(0.8)*"Brazil", (3,3.5));<br />
label(scale(0.8)*"France", (3,2.5));<br />
label(scale(0.8)*"Peru", (3,1.5));<br />
label(scale(0.8)*"Spain", (3,0.5));<br />
label(scale(0.9)*"Juan's Stamp Collection", (9,0), S);<br />
label(scale(0.9)*"Number of Stamps by Decade", (9,5), N);</asy><br />
<br />
==Problem==<br />
In dollars and cents, how much did his South American stampes issued before the ’70s cost him?<br />
<br />
<math>\text{(A)}\ \$0.40 \qquad \text{(B)}\ \$1.06 \qquad \text{(C)}\ \$1.80 \qquad \text{(D)}\ \$2.38 \qquad \text{(E)}\ \$2.64</math><br />
<br />
==Solution==<br />
Brazil 50s and 60s total 11 stamps with each 6 cents, and Peru 50s and 60s total 10 stamps with each 4 cents. So total <math>11*0.06+10*0.04 = \boxed{\text{(B)}\ \$1.06}</math> <br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_10&diff=1219172015 AMC 10B Problems/Problem 102020-05-02T20:36:39Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>==Problem==<br />
What are the sign and units digit of the product of all the odd negative integers strictly greater than <math>-2015</math>?<br />
<br><br />
<math>\textbf{(A) }</math> It is a negative number ending with a 1. <br><br />
<math>\textbf{(B) }</math> It is a positive number ending with a 1. <br><br />
<math>\textbf{(C) }</math> It is a negative number ending with a 5. <br><br />
<math>\textbf{(D) }</math> It is a positive number ending with a 5. <br><br />
<math>\textbf{(E) }</math> It is a negative number ending with a 0. <br><br />
<br />
==Solution==<br />
Since <math>-5>-2015</math>, the product must end with a <math>5</math>.<br />
<br />
The multiplicands are the odd negative integers from <math>-1</math> to <math>-2013</math>. There are <math>\frac{|-2013+1|}2+1=1006+1</math> of these numbers. Since <math>(-1)^{1007}=-1</math>, the product is negative.<br />
<br />
Therefore, the answer must be <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math><br />
<br />
Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_4&diff=1219162015 AMC 10B Problems/Problem 42020-05-02T20:31:05Z<p>Technetiumdolomite: /* Problem 4 */</p>
<hr />
<div>==Problem 4==<br />
<br />
Four siblings ordered an extra large pizza. Alex ate <math>\frac15</math>, Beth <math>\frac13</math>, and Cyril <math>\frac14</math> of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed?<br />
<br />
<math>\textbf{(A) } \text{Alex, Beth, Cyril, Dan}</math> <br><br />
<math>\textbf{(B) } \text{Beth, Cyril, Alex, Dan}</math> <br><br />
<math>\textbf{(C) } \text{Beth, Cyril, Dan, Alex}</math> <br><br />
<math>\textbf{(D) } \text{Beth, Dan, Cyril, Alex}</math> <br><br />
<math>\textbf{(E) } \text{Dan, Beth, Cyril, Alex}</math> <br><br />
<br />
==Solution==<br />
<br />
Let the pizza have <math>60</math> slices, since the least common multiple of <math>(5,3,4)=60</math>. Therefore, Alex ate <math>\frac{1}{5}\times60=12</math> slices, Beth ate <math>\frac{1}{3}\times60=20</math> slices, and Cyril ate <math>\frac{1}{4}\times60=15</math> slices. Dan must have eaten <math>60-(12+20+15)=13</math> slices. In decreasing order, we see the answer is <math>\boxed{\textbf{(C) }\text{Beth, Cyril, Dan, Alex}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_9&diff=1214242019 AMC 10B Problems/Problem 92020-04-21T19:20:34Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The function <math>f</math> is defined by <cmath>f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|</cmath>for all real numbers <math>x</math>, where <math>\lfloor r \rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math>. What is the range of <math>f</math>?<br />
<br />
<math>\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\}</math> <br> <math>\textbf{(E) } \text{The set of nonnegative integers} </math><br />
<br />
==Solution 1==<br />
<br />
There are four cases we need to consider here.<br />
<br />
'''Case 1''': <math>x</math> is a positive integer. Without loss of generality, assume <math>x=1</math>. Then <math>f(1) = 1 - 1 = 0</math>.<br />
<br />
'''Case 2''': <math>x</math> is a positive fraction. Without loss of generality, assume <math>x=\frac{1}{2}</math>. Then <math>f\left(\frac{1}{2}\right) = 0 - 0 = 0</math>.<br />
<br />
'''Case 3''': <math>x</math> is a negative integer. Without loss of generality, assume <math>x=-1</math>. Then <math>f(-1) = 1 - 1 = 0</math>.<br />
<br />
'''Case 4''': <math>x</math> is a negative fraction. Without loss of generality, assume <math>x=-\frac{1}{2}</math>. Then <math>f\left(-\frac{1}{2}\right) = 0 - 1 = -1</math>.<br />
<br />
Thus the range of the function <math>f</math> is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.<br />
<br />
~IronicNinja, edited by someone else hehe<br />
<br />
==Solution 2==<br />
<br />
It is easily verified that when <math>x</math> is an integer, <math>f(x)</math> is zero. We therefore need only to consider the case when <math>x</math> is not an integer.<br />
<br />
When <math>x</math> is positive, <math>\lfloor x\rfloor \geq 0</math>, so <br />
<cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\<br />
&=\lfloor x\rfloor-\lfloor x\rfloor \\<br />
&=0\end{split}</cmath><br />
<br />
When <math>x</math> is negative, let <math>x=-a-b</math> be composed of integer part <math>a</math> and fractional part <math>b</math> (both <math>\geq 0</math>):<br />
<cmath>\begin{split}f(x)&=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor| \\<br />
&=\lfloor a+b\rfloor-|-a-1| \\<br />
&=a-(a+1)=-1\end{split}</cmath><br />
<br />
Thus, the range of f is <math>\boxed{\textbf{(A) } \{-1, 0\}}</math>.<br />
<br />
''Note'': One could solve the case of <math>x</math> as a negative non-integer in this way:<br />
<cmath>\begin{split}f(x)&=\lfloor|x|\rfloor-|\lfloor x\rfloor| \\<br />
&=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1| \\<br />
&=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = -1\end{split}</cmath><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_15&diff=1210932008 AMC 8 Problems/Problem 152020-04-17T21:54:13Z<p>Technetiumdolomite: /* Solution */</p>
<hr />
<div>==Problem==<br />
In Theresa's first <math>8</math> basketball games, she scored <math>7, 4, 3, 6, 8, 3, 1</math> and <math>5</math> points. In her ninth game, she scored fewer than <math>10</math> points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than <math>10</math> points and her points-per-game average for the <math>10</math> games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?<br />
<br />
<math>\textbf{(A)}\ 35\qquad<br />
\textbf{(B)}\ 40\qquad<br />
\textbf{(C)}\ 48\qquad<br />
\textbf{(D)}\ 56\qquad<br />
\textbf{(E)}\ 72</math><br />
<br />
==Solution==<br />
The total number of points from the first <math>8</math> games is <math>7+4+3+6+8+3+1+5=37</math>. We have to make this a multiple of <math>9</math> by scoring less than <math>10</math> points. The closest multiple of <math>9</math> is <math>45</math>. <math>45-37=8</math> now we have to add a number to get a multiple of 10. The next multiple is <math>50</math> we added <math>5</math>, multiplying these together you get <math>8\cdot5</math> is <math>40</math>. The answer is <math>\boxed{(B)\ 40}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2008|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_22&diff=1210732019 AMC 10B Problems/Problem 222020-04-17T01:52:03Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #22]] and [[2019 AMC 12B Problems|2019 AMC 12B #19]]}}<br />
<br />
==Problem==<br />
<br />
Raashan, Sylvia, and Ted play the following game. Each starts with <math> \$1</math>. A bell rings every <math>15</math> seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives <math>\$1</math> to that player. What is the probability that after the bell has rung <math>2019</math> times, each player will have <math>\$1</math>? (For example, Raashan and Ted may each decide to give <math>\$1</math> to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have <math>\$0</math>, Sylvia will have <math>\$2</math>, and Ted will have <math>\$1</math>, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their <math> \$1</math> to, and the holdings will be the same at the end of the second round.)<br />
<br />
<math>\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}</math><br />
<br />
==Solution 1==<br />
On the first turn, each player starts off with <math>\$1</math>. Each turn after that, there are only two possibilities: either everyone stays at <math>\$1</math>, which we will write as <math>(1-1-1)</math>, or the distribution of money becomes <math>\$2-\$1-\$0</math> in some order, which we write as <math>(2-1-0)</math>. (<math>(3-0-0)</math> cannot be achieved since either(1)the person cannot give money to himself or (2)there are a maximum of 2 dollars being distributed and the person has nothing to start with). We will consider these two states separately.<br />
<br />
In the <math>(1-1-1)</math> state, each person has two choices for whom to give their dollar to, meaning there are <math>2^3=8</math> possible ways that the money can be rearranged. Note that there are only two ways that we can reach <math>(1-1-1)</math> again: <br />
<br />
1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.<br />
<br />
2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.<br />
<br />
Thus, the probability of staying in the <math>(1-1-1)</math> state is <math>\frac{1}{4}</math>, while the probability of going to the <math>(2-1-0)</math> state is <math>\frac{3}{4}</math> (we can check that the 6 other possibilities lead to <math>(2-1-0)</math>)<br />
<br />
<br />
In the <math>(2-1-0)</math> state, we will label the person with <math>\$2</math> as person A, the person with <math>\$1</math> as person B, and the person with <math>\$0</math> as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total <math>2\cdot 2 = 4</math> ways the money can be redistributed. The only way that the distribution can return to <math>(1-1-1)</math> is if A gives <math>\$1</math> to B, and B gives <math>\$1</math> to C. We check the other possibilities to find that they all lead back to <math>(2-1-0)</math>. Thus, the probability of going to the <math>(1-1-1)</math> state is <math>\frac{1}{4}</math>, while the probability of staying in the <math>(2-1-0)</math> state is <math>\frac{3}{4}</math>.<br />
<br />
No matter which state we are in, the probability of going to the <math>(1-1-1)</math> state is always <math>\frac{1}{4}</math>. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a <math>\frac{1}{4}</math> probability of going to the <math>(1-1-1)</math> state after the 2019th bell ring. Thus, our answer is simply <math>\boxed{\textbf{(B) } \frac{1}{4}}</math>.<br />
<br />
==Solution 2 (Symmetry)==<br />
After the first ring, either nothing changes, or someone has <math>\$2</math>. No one can have <math>\$3</math>, since in that hypothetical round, that person would have to give away <math>\$1</math>. <br />
<br />
<br />
Thus, the outcome is either <math>1-1-1</math> or six symmetrical cases where one person gets <math>\$2</math> (e.g. a <math>1-2-0</math> or <math>2-1-0</math> split). There are two ways for the three people to exchange dollars to get to the same <math>1-1-1</math> result. As such, there are 8 overall possibilities (which make sense, since each person has 2 choices when giving away his or her dollar, resulting in <math>2^3</math> total possibilities). As such, from the <math>1-1-1</math> case, there is a <math>1/4</math> chance of returning to <math>1-1-1</math>.<br />
<br />
<br />
Without loss of generality, take the <math>1-2-0</math> case. Only 2 people can give money, so there are now <math>2^2</math> possible outcomes after the bell rings. It either decomposes back into <math>1-1-1</math>, remains unchanged, turns into <math>0-1-2</math>, or turns into <math>0-2-1</math>. As such, from the <math>1-1-1</math> case, there is a <math>1/4</math> chance of returning to <math>1-1-1</math>. Notice that this works for any of the 6 cases.<br />
<br />
<br />
Since the starting state has a <math>1/4</math> chance of remaining unchanged, and each of the different 6 symmetric states all also have a <math>1/4</math> chance of reverting back to <math>1-1-1</math>, the chance of it being 1-1-1 after any state is always <math>\boxed{\textbf{(B) } \frac{1}{4}}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/XT440PjAFmQ<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=21|num-a=23}}<br />
{{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_23&diff=1210262017 AMC 10B Problems/Problem 232020-04-16T02:36:08Z<p>Technetiumdolomite: </p>
<hr />
<div>==Problem 23==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
<br />
==Solution 1==<br />
We only need to find the remainders of N when divided by 5 and 9 to determine the answer.<br />
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.<br />
The remainder when <math>N</math> is divided by <math>9</math> is <math>1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which has a remainder of 0 mod 9. Solving these modular congruence using CRT(Chinese Remainder Theorem) we ge the remainder to be <math>9</math>(mod <math>45</math>). Therefore, the answer is <math>\boxed{\textbf{(C) } 9}</math>.<br />
<br />
==Alternative Ending to Solution 1==<br />
Once we find our 2 modular congruences, we can narrow our options down to <math>{C}</math> and <math>{D}</math> because the remainder when <math>N</math> is divided by <math>45</math> should be a multiple of 9 by our modular congruence that states <math>N</math> has a remainder of <math>0</math> when divided by <math>9</math>. Also, our other modular congruence states that the remainder when divided by <math>45</math> should have a remainder of <math>4</math> when divided by <math>5</math>. Out of options <math>C</math> and <math>D</math>, only <math>\boxed{\textbf{(C) } 9}</math> satisfies that the remainder when <math>N</math> is divided by 45 <math>\equiv 4 \text{ (mod 5)}</math>.<br />
==Solution2==<br />
The same way, you can get N=4(Mod 5) and 0(Mod 9). By The Chinese remainder Theorem, the answer come out to be 9-(C)<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_2&diff=1210252017 AMC 10B Problems/Problem 22020-04-16T02:21:31Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Sofia ran <math>5</math> laps around the <math>400</math>-meter track at her school. For each lap, she ran the first <math>100</math> meters at an average speed of <math>4</math> meters per second and the remaining <math>300</math> meters at an average speed of <math>5</math> meters per second. How much time did Sofia take running the <math>5</math> laps? <br />
<br />
<math>\qquad\textbf{(A)}\ \text{5 minutes and 35 seconds}</math> <br><br />
<math>\qquad\textbf{(B)}\ \text{6 minutes and 40 seconds}</math> <br><br />
<math>\qquad\textbf{(C)}\ \text{7 minutes and 5 seconds}</math> <br><br />
<math>\qquad\textbf{(D)}\ \text{7 minutes and 25 seconds}</math> <br><br />
<math>\qquad\textbf{(E)}\ \text{8 minutes and 10 seconds}</math><br />
<br />
==Solution==<br />
<br />
If Sofia ran the first <math>100</math> meters of each lap at <math>4</math> meters per second and the remaining <math>300</math> meters of each lap at <math>5</math> meters per second, then she took <math>\frac{100}{4}+\frac{300}{5}=25+60=85</math> seconds for each lap. Because she ran <math>5</math> laps, she took a total of <math>5 \cdot 85=425</math> seconds, or <math>7</math> minutes and <math>5</math> seconds. The answer is <math>\boxed{\textbf{(C)}\ \text{7 minutes and 5 seconds}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_20&diff=1209912010 AMC 8 Problems/Problem 202020-04-14T21:53:15Z<p>Technetiumdolomite: </p>
<hr />
<div>==Problem==<br />
In a room, <math>2/5</math> of the people are wearing gloves, and <math>3/4</math> of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove? <br />
<br />
<math> \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20 </math><br />
<br />
==Solution==<br />
Let <math>x</math> be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of <math>\text{lcm}(4,5)=20</math>. Since we are trying to find the minimum <math>x</math>, we must use the smallest possible value for the number of people in the room. Similarly, we can assume that there are no people present who are wearing neither of the two items since this would unnecessarily increase the number of people in the room. Thus, we can say that there are <math>20</math> people in the room, all of which are wearing at least a hat or a glove.<br />
<br />
It follows that there are <math>\frac{2}{5}\cdot 20 = 8</math> people wearing gloves and <math>\frac{3}{4}\cdot 20 = 15</math> people wearing hats. Then by applying the Principle of Inclusion Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is <math>8+15-x = 23-x</math>. Since we know that this equals <math>20</math>, it follows that <math>23-x = 20</math>, which implies that <math>x=3</math>. Thus, <math>\boxed{\textbf{(A)}\ 3}</math> is the correct answer.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_7&diff=1209352011 AMC 8 Problems/Problem 72020-04-14T01:29:39Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>==Problem==<br />
Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially '''bolded'''. What percent of the total area is partially bolded?<br />
<asy><br />
import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; <br />
pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); <br />
draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); <br />
draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); <br />
dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<math> \textbf{(A) }12\frac{1}{2} \qquad\textbf{(B) }20 \qquad\textbf{(C) }25 \qquad\textbf{(D) }33\frac{1}{3} \qquad\textbf{(E) }37\frac{1}{2} </math><br />
<br />
==Solution==<br />
Assume that the area of each square is <math>1</math>. Then, the area of the bolded region in the top left square is <math>\frac{1}{4}</math>. The area of the top right bolded region is <math>\frac{1}{8}</math>. The area of the bottom left bolded region is <math>\frac{3}{8}</math>. And the area of the bottom right bolded region is <math>\frac{1}{4}</math>. Add the four fractions: <math>\frac{1}{4} + \frac{1}{8} + \frac{3}{8} + \frac{1}{4} = 1</math>. The four squares together have an area of <math>4</math>, so the percentage bolded is <math>\frac{1}{4} \cdot 100 = \boxed{\textbf{(C)}\ 25}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_1&diff=1196392018 AMC 8 Problems/Problem 12020-03-17T20:41:16Z<p>Technetiumdolomite: /* Solution 1 */</p>
<hr />
<div>==Problem 1==<br />
An amusement park has a collection of scale models, with a ratio <math> 1: 20</math>, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its duplicate to the nearest whole number?<br />
<br />
<math>\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math><br />
<br />
==Solution 1==<br />
<br />
<br />
You can set up a ratio: <math>\frac{1}{20}=\frac{x}{289}</math>. Cross multiplying, you get <math>20x=289</math>. You divide by <math>20</math> on each side to get <math>x=14.45</math>. The closest integer is <math>\boxed{\textbf{(A)}14}</math><br />
<br />
==Solution 2==<br />
<br />
<br />
You can just do <math>\frac{289}{20}</math> and round your answer to get <math>\boxed{\textbf{(A)}14}</math>.<br />
It is basically Solution 1 without the ratio calculation, which might not be necessary.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2018|before=First Problem|num-a=2}}<br />
<br />
{{MAA Notice}}</div>Technetiumdolomitehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&diff=1196382018 AMC 8 Problems/Problem 232020-03-17T20:33:59Z<p>Technetiumdolomite: /* Problem */</p>
<hr />
<div>==Problem==<br />
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br />
<br />
<asy><br />
size(3cm);<br />
pair A[];<br />
for (int i=0; i<9; ++i) {<br />
A[i] = rotate(22.5+45*i)*(1,0);<br />
}<br />
filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black);<br />
for (int i=0; i<8; ++i) { dot(A[i]); }<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}</math><br />
<br />
==Solution==<br />
===Solution 1===<br />
We will use constructive counting to solve this. There are <math>2</math> cases: Either all <math>3</math> points are adjacent, or exactly <math>2</math> points are adjacent.<br />
<br />
If all <math>3</math> points are adjacent, then we have <math>8</math> choices. If we have exactly <math>2</math> adjacent points, then we will have <math>8</math> places to put the adjacent points and also <math>4</math> places to put the remaining point, so we have <math>8\cdot4</math> choices. The total amount of choices is <math>{8 \choose 3} = 8\cdot7</math>.<br />
Thus our answer is <math>\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}</math><br />
<br />
===Solution 2 ===<br />
We can decide <math>2</math> adjacent points with <math>8</math> choices. The remaining point will have <math>6</math> choices. However, we have counted the case with <math>3</math> adjacent points twice, so we need to subtract this case once. The case with the <math>3</math> adjacent points has <math>8</math> arrangements, so our answer is <math>\frac{8\cdot6-8}{{8 \choose 3 }}</math><math>=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}</math><br />
<br />
===Solution 3 (Stars and Bars)===<br />
Let <math>1</math> point of the triangle be fixed at the top. Then, there are <math>{7 \choose 2} = 21</math> ways to chose the other 2 points. There must be <math>3</math> spaces in the points and <math>3</math> points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and <math>2</math> extra points (k) distributed so by the stars and bars formula, <math>{n+k-1 \choose k}</math>, there are <math>{4 \choose 2} = 6</math> ways to arrange the bars and stars. Thus, the probability is <math>\frac{(21 - 6)}{21} = \frac{5}{7}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2018|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Technetiumdolomite