https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sootommylee&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T04:55:51ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_19&diff=2038432007 AMC 8 Problems/Problem 192023-11-16T05:12:14Z<p>Sootommylee: </p>
<hr />
<div>== Problem ==<br />
<br />
Pick two consecutive positive integers whose sum is less than <math>100</math>. Square both<br />
of those integers and then find the difference of the squares. Which of the<br />
following could be the difference?<br />
<br />
<math>\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131</math><br />
<br />
== Solution 1 ==<br />
<br />
Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> (x+1)+x<100</math>. <math> (x+1)^2-x^2=x^2+2x+1-x^2=2x+1 </math>. <math> 2x+1 </math> is obviously odd, so only answer choices C and E need to be considered.<br />
<br />
<math> 2x+1=131 </math> contradicts the fact that <math> 2x+1<100 </math>, so the answer is <math> \boxed{\mathrm{(C)} 79} </math><br />
<br />
==Solution 2==<br />
Since for two consecutive numbers <math>a</math> and <math>b</math>, the difference between their squares are <math>a^2-b^2=(a+b)(a-b)</math>, which equals to <math>a+b</math>, because <math>a</math> and <math>b</math> are consecutive. And because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and that the sum of <math>a</math> and <math>b</math> is less than 100, you can eliminate all answers expect for <math>\boxed{\mathrm{(C)} 79}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/8FGl2vtO1vs Soo, DRMS, NM<br />
<br />
==Video Solution by WhyMath==<br />
https://youtu.be/BrEqmDq82rw<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_23&diff=2038412006 AMC 8 Problems/Problem 232023-11-16T05:07:22Z<p>Sootommylee: /* Video Solution by SOO */</p>
<hr />
<div>==Problem==<br />
<br />
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? <br />
<br />
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are<br />
<math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when<br />
divided by <math>5</math> are <math>3,8,13,18,23,28,33, \cdots</math> So <math>28</math> is the smallest possible number<br />
of coins that meets both conditions. Because <math>4\cdot 7 = 28</math>, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left<br />
when they are divided among seven people.<br />
<br />
===Solution 2===<br />
<br />
If there were two more coins in the box, the number of coins would be divisible<br />
by both <math>6</math> and <math>5</math>. The smallest number that is divisible by <math>6</math> and <math>5</math> is <math>30</math>, so the<br />
smallest possible number of coins in the box is <math>28</math> and the remainder when divided by <math>7</math> is <math>\boxed{\textbf{(A)}\ 0}</math>.<br />
<br />
===Solution 3===<br />
<br />
We can set up a system of modular congruencies:<br />
<cmath>g\equiv 4 \pmod{6}</cmath><br />
<cmath>g\equiv 3 \pmod{5}</cmath><br />
We can use the division algorithm to say <math>g=6n+4</math> <math>\Rightarrow</math> <math>6n\equiv 4 \pmod{5}</math> <math>\Rightarrow</math> <math>n\equiv 4 \pmod{5}</math>. If we plug the division algorithm in again, we get <math>n=5q+4</math>. This means that <math>g=30q+28</math>, which means that <math>g\equiv 28 \pmod{30}</math>. From this, we can see that <math>28</math> is our smallest possible integer satisfying <math>g\equiv 28 \pmod{30}</math>. <math>28</math> <math>\div</math> <math>7=4</math>, making our remainder <math>0</math>. This means that there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left over when equally divided amongst <math>7</math> people.<br />
<br />
~Champion1234<br />
<br />
==Video Solution==<br />
https://youtu.be/Gxfjwxl3Sbo Soo, DRMS, NM<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/g1PLxYVZE_U<br />
-Happytwin<br />
<br />
https://www.youtube.com/watch?v=uMBev3FUoTs ~David<br />
<br />
==See Also==<br />
{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_23&diff=2038402006 AMC 8 Problems/Problem 232023-11-16T05:07:03Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? <br />
<br />
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
The counting numbers that leave a remainder of <math>4</math> when divided by <math>6</math> are<br />
<math>4, 10, 16, 22, 28, 34, \cdots</math> The counting numbers that leave a remainder of <math>3</math> when<br />
divided by <math>5</math> are <math>3,8,13,18,23,28,33, \cdots</math> So <math>28</math> is the smallest possible number<br />
of coins that meets both conditions. Because <math>4\cdot 7 = 28</math>, there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left<br />
when they are divided among seven people.<br />
<br />
===Solution 2===<br />
<br />
If there were two more coins in the box, the number of coins would be divisible<br />
by both <math>6</math> and <math>5</math>. The smallest number that is divisible by <math>6</math> and <math>5</math> is <math>30</math>, so the<br />
smallest possible number of coins in the box is <math>28</math> and the remainder when divided by <math>7</math> is <math>\boxed{\textbf{(A)}\ 0}</math>.<br />
<br />
===Solution 3===<br />
<br />
We can set up a system of modular congruencies:<br />
<cmath>g\equiv 4 \pmod{6}</cmath><br />
<cmath>g\equiv 3 \pmod{5}</cmath><br />
We can use the division algorithm to say <math>g=6n+4</math> <math>\Rightarrow</math> <math>6n\equiv 4 \pmod{5}</math> <math>\Rightarrow</math> <math>n\equiv 4 \pmod{5}</math>. If we plug the division algorithm in again, we get <math>n=5q+4</math>. This means that <math>g=30q+28</math>, which means that <math>g\equiv 28 \pmod{30}</math>. From this, we can see that <math>28</math> is our smallest possible integer satisfying <math>g\equiv 28 \pmod{30}</math>. <math>28</math> <math>\div</math> <math>7=4</math>, making our remainder <math>0</math>. This means that there are <math>\boxed{\textbf{(A)}\ 0}</math> coins left over when equally divided amongst <math>7</math> people.<br />
<br />
~Champion1234<br />
<br />
==Video Solution by SOO==<br />
https://youtu.be/Gxfjwxl3Sbo Soo, DRMS, NM<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/g1PLxYVZE_U<br />
-Happytwin<br />
<br />
https://www.youtube.com/watch?v=uMBev3FUoTs ~David<br />
<br />
==See Also==<br />
{{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_25&diff=2038392005 AMC 8 Problems/Problem 252023-11-16T04:53:34Z<p>Sootommylee: </p>
<hr />
<div>== Problem 25 ==<br />
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?<br />
<br />
<asy><br />
pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2);<br />
draw(a--d--b--c--cycle);<br />
draw(circle(o, 2.5));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
Let the region within the circle and square be <math>a</math>. In other words, it is the area inside the circle <math>\textbf{and}</math> the square. Let <math>r</math> be the radius. We know that the area of the circle minus <math>a</math> is equal to the area of the square, minus <math>a</math> . <br />
<br />
We get:<br />
<br />
<math>\pi r^2 -a=4-a</math><br />
<br />
<math>r^2=\frac{4}{\pi}</math><br />
<br />
<math>r=\frac{2}{\sqrt{\pi}}</math><br />
<br />
So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>.<br />
<br />
=== Solution 2 ===<br />
We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.<br />
<br />
<math>\pi r^2=4</math><br />
<br />
<math>r^2=\frac{4}{\pi}</math><br />
<br />
<math>r=\frac{2}{\sqrt{\pi}}</math><br />
<br />
So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/dWxzg4zmGOs Soo, DRMS, NM<br />
<br />
==Video Solution by OmegaLearn==<br />
https://youtu.be/abSgjn4Qs34?t=2763<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=xdeWCR666q8 ~David<br />
<br />
== See Also ==<br />
{{AMC8 box|year=2005|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_22&diff=2038322005 AMC 8 Problems/Problem 222023-11-16T04:39:38Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs 50% more than the small size and contains 20% less detergent than the large size. The large size contains twice as much detergent as the small size and costs 30% more than the medium size. Rank the three sizes from best to worst buy.<br />
<br />
<math>\text{(A)}\, SML \qquad \text{(B)}\, LMS \qquad \text{(C)}\, MSL \qquad \text{(D)}\, LSM \qquad \text{(E)}\, MLS</math><br />
<br />
==Solution==<br />
Suppose the small size costs <math>\textdollar 1</math> and the large size has <math>10</math> oz. The medium size then costs <math>\textdollar 1.50</math> and has <math>8</math> oz. The small size has <math>5</math> oz and the large size costs <math>\textdollar 1.95</math>. The small, medium, and large size cost respectively, <math>0.200, 0.188, 0.195</math> dollars per oz. The sizes from best to worst buy are <math>\boxed{\textbf{(E)}\ \text{MLS}}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/cXcYrjJ0eL0 Soo, DRMS, NM<br />
<br />
==Video solution==<br />
<br />
https://www.youtube.com/watch?v=65cfokoNyfA ~David<br />
<br />
==See Also==<br />
{{AMC8 box|year=2005|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_12&diff=1997242000 AMC 8 Problems/Problem 122023-10-16T04:55:26Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?<br />
<br />
<asy><br />
draw((0,0)--(6,0)--(6,1)--(5,1)--(5,2)--(0,2)--cycle);<br />
draw((0,1)--(5,1));<br />
draw((1,1)--(1,2));<br />
draw((3,1)--(3,2));<br />
draw((2,0)--(2,1));<br />
draw((4,0)--(4,1));<br />
<br />
</asy><br />
<br />
<math> \text{(A)}\ 344\qquad\text{(B)}\ 347\qquad\text{(C)}\ 350\qquad\text{(D)}\ 353\qquad\text{(E)}\ 356 </math><br />
<br />
==Solution==<br />
<br />
Since the bricks are <math>1</math> foot high, there will be <math>7</math> rows. To minimize the number of blocks used, rows <math>1, 3, 5,</math> and <math>7</math> will look like the bottom row of the picture, which takes <math>\frac{100}{2} = 50</math> bricks to construct. Rows <math>2, 4, </math> and <math>6</math> will look like the upper row pictured, which has <math>49</math> 2-foot bricks in the middle, and <math>2</math> 1-foot bricks on each end for a total of <math>51</math> bricks.<br />
<br />
Four rows of <math>50</math> bricks and three rows of <math>51</math> bricks totals <math>4\cdot 50 + 3\cdot 51 = 200 + 153 = 353</math> bricks, giving the answer <math>\boxed{D}.</math><br />
<br />
==Video Solution by Soo==<br />
https://youtu.be/EeHv96Ry_Ww - Soo, DRMS, NM<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_24&diff=1997231999 AMC 8 Problems/Problem 242023-10-16T04:46:25Z<p>Sootommylee: /* =Video Solution by Soo */</p>
<hr />
<div>==Problem==<br />
<br />
When <math>1999^{2000}</math> is divided by <math>5</math>, the remainder is <br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4</math><br />
<br />
<br />
==Solution 1==<br />
Note that the units digits of the powers of 9 have a pattern: <math>9^1 = {\bf 9}</math>, <math>9^2 = 8{\bf 1}</math>, <math>9^3 = 72{\bf 9}</math>, <math>9^4 = 656{\bf 1}</math>, and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of <math>9</math>, the number ends in a <math>1</math>. Since the exponent is even, the final digit is <math>1</math>. Note that all natural numbers that end in <math>1</math> have a remainder of <math>1</math> when divided by <math>5</math>. So, our answer is <math>\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
<br />
Write <math>1999</math> as <math>2000-1</math>. We are taking <math>(2000-1)^{2000} \mod{10}</math>. Using the binomial theorem, we see that ALL terms in this expansion are divisible by <math>2000</math> except for the very last term, which is just <math>(-1)^{2000}</math>. This is clear because the binomial expansion is just choosing how many <math>2000</math>s and how many <math>-1</math>s there are for each term. Using this, we can take the entire polynomial <math>\mod{10}</math>, which leaves just <math>(-1)^{2000}=\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Solution 3==<br />
As <math>1999 \equiv -1 \pmod{5}</math>, we have <math>1999^{2000} \equiv (-1)^{2000} \equiv 1 \pmod{5}</math>. Thus, the answer is <math>\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Solution 4==<br />
A sum/product of any two natural numbers has the same remainder, when divided by <math>n \in \mathbb{N}</math>, as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as <math>1999 \cdot 1999 \cdot \cdot \cdot 1999</math>. <br />
<br />
Using the fact stated in the first sentence, we see that the remainder of <math>1999</math>, when divided by <math>5</math>, is <math>4</math>. The problem can now be written viewed as finding the remainder of <math>4^{2000}</math> when it is divided by <math>5</math>, which is already much simpler. <br />
<br />
Now, toying with the simplified problem a little, see notice that powers of <math>4</math> alternate from ending in <math>4</math> and <math>6</math>. We notice that even powers of <math>4</math> always end in <math>6</math>, and also that <math>2000</math> is even. Thus <math>4^{2000}</math> must end in <math>6</math>, which, when divided by <math>5</math> gives a remainder of <math>\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Video Solution by Soo==<br />
https://youtu.be/jCHiQMK7k3w - Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=1999|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_24&diff=1997221999 AMC 8 Problems/Problem 242023-10-16T04:46:10Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
When <math>1999^{2000}</math> is divided by <math>5</math>, the remainder is <br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4</math><br />
<br />
<br />
==Solution 1==<br />
Note that the units digits of the powers of 9 have a pattern: <math>9^1 = {\bf 9}</math>, <math>9^2 = 8{\bf 1}</math>, <math>9^3 = 72{\bf 9}</math>, <math>9^4 = 656{\bf 1}</math>, and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of <math>9</math>, the number ends in a <math>1</math>. Since the exponent is even, the final digit is <math>1</math>. Note that all natural numbers that end in <math>1</math> have a remainder of <math>1</math> when divided by <math>5</math>. So, our answer is <math>\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Solution 2==<br />
<br />
Write <math>1999</math> as <math>2000-1</math>. We are taking <math>(2000-1)^{2000} \mod{10}</math>. Using the binomial theorem, we see that ALL terms in this expansion are divisible by <math>2000</math> except for the very last term, which is just <math>(-1)^{2000}</math>. This is clear because the binomial expansion is just choosing how many <math>2000</math>s and how many <math>-1</math>s there are for each term. Using this, we can take the entire polynomial <math>\mod{10}</math>, which leaves just <math>(-1)^{2000}=\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Solution 3==<br />
As <math>1999 \equiv -1 \pmod{5}</math>, we have <math>1999^{2000} \equiv (-1)^{2000} \equiv 1 \pmod{5}</math>. Thus, the answer is <math>\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Solution 4==<br />
A sum/product of any two natural numbers has the same remainder, when divided by <math>n \in \mathbb{N}</math>, as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as <math>1999 \cdot 1999 \cdot \cdot \cdot 1999</math>. <br />
<br />
Using the fact stated in the first sentence, we see that the remainder of <math>1999</math>, when divided by <math>5</math>, is <math>4</math>. The problem can now be written viewed as finding the remainder of <math>4^{2000}</math> when it is divided by <math>5</math>, which is already much simpler. <br />
<br />
Now, toying with the simplified problem a little, see notice that powers of <math>4</math> alternate from ending in <math>4</math> and <math>6</math>. We notice that even powers of <math>4</math> always end in <math>6</math>, and also that <math>2000</math> is even. Thus <math>4^{2000}</math> must end in <math>6</math>, which, when divided by <math>5</math> gives a remainder of <math>\boxed{\text{(B)}\ 1}</math>.<br />
<br />
==Video Solution by Soo=<br />
https://youtu.be/jCHiQMK7k3w - Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=1999|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_24&diff=1997212008 AMC 8 Problems/Problem 242023-10-16T04:30:34Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Ten tiles numbered <math>1</math> through <math>10</math> are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math>\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\ \frac{1}{5}\qquad\textbf{(E)}\ \frac{7}{30}</math><br />
<br />
<br />
== Video Solution ==<br />
https://www.youtube.com/watch?v=ZqPFm9cU0MY ~David<br />
<br />
== Video Solution by Soo ==<br />
https://youtu.be/IUh28kMLLVE - Soo, DRMS, NM<br />
<br />
==Solution==<br />
The numbers can at most multiply to be <math>60</math>. The squares less than <math>60</math> are <math>1,4,9,16,25,36,</math> and <math>49</math>. The possible pairs are <math>(1,1),(1,4),(2,2),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6),</math> and <math>(9,4)</math>. There are <math>11</math> choices and <math>60</math> possibilities giving a probability of <math>\boxed{\textbf{(C)}\ \frac{11}{60}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2008|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_22&diff=1997202008 AMC 8 Problems/Problem 222023-10-16T04:14:42Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>For how many positive integer values of <math>n</math> are both <math>\frac{n}{3}</math> and <math>3n</math> three-digit whole numbers?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math>\textbf{(A)}\ 12\qquad<br />
\textbf{(B)}\ 21\qquad<br />
\textbf{(C)}\ 27\qquad<br />
\textbf{(D)}\ 33\qquad<br />
\textbf{(E)}\ 34</math><br />
<br />
==Solution 1==<br />
Instead of finding n, we find <math>x=\frac{n}{3}</math>. We want <math>x</math> and <math>9x</math> to be three-digit whole numbers. The smallest three-digit whole number is <math>100</math>, so that is our minimum value for <math>x</math>, since if <math>x \in \mathbb{Z^+}</math>, then <math>9x \in \mathbb{Z^+}</math>. The largest three-digit whole number divisible by <math>9</math> is <math>999</math>, so our maximum value for <math>x</math> is <math>\frac{999}{9}=111</math>. There are <math>12</math> whole numbers in the closed set <math>\left[100,111\right]</math> , so the answer is <math>\boxed{\textbf{(A)}\ 12}</math>.<br />
<br />
- ColtsFan10<br />
<br />
==Solution 2==<br />
<br />
We can set the following inequalities up to satisfy the conditions given by the question,<br />
<math>100 \leq \frac{n}{3} \leq 999</math>,<br />
and <br />
<math>100 \leq 3n \leq 999</math>.<br />
Once we simplify these and combine the restrictions, we get the inequality, <math>300 \leq n \leq 333</math>.<br />
Now we have to find all multiples of 3 in this range for <math>\frac{n}{3}</math> to be an integer. We can compute this by setting <math>\frac{n}<br />
{3}=x</math>, where <math>x \in \mathbb{Z^+}</math>. Substituting <math>x</math> for <math>n</math> in the previous inequality, we get, <math>100 \leq x \leq 111</math>, and there are <math>111-100+1</math> integers in this range giving us the answer, <math>\boxed{\textbf{(A)}\ 12}</math>.<br />
<br />
- kn07<br />
<br />
==<br />
<br />
<br />
<br />
==Solution 3==<br />
<br />
So we know the largest <math>3</math> digit number is <math>999</math> and the lowest is <math>100</math>. This means <math>\dfrac{n}{3} \ge 100 \rightarrow n \ge 300</math> but <math>3n \le 999 \rightarrow n \le 333</math>. So we have the set <math>{300, 301, 302, \cdots, 330, 331, 332, 333}</math> for <math>n</math>. Now we have to find the multiples of <math>3</math> suitable for <math>n</math>, or else <math>\dfrac{n}{3}</math> will be a decimal. Only numbers <math>{300, 303, \cdots, 333}</math> are counted. We can divide by <math>3</math> to make the difference <math>1</math> again, getting <math>{100, 101 \cdots , 111}</math>. Due to it being inclusive, we have <math>111-100+1 =\boxed{\textbf{(A) } 12}</math><br />
<br />
==Video Solution by OmegaLearn==<br />
https://youtu.be/rQUwNC0gqdg?t=230<br />
<br />
==Video Solution 2==<br />
https://youtu.be/TFAp4X-OeE4 - Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2008|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_25&diff=1997182006 AMC 8 Problems/Problem 252023-10-16T04:00:39Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? <br />
<br />
<asy><br />
path card=((0,0)--(0,3)--(2,3)--(2,0)--cycle);<br />
draw(card, linewidth(1));<br />
draw(shift(2.5,0)*card, linewidth(1));<br />
draw(shift(5,0)*card, linewidth(1));<br />
label("$44$", (1,1.5));<br />
label("$59$", shift(2.5,0)*(1,1.5));<br />
label("$38$", shift(5,0)*(1,1.5));</asy><br />
<br />
<math> \textbf{(A)}\ 13\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 17 </math><br />
<br />
==Solution==<br />
<br />
Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is <math>59+2=61</math>. Thus, the first card's hidden number is <math>61-44=17</math>, and the last card's hidden number is <math>61-38=23</math>. <br />
<br />
Since the sum of the hidden primes is <math>2+17+23=42</math>, the average of the primes is <math>\dfrac{42}{3}=\boxed{\textbf{(B)} 14}</math>.<br />
<br />
<br />
==Video solution==<br />
<br />
https://www.youtube.com/watch?v=I8E7XUYlIFI ~David<br />
<br />
==Video Solution 2==<br />
<br />
https://youtu.be/NK4DtD_8kbA - Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2006|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_24&diff=1997172007 AMC 8 Problems/Problem 242023-10-16T03:42:50Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
A bag contains four pieces of paper, each labeled with one of the digits <math>1</math>, <math>2</math>, <math>3</math> or <math>4</math>, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of <math>3</math>?<br />
<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution==<br />
The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4). The number of ways to choose three digits out of four is 4. Therefore, the probability is <math>\boxed{\textbf{(C)}\ \frac{1}{2}}</math>.<br />
<br />
==Solution 2==<br />
The number of ways to form a 3-digit number is <math>4 \cdot 3 \cdot 2 = 24</math>. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is <math>3! + 3! = 12</math>. <br />
Therefore, the probability is <math>\frac{12}{24} = \boxed{\frac{1}{2}}</math>.<br />
<br />
~abc2142<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/hwc11K02cEc - Happytwin<br />
<br />
==Video Solution 2==<br />
<br />
https://youtu.be/cUxFS9l-Pb4 - Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_24&diff=1950612003 AMC 8 Problems/Problem 242023-07-02T05:22:55Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
A ship travels from point <math>A</math> to point <math>B</math> along a semicircular path, centered at Island <math>X</math>. Then it travels along a straight path from <math>B</math> to <math>C</math>. Which of these graphs best shows the ship's distance from Island <math>X</math> as it moves along its course?<br />
<br />
<asy>size(150);<br />
pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br />
draw(Arc(X, 5, 180, 360)^^B--C);<br />
dot(X);<br />
label("$X$", X, NE);<br />
label("$C$", C, N);<br />
label("$B$", B, E);<br />
label("$A$", A, W);<br />
</asy><br />
<br />
<center><br />
[[Image:2003amc8prob24ans.png|800px]]<br />
</center><br />
<br />
==Solution==<br />
<br />
The distance from Island <math>\text{X}</math> to any point on the semicircle will always be constant. On the graph, this will represent a straight line. The distance between Island <math>\text{X}</math> and line <math>\text{BC}</math> will not be constant though. We can easily prove that the distance between <math>\text{X}</math> and line <math>\text{BC}</math> will represent a curve. As the ship travels from <math>B</math> to <math>C</math>, the distance between the ship and Island <math>X</math> will first decrease until it reaches the point <math>Y</math> so that <math>\overline{XY}</math> is perpendicular to <math>\overline{BC}</math>, and then increase afterwards. Hence the answer choice that fits them all is <math>\boxed{\text{(B)}}</math>.<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=ibhy_qcQXiw ~David<br />
<br />
==Video Solution 2==<br />
https://youtu.be/TlCTRKo5mQ4 Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2003|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_11&diff=1950602001 AMC 8 Problems/Problem 112023-07-02T05:18:24Z<p>Sootommylee: </p>
<hr />
<div>== Problem ==<br />
Points <math>A</math>, <math>B</math>, <math>C</math> and <math>D</math> have these coordinates: <math>A(3,2)</math>, <math>B(3,-2)</math>, <math>C(-3,-2)</math> and <math>D(-3, 0)</math>. The area of quadrilateral <math>ABCD</math> is<br />
<br />
<asy><br />
for (int i = -4; i <= 4; ++i)<br />
{<br />
for (int j = -4; j <= 4; ++j)<br />
{<br />
dot((i,j));<br />
}<br />
}<br />
<br />
draw((0,-4)--(0,4),linewidth(1));<br />
draw((-4,0)--(4,0),linewidth(1));<br />
for (int i = -4; i <= 4; ++i)<br />
{<br />
draw((i,-1/3)--(i,1/3),linewidth(0.5));<br />
draw((-1/3,i)--(1/3,i),linewidth(0.5));<br />
}<br />
</asy><br />
<br />
<math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24</math><br />
<br />
== Solution 1 ==<br />
<br />
<asy><br />
for (int i = -4; i <= 4; ++i)<br />
{<br />
for (int j = -4; j <= 4; ++j)<br />
{<br />
dot((i,j));<br />
}<br />
}<br />
<br />
draw((0,-4)--(0,4),linewidth(1));<br />
draw((-4,0)--(4,0),linewidth(1));<br />
for (int i = -4; i <= 4; ++i)<br />
{<br />
draw((i,-1/3)--(i,1/3),linewidth(0.5));<br />
draw((-1/3,i)--(1/3,i),linewidth(0.5));<br />
}<br />
{<br />
draw((3,2)--(3,-2)--(-3,-2)--(-3,0)--cycle,linewidth(1));<br />
}<br />
<br />
label("$A$",(3,2),NE);<br />
label("$B$",(3,-2),SE);<br />
label("$C$",(-3,-2),SW);<br />
label("$D$",(-3,0),NW);<br />
</asy><br />
<br />
<br />
This quadrilateral is a trapezoid, because <math> AB\parallel CD </math> but <math> BC </math> is not parallel to <math> AD </math>. The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are <math> AB </math> and <math> CD </math>, which have lengths <math> 2 </math> and <math> 4 </math>, respectively, so the length of the median is <math> \frac{2+4}{2}=3 </math>. <math> CB </math> is perpendicular to the bases, so it is the height, and has length <math> 6 </math>. Therefore, the area of the trapezoid is <math> (3)(6)=18, \boxed{\text{C}} </math><br />
<br />
== Solution 2 ==<br />
Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle. Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle.<br />
<br />
<math>A_{trap} = A_{tri} + A_{rect}</math><br />
<br />
<math>A_{trap} = \frac{1}{2}bh + lw</math><br />
<br />
<math>A_{trap} = \frac{1}{2}\cdot 6 \cdot 2 + 6\cdot 2</math><br />
<br />
<math>A_{trap} = 6 + 12 = 18 \rightarrow \boxed{C}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/5gldUJaZZCg Soo, DRMS, NM<br />
<br />
<br />
== See Also ==<br />
{{AMC8 box|year=2001|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_19&diff=1950592008 AMC 8 Problems/Problem 192023-07-02T04:41:17Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Eight points are spaced around at intervals of one unit around a <math>2 \times 2</math> square, as shown. Two of the <math>8</math> points are chosen at random. What is the probability that the two points are one unit apart?<br />
<asy><br />
size((50));<br />
dot((5,0));<br />
dot((5,5));<br />
dot((0,5));<br />
dot((-5,5));<br />
dot((-5,0));<br />
dot((-5,-5));<br />
dot((0,-5));<br />
dot((5,-5));<br />
</asy><br />
<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7} </math><br />
<br />
==Solution 1==<br />
The two points are one unit apart at <math>8</math> places around the edge of the square. There are <math>8 \choose 2</math><math> = 28</math> ways to choose two points. The probability is<br />
<br />
<cmath>\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}</cmath><br />
<br />
==Solution 2==<br />
Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is <math>\boxed{\textbf{(B)}\ \frac27}</math><br />
<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=ZVD_PZYjonQ ~David<br />
<br />
==Video Solution 2==<br />
https://youtu.be/PeI8YSHCdlM Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2008|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_14&diff=1950582008 AMC 8 Problems/Problem 142023-07-02T04:34:35Z<p>Sootommylee: </p>
<hr />
<div>== Problem ==<br />
Three <math>\text{A's}</math>, three <math>\text{B's}</math>, and three <math>\text{C's}</math> are placed in the nine spaces so that each row and column contain one of each letter. If <math>\text{A}</math> is placed in the upper left corner, how many arrangements are possible?<br />
<br />
<asy><br />
size((80));<br />
draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0));<br />
draw((3,0)--(3,9));<br />
draw((6,0)--(6,9));<br />
draw((0,3)--(9,3));<br />
draw((0,6)--(9,6));<br />
label("A", (1.5,7.5));<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math><br />
<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=8qzMymleTIg ~David<br />
<br />
==Video Solution 2==<br />
https://youtu.be/1m_c_iMvxKo Soo, DRMS, NM<br />
<br />
== Solution ==<br />
There are <math>2</math> ways to place the remaining <math>\text{As}</math>, <math>2</math> ways to place the remaining <math>\text{Bs}</math>, and <math>1</math> way to place the remaining <math>\text{Cs}</math> for a total of <math>(2)(2)(1) = \boxed{\textbf{(C)}\ 4}</math>.<br />
<br />
== See Also ==<br />
{{AMC8 box|year=2008|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_17&diff=1950572007 AMC 8 Problems/Problem 172023-07-02T04:28:13Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
A mixture of <math>30</math> liters of paint is <math>25\%</math> red tint, <math>30\%</math> yellow<br />
tint and <math>45\%</math> water. Five liters of yellow tint are added to<br />
the original mixture. What is the percent of yellow tint<br />
in the new mixture?<br />
<br />
<math>\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50</math><br />
<br />
==Solution==<br />
Since <math>30\%</math> of the original <math>30</math> liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are <math>9+5=14</math> liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This gives <math>40\%</math> of yellow tint in the new mixture, which is <math>\boxed{\textbf{(C) 40}}</math>.<br />
<br />
==Video Solution by WhyMath==<br />
https://youtu.be/5MfwBvLHUCw<br />
<br />
~savannahsolver<br />
<br />
==Video Solution==<br />
https://youtu.be/StRqfN7yYE8 Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_15&diff=1950562007 AMC 8 Problems/Problem 152023-07-02T04:20:22Z<p>Sootommylee: </p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>a, b</math> and <math>c</math> be numbers with <math>0 < a < b < c</math>. Which of the following is<br />
impossible?<br />
<br />
<math>\mathrm{(A)} \ a + c < b \qquad \mathrm{(B)} \ a \cdot b < c \qquad \mathrm{(C)} \ a + b < c \qquad \mathrm{(D)} \ a \cdot c < b \qquad \mathrm{(E)}\frac{b}{c} = a</math><br />
<br />
== Solution ==<br />
<br />
According to the given rules, every number needs to be positive. Since <math>c</math> is always greater than <math>b</math>, adding a positive number (<math>a</math>) to <math>c</math> will always make it greater than <math>b</math>.<br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(A)}\ a+c<b}</math><br />
<br />
<br />
==Solution 2==<br />
<br />
We can test numbers into the inequality we’re given. The simplest is <math>0<1<2<3</math>. We can see that <math>3+1>2</math>, so <math>\boxed{\textbf{(A)} }</math> is correct.<br />
<br />
—jason.ca<br />
<br />
==Video Solution by WhyMath==<br />
https://youtu.be/UdzJetT-XOY<br />
<br />
~savannahsolver<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=_ZHS4M7kpnE<br />
<br />
==Video Solution 2==<br />
https://youtu.be/GxR1giTQeD0 Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_15&diff=1950552007 AMC 8 Problems/Problem 152023-07-02T04:20:07Z<p>Sootommylee: </p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>a, b</math> and <math>c</math> be numbers with <math>0 < a < b < c</math>. Which of the following is<br />
impossible?<br />
<br />
<math>\mathrm{(A)} \ a + c < b \qquad \mathrm{(B)} \ a \cdot b < c \qquad \mathrm{(C)} \ a + b < c \qquad \mathrm{(D)} \ a \cdot c < b \qquad \mathrm{(E)}\frac{b}{c} = a</math><br />
<br />
== Solution ==<br />
<br />
According to the given rules, every number needs to be positive. Since <math>c</math> is always greater than <math>b</math>, adding a positive number (<math>a</math>) to <math>c</math> will always make it greater than <math>b</math>.<br />
<br />
Therefore, the answer is <math>\boxed{\textbf{(A)}\ a+c<b}</math><br />
<br />
<br />
==Solution 2==<br />
<br />
We can test numbers into the inequality we’re given. The simplest is <math>0<1<2<3</math>. We can see that <math>3+1>2</math>, so <math>\boxed{\textbf{(A)} }</math> is correct.<br />
<br />
—jason.ca<br />
<br />
==Video Solution by WhyMath==<br />
https://youtu.be/UdzJetT-XOY<br />
<br />
~savannahsolver<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=_ZHS4M7kpnE<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/GxR1giTQeD0 Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_15&diff=1950542005 AMC 8 Problems/Problem 152023-07-02T04:11:58Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
How many different isosceles triangles have integer side lengths and perimeter 23?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11</math><br />
<br />
==Solution==<br />
Let <math>b</math> be the base of the isosceles triangles, and let <math>a</math> be the lengths of the other legs. From this, <math>2a+b=23</math> and <math>b=23-2a</math>. From triangle inequality, <math>2a>b</math>, then plug in the value from the previous equation to get <math>2a>23-2a</math> or <math>a>5.75</math>. The maximum value of <math>a</math> occurs when <math>b=1</math>, in which from the first equation <math>a=11</math>. Thus, <math>a</math> can have integer side lengths from <math>6</math> to <math>11</math>, and there are <math>\boxed{\textbf{(C)}\ 6}</math> triangles.<br />
<br />
==Video Solution==<br />
https://youtu.be/DGUZn2BMuiE Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2005|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_17&diff=1950532004 AMC 8 Problems/Problem 172023-07-02T04:00:53Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Three friends have a total of <math>6</math> identical pencils, and each one has at least one pencil. In how many ways can this happen?<br />
<br />
<math>\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math><br />
<br />
==Solution 1==<br />
For each person to have at least one pencil, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use [[Ball-and-urn]] to find the number of possibilities is <math>\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.<br />
<br />
Solution by [[User:phoenixfire|phoenixfire]]<br />
Minor Edits by [[User:Yuvag|Yuvag]] : "... is <math>\binom{3+3-1}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>." to "... is <math>\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>."<br />
<br />
All credit still goes to phoenixfire.<br />
<br />
==Solution 2==<br />
Like in solution 1, for each person to have at least one pencil, assign one of the pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use number of non-negetive integral soutions. <br />
Let the three friends be <math>a, b, c</math> repectively.<br />
<br />
<math>a + b + c = 3</math><br />
The total being 3 and 2 plus signs, which implies<br />
<math>\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.<br />
<br />
Solution by [[User:phoenixfire|phoenixfire]]<br />
<br />
Minor Edits by [[User:Yuvag|Yuvag]] : "<math>\binom{3+2}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>." to "<math>\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.<br />
<br />
All credit still goes to phoenixfire.<br />
<br />
==Solution 3==<br />
For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the [[Ball-and-urn]] counting technique, shown below:<br />
<br />
<br />
<br />
for n = number of items, and s = slots:<br />
<br />
<br />
<math>\binom{n-1}{s-1}</math><br />
<br />
<br />
Now we can plug in our values, <br />
<br />
number of items = 6, and slots = 3:<br />
<br />
<br />
<math>\binom{6-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.<br />
<br />
<br />
<br />
<br />
Solution by [[User:Yuvag|Yuvag]]<br />
<br />
==Solution 4==<br />
Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups use casework. <br />
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.<br />
<br />
<math>a + b + c = 3</math>,<br />
<br />
<br />
Case <math>1:a=0</math>,<br />
<br />
<math>b + c = 3</math>,<br />
<br />
<math>b = 0,1,2,3</math> ,<br />
<br />
<math>c = 3,2,1,0</math>,<br />
<br />
<math>\boxed{\textbf\ 4}</math> solutions.<br />
<br />
<br />
Case <math>2:a=1</math>,<br />
<br />
<math>1 + b + c = 3</math>,<br />
<br />
<math>b + c = 2</math>,<br />
<br />
<math>b = 0,1,2</math> ,<br />
<br />
<math>c = 2,1,0</math> ,<br />
<br />
<math>\boxed{\textbf\ 3}</math> solutions.<br />
<br />
<br />
Case <math>3:a= 2</math>,<br />
<br />
<math>2 + b + c = 3</math>,<br />
<br />
<math>b + c = 1</math>,<br />
<br />
<math>b = 0,1</math>,<br />
<br />
<math>c = 1,0</math>,<br />
<br />
<math>\boxed{\textbf\ 2}</math> solutions.<br />
<br />
<br />
Case <math>4:a = 3</math>,<br />
<br />
<math>3 + b + c = 3</math>,<br />
<br />
<math>b + c = 0</math>,<br />
<br />
<math>b = 0</math>,<br />
<br />
<math>c = 0</math>,<br />
<br />
<math>\boxed{\textbf\ 1}</math> solution.<br />
<br />
Therefore there will be a total of 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.<br />
Solution by [[User:phoenixfire|phoenixfire]]<br />
<br />
<br />
==Video Solution==<br />
https://youtu.be/FUnwTLP7gr0 Soo, DRMS, NM<br />
<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_16&diff=1950522003 AMC 8 Problems/Problem 162023-07-02T03:51:24Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has <math>4</math> seats: <math>1</math> Driver seat, <math>1</math> front passenger seat, and <math>2</math> back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24 </math><br />
==Solution==<br />
There are only <math>2</math> people who can go in the driver's seat--Bonnie and Carlo. Any of the <math>3</math> remaining people can go in the front passenger seat. There are <math>2</math> people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are <math>2\cdot3\cdot2</math> or <math>12</math> ways. The answer is then <math>\boxed{\textbf{(D)}\ 12}</math>.<br />
<br />
==Solution 2 (Quick)==<br />
If there weren't any extra requirements, there would be 24 combinations. However, there are only 2, which is half of 4, ways to put the people. Therefore, half of 24 is <math>\boxed{\textbf{(D)}\ 12}</math>.<br />
<br />
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]]<br />
<br />
==Video Solution==<br />
https://youtu.be/GeAxPKwsL5E Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2003|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_16&diff=1950512003 AMC 8 Problems/Problem 162023-07-02T03:50:26Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has <math>4</math> seats: <math>1</math> Driver seat, <math>1</math> front passenger seat, and <math>2</math> back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24 </math><br />
==Solution==<br />
There are only <math>2</math> people who can go in the driver's seat--Bonnie and Carlo. Any of the <math>3</math> remaining people can go in the front passenger seat. There are <math>2</math> people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are <math>2\cdot3\cdot2</math> or <math>12</math> ways. The answer is then <math>\boxed{\textbf{(D)}\ 12}</math>.<br />
<br />
==Solution 2 (Quick)==<br />
If there weren't any extra requirements, there would be 24 combinations. However, there are only 2, which is half of 4, ways to put the people. Therefore, half of 24 is <math>\boxed{\textbf{(D)}\ 12}</math>.<br />
<br />
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]]<br />
<br />
==Video Solution==<br />
https://youtu.be/GeAxPKwsL5E Soo , DRMS<br />
<br />
==See Also==<br />
{{AMC8 box|year=2003|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_17&diff=1789402013 AMC 8 Problems/Problem 172022-10-09T03:50:21Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?<br />
<br />
<math>\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350</math><br />
<br />
==Solution 1==<br />
The mean of these numbers is <math>\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5</math>. Therefore the numbers are <math>333, 334, 335, 336, 337, 338</math>, so the answer is <math>\boxed{\textbf{(B)}\ 338}</math><br />
<br />
==Solution 2==<br />
Let the <math>4^{\text{th}}</math> number be <math>x</math>. Then our desired number is <math>x+2</math>. <br />
<br />
Our integers are <math>x-3,x-2,x-1,x,x+1,x+2</math>, so we have that <math>6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{\textbf{(B)}\ 338}</math>.<br />
<br />
==Solution 3==<br />
Let the first term be <math>x</math>. Our integers are <math>x,x+1,x+2,x+3,x+4,x+5</math>. We have, <math>6x+15=2013\implies x=333\implies x+5=\boxed{\textbf{(B)}\ 338}</math><br />
<br />
==Solution 4==<br />
Since there are <math>6</math> numbers, we divide <math>2013</math> by <math>6</math> to find the mean of the numbers. <math>\frac{2013}{6} = 335 \frac{1}{2}</math>.<br />
Then, <math>335 \frac{1}{2} + \frac{1}{2} = 336</math> (the fourth number). Fifth: <math>337</math>; Sixth: <math>\boxed {338}</math>.<br />
<br />
==Solution 5==<br />
Let the <math>6th</math> number be <math>x</math>. Then our list is: <math>x-6+x-5+x-4+x-3+x-x-1=2013</math>. Simplifying this gets you <math>6x-21=2013\implies 6x=2034</math>, which means that <math>x = \boxed{\textbf{(B)}338}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/E7DZ1oILqE0 ~savannahsolver<br />
<br />
<br />
== Video Solution 2 ==<br />
https://youtu.be/AvOXHvvTlio Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2013|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_21&diff=1789392011 AMC 8 Problems/Problem 212022-10-09T03:44:59Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Students guess that Norb's age is <math>24, 28, 30, 32, 36, 38, 41, 44, 47</math>, and <math>49</math>. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?<br />
<br />
<math> \textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48 </math><br />
<br />
==Solution==<br />
If at least half the guesses are too low, then Norb's age must be greater than <math>36.</math><br />
<br />
If two of the guesses are off by one, then his age is in between two guesses whose difference is <math>2.</math> It could be <math>31,37,</math> or <math>48,</math> but because his age is greater than <math>36</math> it can only be <math>37</math> or <math>48.</math><br />
<br />
Lastly, Norb's age is a prime number so the answer must be <math>\boxed{\textbf{(C)}\ 37}</math><br />
<br />
==Soultion 2 (Alternative approach)==<br />
Since two guesses are off by one, we know that both <math>x+1</math> and <math>x-1</math> are in the list where <math>x</math> is the age of Norb. Now, we know that <math>x+1</math> and <math>x-1</math> are <math>28</math> and <math>30</math>, <math>30</math> and <math>32</math>, and <math>36</math> and <math>38</math>. From these values, we know that <math>x</math> must be <math>29</math>, <math>31</math>, and <math>37</math>. Since half of the guesses are too low, <math>24, 28, 30, 32,</math> and <math>36</math> are all too low so we can eliminate all numbers in our list lesser than or equal to <math>36</math>. Therefore, our list has only <math>37</math> left so the answer is <math>\boxed{\textbf{(C)}\ 37}</math>. <br />
~ rabbit317<br />
<br />
== Video Solution ==<br />
https://youtu.be/HISL2-N5NVg?t=3886<br />
<br />
~ pi_is_3.14<br />
<br />
<br />
== Video Solution 2 ==<br />
https://youtu.be/lhBDgiYKpgs. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_20&diff=1789372011 AMC 8 Problems/Problem 202022-10-09T03:37:34Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
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 />
==Video Solution==<br />
https://youtu.be/SiP_0xLVFSo<br />
<br />
<br />
== Video Solution 2 ==<br />
https://youtu.be/_K32SZW_Jko. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_18&diff=1789352011 AMC 8 Problems/Problem 182022-10-09T03:31:43Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
A fair <math>6</math> sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?<br />
<br />
<math> \textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56 </math><br />
<br />
==Solution==<br />
There are <math>6\cdot6=36</math> ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining <math>36-6=30</math> ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are <math>30/2=15</math> ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is<br />
<br />
<cmath>\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}</cmath><br />
<br />
<br />
== Video Solution ==<br />
https://youtu.be/lfrofNkOol0. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_17&diff=1789332011 AMC 8 Problems/Problem 172022-10-09T02:46:30Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Let <math>w</math>, <math>x</math>, <math>y</math>, and <math>z</math> be whole numbers. If <math>2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588</math>, then what does <math>2w + 3x + 5y + 7z</math> equal?<br />
<br />
<math> \textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56 </math><br />
<br />
==Solution==<br />
The [[prime factorization]] of <math>588</math> is <math>2^2\cdot3\cdot7^2.</math> We can see <math>w=2, x=1,</math> and <math>z=2.</math> Because <math>5^0=1, y=0.</math><br />
<br />
<cmath>2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}</cmath><br />
<br />
==Video Solution==<br />
https://youtu.be/PxBKpg-HKu8<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/5vpKkAue8Is. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_20&diff=1789322000 AMC 8 Problems/Problem 202022-10-09T02:11:53Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of &#36;<math>1.02</math>, with at least one coin of each type. How many dimes must you have?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math>\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5</math><br />
<br />
==Solution==<br />
Since you have one coin of each type, <math>1 + 5 + 10 + 25 = 41</math> cents are already determined, leaving you with a total of <math>102 - 41 = 61</math> cents remaining for <math>5</math> coins.<br />
<br />
You must have <math>1</math> more penny. If you had more than <math>1</math> penny, you must have at least <math>6</math> pennies to leave a multiple of <math>5</math> for the nickels, dimes, and quarters. But you only have <math>5</math> more coins to assign.<br />
<br />
Now you have <math>61 - 1 = 60</math> cents remaining for <math>4</math> coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves <math>50</math> cents in <math>3</math> nickels or quarters, which is impossible. If you have two dimes, that leaves <math>40</math> cents for <math>2</math> nickels or quarters, which is again impossible. If you have three dimes, that leaves <math>30</math> cents for <math>1</math> nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough.<br />
<br />
Therefore, you must have no more dimes to assign, and the <math>60</math> cents in <math>4</math> coins must be divided between the quarters and nickels. We quickly see that <math>2</math> nickels and <math>2</math> quarters work. Thus, the total count is <math>2</math> quarters, <math>2</math> nickels, <math>1</math> penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total <math>2 + 2 + 1 + 4 = 9</math> coins, and a total of <math>2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102</math> cents.<br />
<br />
There is only <math>1</math> dime in that combo, so the answer is <math>\boxed{A}</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/HISL2-N5NVg?t=1409<br />
<br />
~ pi_is_3.14<br />
<br />
<br />
== Video Solution 2 ==<br />
https://youtu.be/Vm-proRV5wI. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2000|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_20&diff=1789312000 AMC 8 Problems/Problem 202022-10-09T02:11:42Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of &#36;<math>1.02</math>, with at least one coin of each type. How many dimes must you have?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math>\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5</math><br />
<br />
==Solution==<br />
Since you have one coin of each type, <math>1 + 5 + 10 + 25 = 41</math> cents are already determined, leaving you with a total of <math>102 - 41 = 61</math> cents remaining for <math>5</math> coins.<br />
<br />
You must have <math>1</math> more penny. If you had more than <math>1</math> penny, you must have at least <math>6</math> pennies to leave a multiple of <math>5</math> for the nickels, dimes, and quarters. But you only have <math>5</math> more coins to assign.<br />
<br />
Now you have <math>61 - 1 = 60</math> cents remaining for <math>4</math> coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves <math>50</math> cents in <math>3</math> nickels or quarters, which is impossible. If you have two dimes, that leaves <math>40</math> cents for <math>2</math> nickels or quarters, which is again impossible. If you have three dimes, that leaves <math>30</math> cents for <math>1</math> nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough.<br />
<br />
Therefore, you must have no more dimes to assign, and the <math>60</math> cents in <math>4</math> coins must be divided between the quarters and nickels. We quickly see that <math>2</math> nickels and <math>2</math> quarters work. Thus, the total count is <math>2</math> quarters, <math>2</math> nickels, <math>1</math> penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total <math>2 + 2 + 1 + 4 = 9</math> coins, and a total of <math>2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102</math> cents.<br />
<br />
There is only <math>1</math> dime in that combo, so the answer is <math>\boxed{A}</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/HISL2-N5NVg?t=1409<br />
<br />
~ pi_is_3.14<br />
<br />
<br />
== Video Solution 2 ==<br />
https://youtu.be/Vm-proRV5wI. Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2000|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_19&diff=1789212013 AMC 8 Problems/Problem 192022-10-08T22:57:13Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?<br />
<br />
<math>\textbf{(A)}\ \text{Hannah, Cassie, Bridget} \qquad \textbf{(B)}\ \text{Hannah, Bridget, Cassie} \\ \qquad \textbf{(C)}\ \text{Cassie, Bridget, Hannah} \qquad \textbf{(D)}\ \text{Cassie, Hannah, Bridget} \\ \qquad \textbf{(E)}\ \text{Bridget, Cassie, Hannah}</math><br />
<br />
==Solution==<br />
If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class. Therefore, Hannah did better than Bridget, so our order is <math>\boxed{\textbf{(D) Cassie, Hannah, Bridget}}</math><br />
<br />
<br />
<br />
Note: It could be said that Cassie did better than Hannah, and Bridget knew this, and she had the same score as Hannah, so Bridget didn't get the highest score. But that is not in the answer choices so...<br />
<br />
==Video Solution==<br />
https://youtu.be/CqmAwiiMeBQ ~savannahsolver<br />
<br />
<br />
== Video Solution 2 ==<br />
https://youtu.be/U3VrL9yKh3g. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2013|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_19&diff=1789202013 AMC 8 Problems/Problem 192022-10-08T22:55:36Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?<br />
<br />
<math>\textbf{(A)}\ \text{Hannah, Cassie, Bridget} \qquad \textbf{(B)}\ \text{Hannah, Bridget, Cassie} \\ \qquad \textbf{(C)}\ \text{Cassie, Bridget, Hannah} \qquad \textbf{(D)}\ \text{Cassie, Hannah, Bridget} \\ \qquad \textbf{(E)}\ \text{Bridget, Cassie, Hannah}</math><br />
<br />
==Solution==<br />
If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class. Therefore, Hannah did better than Bridget, so our order is <math>\boxed{\textbf{(D) Cassie, Hannah, Bridget}}</math><br />
<br />
<br />
<br />
Note: It could be said that Cassie did better than Hannah, and Bridget knew this, and she had the same score as Hannah, so Bridget didn't get the highest score. But that is not in the answer choices so...<br />
<br />
==Video Solution==<br />
https://youtu.be/CqmAwiiMeBQ ~savannahsolver<br />
<br />
<br />
== Video Solution 2 ==<br />
https://youtu.be/U3VrL9yKh3g. Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2013|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=1789192011 AMC 8 Problems/Problem 252022-10-08T22:48:54Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
A circle with radius <math>1</math> is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br />
<br />
<asy><br />
filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black);<br />
filldraw(Circle((0,0),1), mediumgray,black);<br />
filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 </math><br />
<br />
==Solution 1==<br />
The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: <math>2 \cdot <br />
2 \cdot \frac{1}{2}=2.</math><br />
<br />
The circle's shaded area is the area of the smaller square subtracted from the area of the circle: <math>\pi - 2.</math><br />
<br />
If you draw the diagonals of the smaller square, you will see that the larger square is split <math>4</math> congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: <math>2.</math><br />
<br />
Approximating <math>\pi</math> to <math>3.14,</math> the ratio of the circle's shaded area to the area between the two squares is about<br />
<br />
<cmath>\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}</cmath><br />
<br />
==Solution 2==<br />
For the ratio of the circle's shaded area to the area between the squares to be <math>1,</math> they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
Note that this solution is not rigorous, because we still should show that the ratio is less than <math>\frac{3}{4}</math>.<br />
<br />
<br />
==Solution 3==<br />
<br />
Set the side length of the bigger square to be <math>8</math>. <br />
Then the area of the big square is <math>8^2 =64</math> and <br />
the area of the small square <math>(4\sqrt{2})^2 = 32</math>. <br />
The difference is <math>32</math>. <br />
The area of the circle is <math>4^2</math> times <math>\pi</math> which is <math>16 \pi</math> or about <math>48</math>. <br />
Knowing the area of the small square is <math>32</math>. <math>48-32</math> is <math>16</math>. <br />
The area of the big square is <math>64</math>. <br />
So <math>32/64</math> is <math>1/2</math>, or <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
<br />
- by goldenn<br />
<br />
- (tex) updated by CasperYC on 26th July 2020<br />
<br />
<br />
== Video Solution ==<br />
https://youtu.be/PR-45YsyBGM Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=1789182011 AMC 8 Problems/Problem 252022-10-08T22:48:37Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
A circle with radius <math>1</math> is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br />
<br />
<asy><br />
filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black);<br />
filldraw(Circle((0,0),1), mediumgray,black);<br />
filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 </math><br />
<br />
==Solution 1==<br />
The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: <math>2 \cdot <br />
2 \cdot \frac{1}{2}=2.</math><br />
<br />
The circle's shaded area is the area of the smaller square subtracted from the area of the circle: <math>\pi - 2.</math><br />
<br />
If you draw the diagonals of the smaller square, you will see that the larger square is split <math>4</math> congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: <math>2.</math><br />
<br />
Approximating <math>\pi</math> to <math>3.14,</math> the ratio of the circle's shaded area to the area between the two squares is about<br />
<br />
<cmath>\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}</cmath><br />
<br />
==Solution 2==<br />
For the ratio of the circle's shaded area to the area between the squares to be <math>1,</math> they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
Note that this solution is not rigorous, because we still should show that the ratio is less than <math>\frac{3}{4}</math>.<br />
<br />
<br />
==Solution 3==<br />
<br />
Set the side length of the bigger square to be <math>8</math>. <br />
Then the area of the big square is <math>8^2 =64</math> and <br />
the area of the small square <math>(4\sqrt{2})^2 = 32</math>. <br />
The difference is <math>32</math>. <br />
The area of the circle is <math>4^2</math> times <math>\pi</math> which is <math>16 \pi</math> or about <math>48</math>. <br />
Knowing the area of the small square is <math>32</math>. <math>48-32</math> is <math>16</math>. <br />
The area of the big square is <math>64</math>. <br />
So <math>32/64</math> is <math>1/2</math>, or <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
<br />
- by goldenn<br />
<br />
- (tex) updated by CasperYC on 26th July 2020<br />
<br />
<br />
== Video Solution ==<br />
https://youtu.be/PR-45YsyBGM Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_19&diff=1789152010 AMC 8 Problems/Problem 192022-10-08T22:21:04Z<p>Sootommylee: </p>
<hr />
<div>== Problem ==<br />
The two circles pictured have the same center <math>C</math>. Chord <math>\overline{AD}</math> is tangent to the inner circle at <math>B</math>, <math>AC</math> is <math>10</math>, and chord <math>\overline{AD}</math> has length <math>16</math>. What is the area between the two circles?<br />
<br />
<asy><br />
unitsize(45);<br />
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1);<br />
draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0));<br />
dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle);<br />
</asy><br />
<br />
<br />
<math> \textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi </math><br />
<br />
== Solution == <br />
Since <math>\triangle ACD</math> is isosceles, <math>CB</math> bisects <math>AD</math>. Thus <math>AB=BD=8</math>. From the Pythagorean Theorem, <math>CB=6</math>. Thus the area between the two circles is<br />
<math>100\pi - 36\pi=64\pi</math> <math>\boxed{\textbf{(C)}\ 64\pi}</math><br />
<br />
Note: The length <math>AC</math> is necessary information, as this tells us the radius of the larger circle. The area of the annulus is <math>\pi(AC^2-BC^2)=\pi AB^2=64\pi</math>.<br />
<br />
<br />
== Video Solution ==<br />
https://youtu.be/Q6rnoQChiyU. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_25&diff=1789112000 AMC 8 Problems/Problem 252022-10-08T21:52:35Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
The area of rectangle <math>ABCD</math> is <math>72</math> units squared. If point <math>A</math> and the midpoints of <math> \overline{BC} </math> and <math> \overline{CD} </math> are joined to form a triangle, the area of that triangle is<br />
<br />
<asy><br />
pair A,B,C,D;<br />
A = (0,8); B = (9,8); C = (9,0); D = (0,0);<br />
draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);</asy><br />
<br />
<math> \text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40 </math><br />
<br />
==Solution 1==<br />
<br />
To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that <math>ABCD</math> can have any dimension. Give the rectangle dimensions of <math>AB = CD = 12</math> and <math>BC = AD= 6</math>, which is the easiest way to avoid fractions. Labelling the right midpoint as <math>M</math>, and the bottom midpoint as <math>N</math>, we know that <math>DN = NC = 6</math>, and <math>BM = MC = 3</math>. <br />
<br />
<math>[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18</math><br />
<br />
<math>[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9</math><br />
<br />
<math>[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18</math><br />
<br />
<math>[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]</math><br />
<br />
<math>[\triangle AMN] = 72 - 18 - 9 - 18</math><br />
<br />
<math>[\triangle AMN] = 27</math>, and the answer is <math>\boxed{B}</math> <br />
<br />
==Solution 2==<br />
<br />
The above answer is fast, but satisfying, and assumes that the area of <math>\triangle AMN</math> is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label <math>AB = CD = l</math> and <math>BC = DA = h</math><br />
<br />
Labelling <math>m</math> and <math>n</math> as the right and lower midpoints respectively, and redoing all the work above, we get:<br />
<br />
<math>[\triangle ABN] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}</math><br />
<br />
<math>[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}</math><br />
<br />
<math>[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}</math><br />
<br />
<math>[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]</math><br />
<br />
<math>[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}</math><br />
<br />
<math>[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27</math>, and the answer is <math>\boxed{B}</math><br />
<br />
<br />
== Video Solution ==<br />
https://youtu.be/yoIO9q_GTig. Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=24|after=Last<br />Question}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_23&diff=1789102000 AMC 8 Problems/Problem 232022-10-08T21:44:30Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
There is a list of seven numbers. The average of the first four numbers is <math>5</math>, and the average of the last four numbers is <math>8</math>. If the average of all seven numbers is <math> 6\frac{4}{7} </math>, then the number common to both sets of four numbers is<br />
<br />
<math> \text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7} </math><br />
<br />
==Solution==<br />
<br />
Remember that if a list of <math>n</math> numbers has an average of <math>k</math>, then the sum <math>S</math> of all the numbers on the list is <math>S = nk</math>.<br />
<br />
So if the average of the first <math>4</math> numbers is <math>5</math>, then the first four numbers total <math>4 \cdot 5 = 20</math>.<br />
<br />
If the average of the last <math>4</math> numbers is <math>8</math>, then the last four numbers total <math>4 \cdot 8 = 32</math>.<br />
<br />
If the average of all <math>7</math> numbers is <math>6\frac{4}{7}</math>, then the total of all seven numbers is <math>7 \cdot 6\frac{4}{7} = 7\cdot 6 + 4 = 46</math>.<br />
<br />
If the first four numbers are <math>20</math>, and the last four numbers are <math>32</math>, then all "eight" numbers are <math>20 + 32 = 52</math>. But that's counting one number twice. Since the sum of all seven numbers is <math>46</math>, then the number that was counted twice is <math>52 - 46 = 6</math>, and the answer is <math>\boxed{(B) 6}</math><br />
<br />
Algebraically, if <math>a + b + c + d = 20</math>, and <math>d + e + f + g = 32</math>, you can add both equations to get <math>a + b + c + 2d + e + f + g = 52</math>. You know that <math>a + b + c + d + e + f + g = 46</math>, so you can subtract that from the last equation to get <math>d = \boxed{(B) 6}</math>, and <math>d</math> is the number that appeared twice.<br />
<br />
Yay! :D<br />
<br />
<br />
== Video Solution ==<br />
https://youtu.be/xdNSN7hYK6k Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_21&diff=1789092000 AMC 8 Problems/Problem 212022-10-08T21:21:47Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is<br />
<br />
<math> \text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution==<br />
Divide it into <math>2</math> cases:<br />
<br />
1) Keiko and Ephriam both get <math>0</math> heads:<br />
This means that they both roll all tails, so there is only <math>1</math> way for this to happen.<br />
<br />
2) Keiko and Ephriam both get <math>1</math> head:<br />
For Keiko, there is only <math>1</math> way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are <math>2</math> choices for when he can flip the head. So, in total there are <math>2 \cdot 1 = 2</math> ways for this case.<br />
<br />
Thus, in total there are <math>3</math> ways that work. Since there are <math>2</math> choices for each coin flip (Heads or Tails), there are <math>2^3 = 8</math> total ways of flipping 3 coins.<br />
<br />
Thus, since all possible coin flips of 3 coins are equally likely, the probability is <math>\boxed{(B) \frac38}</math>.<br />
<br />
~pi_is_3.14<br />
<br />
==Solution 2==<br />
<br />
Let <math>K(n)</math> be the probability that Keiko gets <math>n</math> heads, and let <math>E(n)</math> be the probability that Ephriam gets <math>n</math> heads.<br />
<br />
<math>K(0) = \frac{1}{2}</math><br />
<br />
<math>K(1) = \frac{1}{2}</math><br />
<br />
<math>K(2) = 0</math> (Keiko only has one penny!)<br />
<br />
<math>E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math><br />
<br />
<math>E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = 2\cdot\frac{1}{4} = \frac{1}{2}</math> (because Ephraim can get HT or TH)<br />
<br />
<math>E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math><br />
<br />
The probability that Keiko gets <math>0</math> heads and Ephriam gets <math>0</math> heads is <math>K(0)\cdot E(0)</math>. Similarly for <math>1</math> head and <math>2</math> heads. Thus, we have:<br />
<br />
<math>P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)</math><br />
<br />
<math>P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0</math><br />
<br />
<math>P = \frac{3}{8}</math><br />
<br />
Thus the answer is <math>\boxed{B}</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/a_Tfeb_6dqE Soo, DRMS, NM<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_19&diff=1789082000 AMC 8 Problems/Problem 192022-10-08T21:13:29Z<p>Sootommylee: /* Video Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Three circular arcs of radius <math>5</math> units bound the region shown. Arcs <math>AB</math> and <math>AD</math> are quarter-circles, and arc <math>BCD</math> is a semicircle. What is the area, in square units, of the region?<br />
<br />
<asy><br />
pair A,B,C,D;<br />
A = (0,0);<br />
B = (-5,5);<br />
C = (0,10);<br />
D = (5,5);<br />
draw(arc((-5,0),A,B,CCW));<br />
draw(arc((0,5),B,D,CW));<br />
draw(arc((5,0),D,A,CCW));<br />
label("$A$",A,S);<br />
label("$B$",B,W);<br />
label("$C$",C,N);<br />
label("$D$",D,E);</asy><br />
<br />
<math> \text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi </math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
Draw two squares: one that has opposing corners at <math>A</math> and <math>B</math>, and one that has opposing corners at <math>A</math> and <math>D</math>. These squares share side <math>\overline{AO}</math>, where <math>O</math> is the center of the large semicircle. <br />
<br />
These two squares have a total area of <math>2 \cdot 5^2</math>, but have two quarter circle "bites" of radius <math>5</math> that must be removed. Thus, the bottom part of the figure has area <br />
<br />
<math>2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2</math><br />
<br />
<math>50 - \frac{25\pi}{2}</math><br />
<br />
This is the area of the part of the figure underneath <math>\overline{BD}</math>. The part of the figure over <math>\overline{BD}</math> is just a semicircle with radius <math>5</math>, which has area of <math>\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}</math><br />
<br />
Adding the two areas gives a total area of <math>50</math>, for an answer of <math>\boxed{C}</math><br />
<br />
=== Solution 2 ===<br />
Draw line <math>\overline{BD}</math>. Then draw <math>\overline {CO}</math>, where <math>O</math> is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length <math>\overline {BD}</math> and height <math>\overline {AO}</math>, which are equal to <math>10</math> and <math>5</math>, respectively. Thus, the total area is <math>50</math>, and the answer is <math>\boxed{C}</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/kLH0ql186UE Soo, DRMS, NM<br />
<br />
== See Also ==<br />
{{AMC8 box|year=2000|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_19&diff=1789072000 AMC 8 Problems/Problem 192022-10-08T21:13:12Z<p>Sootommylee: </p>
<hr />
<div>== Problem ==<br />
Three circular arcs of radius <math>5</math> units bound the region shown. Arcs <math>AB</math> and <math>AD</math> are quarter-circles, and arc <math>BCD</math> is a semicircle. What is the area, in square units, of the region?<br />
<br />
<asy><br />
pair A,B,C,D;<br />
A = (0,0);<br />
B = (-5,5);<br />
C = (0,10);<br />
D = (5,5);<br />
draw(arc((-5,0),A,B,CCW));<br />
draw(arc((0,5),B,D,CW));<br />
draw(arc((5,0),D,A,CCW));<br />
label("$A$",A,S);<br />
label("$B$",B,W);<br />
label("$C$",C,N);<br />
label("$D$",D,E);</asy><br />
<br />
<math> \text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi </math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
Draw two squares: one that has opposing corners at <math>A</math> and <math>B</math>, and one that has opposing corners at <math>A</math> and <math>D</math>. These squares share side <math>\overline{AO}</math>, where <math>O</math> is the center of the large semicircle. <br />
<br />
These two squares have a total area of <math>2 \cdot 5^2</math>, but have two quarter circle "bites" of radius <math>5</math> that must be removed. Thus, the bottom part of the figure has area <br />
<br />
<math>2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2</math><br />
<br />
<math>50 - \frac{25\pi}{2}</math><br />
<br />
This is the area of the part of the figure underneath <math>\overline{BD}</math>. The part of the figure over <math>\overline{BD}</math> is just a semicircle with radius <math>5</math>, which has area of <math>\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}</math><br />
<br />
Adding the two areas gives a total area of <math>50</math>, for an answer of <math>\boxed{C}</math><br />
<br />
=== Solution 2 ===<br />
Draw line <math>\overline{BD}</math>. Then draw <math>\overline {CO}</math>, where <math>O</math> is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length <math>\overline {BD}</math> and height <math>\overline {AO}</math>, which are equal to <math>10</math> and <math>5</math>, respectively. Thus, the total area is <math>50</math>, and the answer is <math>\boxed{C}</math>.<br />
<br />
== Video Solution 1 ==<br />
https://youtu.be/kLH0ql186UE Soo, DRMS, NM<br />
<br />
== See Also ==<br />
{{AMC8 box|year=2000|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_14&diff=1737302009 AMC 8 Problems/Problem 142022-05-02T05:28:12Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
Austin and Temple are <math> 50</math> miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging <math> 60</math> miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged <math> 40</math> miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?<br />
<br />
<math> \textbf{(A)}\ 46 \qquad<br />
\textbf{(B)}\ 48 \qquad<br />
\textbf{(C)}\ 50 \qquad<br />
\textbf{(D)}\ 52 \qquad<br />
\textbf{(E)}\ 54</math><br />
<br />
==Solution==<br />
The way to Temple took <math>\frac{50}{60}=\frac56</math> hours, and the way back took <math>\frac{50}{40}=\frac54</math> for a total of <math>\frac56 + \frac54 = \frac{25}{12}</math> hours. The trip is <math>50\cdot2=100</math> miles. The average speed is <math>\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}</math> miles per hour.<br />
<br />
==Solution 2==<br />
We calculate the harmonic mean of Austin and Temple. <br />
Plugging in, we have: <math>\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}</math> miles per hour.<br />
<br />
<br />
==Video Solution==<br />
https://youtu.be/CnWuadfT3xA Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
2015 Problem 17<br />
{{AMC8 box|year=2009|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_14&diff=1737292009 AMC 8 Problems/Problem 142022-05-02T05:28:01Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
Austin and Temple are <math> 50</math> miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging <math> 60</math> miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged <math> 40</math> miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?<br />
<br />
<math> \textbf{(A)}\ 46 \qquad<br />
\textbf{(B)}\ 48 \qquad<br />
\textbf{(C)}\ 50 \qquad<br />
\textbf{(D)}\ 52 \qquad<br />
\textbf{(E)}\ 54</math><br />
<br />
==Solution==<br />
The way to Temple took <math>\frac{50}{60}=\frac56</math> hours, and the way back took <math>\frac{50}{40}=\frac54</math> for a total of <math>\frac56 + \frac54 = \frac{25}{12}</math> hours. The trip is <math>50\cdot2=100</math> miles. The average speed is <math>\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}</math> miles per hour.<br />
<br />
==Solution 2==<br />
We calculate the harmonic mean of Austin and Temple. <br />
Plugging in, we have: <math>\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}</math> miles per hour.<br />
<br />
<br />
==Video Solution==<br />
https://youtu.be/CnWuadfT3xA Soo, DRMS, NM<br />
<br />
==See Also==<br />
2015 Problem 17<br />
{{AMC8 box|year=2009|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_18&diff=1737282013 AMC 8 Problems/Problem 182022-05-02T05:05:46Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
<br />
Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?<br />
<br />
<asy>import three;<br />
currentprojection=orthographic(-8,15,15);<br />
triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P;<br />
A = (0,0,0);<br />
B = (0,10,0);<br />
C = (12,10,0);<br />
D = (12,0,0);<br />
E = (0,0,5);<br />
F = (0,10,5);<br />
G = (12,10,5);<br />
H = (12,0,5);<br />
I = (1,1,1);<br />
J = (1,9,1);<br />
K = (11,9,1);<br />
L = (11,1,1);<br />
M = (1,1,5);<br />
N = (1,9,5);<br />
O = (11,9,5);<br />
P = (11,1,5);<br />
//outside box far<br />
draw(surface(A--B--C--D--cycle),white,nolight);<br />
draw(A--B--C--D--cycle);<br />
draw(surface(E--A--D--H--cycle),white,nolight);<br />
draw(E--A--D--H--cycle);<br />
draw(surface(D--C--G--H--cycle),white,nolight);<br />
draw(D--C--G--H--cycle);<br />
//inside box far<br />
draw(surface(I--J--K--L--cycle),white,nolight);<br />
draw(I--J--K--L--cycle);<br />
draw(surface(I--L--P--M--cycle),white,nolight);<br />
draw(I--L--P--M--cycle);<br />
draw(surface(L--K--O--P--cycle),white,nolight);<br />
draw(L--K--O--P--cycle);<br />
//inside box near<br />
draw(surface(I--J--N--M--cycle),white,nolight);<br />
draw(I--J--N--M--cycle);<br />
draw(surface(J--K--O--N--cycle),white,nolight);<br />
draw(J--K--O--N--cycle);<br />
//outside box near<br />
draw(surface(A--B--F--E--cycle),white,nolight);<br />
draw(A--B--F--E--cycle);<br />
draw(surface(B--C--G--F--cycle),white,nolight);<br />
draw(B--C--G--F--cycle);<br />
//top<br />
draw(surface(E--H--P--M--cycle),white,nolight);<br />
draw(surface(E--M--N--F--cycle),white,nolight);<br />
draw(surface(F--N--O--G--cycle),white,nolight);<br />
draw(surface(O--G--H--P--cycle),white,nolight);<br />
draw(M--N--O--P--cycle);<br />
draw(E--F--G--H--cycle);<br />
<br />
label("10",(A--B),SE);<br />
label("12",(C--B),SW);<br />
label("5",(F--B),W);</asy><br />
<br />
<math>\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600</math><br />
<br />
==Solution 1==<br />
<br />
There are <math>10 \cdot 12 = 120</math> cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are <math>9 + 11 + 9 + 11 = 40</math> cubes. Hence, the answer is <math>120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can just calculate the volume of the prism that was cut out of the original <math>12\times 10\times 5</math> box. Each interior side of the fort will be <math>2</math> feet shorter than each side of the outside. Since the floor is <math>1</math> foot, the height will be <math>4</math> feet. So the volume of the interior box is <math>10\times 8\times 4=320\text{ ft}^3</math>.<br />
<br />
The volume of the original box is <math>12\times 10\times 5=600\text{ ft}^3</math>. Therefore, the number of blocks contained in the fort is <math>600-320=\boxed{\textbf{(B)}\ 280}</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/FDgcLW4frg8?t=5261<br />
<br />
~ pi_is_3.14<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/WlMUXUloTFM Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
(Other problems)<br />
{{AMC8 box|year=2013|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_19&diff=1737272012 AMC 8 Problems/Problem 192022-05-02T04:59:02Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?<br />
<br />
<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12 </math><br />
<br />
==Solution 1==<br />
<br />
6 are blue and green- b+g=6<br />
<br />
8 are red and blue- r+b=8<br />
<br />
4 are red and green- r+g=4<br />
<br />
<br />
We can do trial and error. Let's make blue 5. That makes green 1 and red 3 because 6-5=1 and 8-5=3. To check this let's plug 1 and 3 into r+g=4 and it does work. Now count the number of marbles- 5+3+1=9. So 9 (C) is the answer.<br />
<br />
==Solution 2==<br />
<br />
We already knew the facts: <math>6</math> are blue and green, meaning <math>b+g=6</math>; <math>8</math> are red and blue, meaning <math>r+b=8</math>; <math>4</math> are red and green, meaning <math>r+g=4</math>. Then we need to add these three equations: <math>b+g+r+b+r+g=2(r+g+b)=6+8+4=18</math>. It gives us all of the marbles are <math>r+g+b = 18/2 = 9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. ---LarryFlora<br />
<br />
==Solution 3 Venn Diagrams==<br />
We may draw three Venn diagrams to represent these three cases, respectively. <br />
<br />
[[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]] <br />
<br />
Let the amount of all the marbles is <math>x</math>, meaning <math>R+G+B = x</math>. <br />
<br />
The Venn diagrams give us the equation: <math>x = (x-6)+(x-8)+(x-4)</math>. <br />
So <math>x = 3x-18</math>, <math>x = 18/2 =9</math>. <br />
Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. ---LarryFlora<br />
<br />
==Solution 4 Venn Diagrams==<br />
We may draw three Venn diagrams to represent these three cases, respectively. <br />
<br />
[[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]] <br />
<br />
Let the amount of all the marbles is <math>x</math>, meaning <math>R+G+B = x</math>. <br />
<br />
Adding the three Venn diagrams, it gives us the equation: <math>x+18 = 3x</math>. <br />
So <math>2x = 18</math>, <math>x = 18/2 =9</math>. <br />
Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. ---LarryFlora<br />
<br />
<br />
==Video Solution==<br />
https://youtu.be/mMph7QH1kX0 Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2012|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_24&diff=1737262012 AMC 8 Problems/Problem 242022-05-02T04:56:42Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?<br />
<br />
<asy><br />
size(0,50);<br />
draw((-1,1)..(-2,2)..(-3,1)..(-2,0)..cycle);<br />
dot((-1,1));<br />
dot((-2,2));<br />
dot((-3,1));<br />
dot((-2,0));<br />
draw((1,0){up}..{left}(0,1));<br />
dot((1,0));<br />
dot((0,1));<br />
draw((0,1){right}..{up}(1,2));<br />
dot((1,2));<br />
draw((1,2){down}..{right}(2,1));<br />
dot((2,1));<br />
draw((2,1){left}..{down}(1,0));</asy><br />
<br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{4-\pi}{\pi}\qquad\textbf{(B)}\hspace{.05in}\frac{1}\pi\qquad\textbf{(C)}\hspace{.05in}\frac{\sqrt2}{\pi}\qquad\textbf{(D)}\hspace{.05in}\frac{\pi-1}{\pi}\qquad\textbf{(E)}\hspace{.05in}\frac{3}\pi </math><br />
<br />
<br />
==Solution==<br />
<asy><br />
dot((0,0),red);<br />
dot((0,2),red);<br />
dot((2,0),red);<br />
dot((2,2),red);<br />
draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,red);<br />
size(0,50);<br />
draw((1,0){up}..{left}(0,1));<br />
dot((1,0));<br />
dot((0,1));<br />
draw((0,1){right}..{up}(1,2));<br />
dot((1,2));<br />
draw((1,2){down}..{right}(2,1));<br />
dot((2,1));<br />
draw((2,1){left}..{down}(1,0));</asy><br />
<br />
Draw a square around the star figure. The side length of this square is <math> 4 </math>, because the side length is the diameter of the circle. The square forms <math>4</math>-quarter circles around the star figure. This is the equivalent of one large circle with radius <math> 2 </math>, meaning that the total area of the quarter circles is <math> 4\pi </math>. The area of the square is <math> 16 </math>. Thus, the area of the star figure is <math> 16 - 4\pi </math>. The area of the circle is <math> 4\pi </math>. Taking the ratio of the two areas, we find the answer is <math> \boxed{\textbf{(A)}\ \frac{4-\pi}{\pi}} </math>.<br />
<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=hZB77LITy5w<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/npKHhh30Eeg Soo, DRMS, NM<br />
<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2012|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_19&diff=1737252009 AMC 8 Problems/Problem 192022-05-02T04:52:16Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Two angles of an isosceles triangle measure <math> 70^\circ</math> and <math> x^\circ</math>. What is the sum of the three possible values of <math> x</math>?<br />
<br />
<math> \textbf{(A)}\ 95 \qquad<br />
\textbf{(B)}\ 125 \qquad<br />
\textbf{(C)}\ 140 \qquad<br />
\textbf{(D)}\ 165 \qquad<br />
\textbf{(E)}\ 180</math><br />
<br />
<br />
==Solution==<br />
<br />
There are 3 cases: where <math> x^\circ</math> is a base angle with the <math> 70^\circ</math> as the other angle, where <math> x^\circ</math> is a base angle with <math> 70^\circ</math> as the vertex angle, and where <math> x^\circ</math> is the vertex angle with <math> 70^\circ</math> as a base angle.<br />
<br />
Case 1: <math> x^\circ</math> is a base angle with the <math> 70^\circ</math> as the other angle:<br />
Here, <math> x=70</math>, since base angles are congruent.<br />
<br />
Case 2: <math> x^\circ</math> is a base angle with <math> 70^\circ</math> as the vertex angle:<br />
Here, the 2 base angles are both <math> x^\circ</math>, so we can use the equation <math> 2x+70=180</math>, which simplifies to <math> x=55</math>.<br />
<br />
Case 3: <math> x^\circ</math> is the vertex angle with <math> 70^\circ</math> as a base angle:<br />
Here, both base angles are <math> 70^\circ</math>, since base angles are congruent. Thus, we can use the equation <math> x+140=180</math>, which simplifies to <math> x=40</math>.<br />
<br />
Adding up all the cases, we get <math>70+55+40=165</math>, so the answer is <math> \boxed{\textbf{(D)}\ 165}</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/FDgcLW4frg8?t=902<br />
<br />
~ pi_is_3.14<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/iB0dcpVREE8 Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2009|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_19&diff=1737242009 AMC 8 Problems/Problem 192022-05-02T04:52:05Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Two angles of an isosceles triangle measure <math> 70^\circ</math> and <math> x^\circ</math>. What is the sum of the three possible values of <math> x</math>?<br />
<br />
<math> \textbf{(A)}\ 95 \qquad<br />
\textbf{(B)}\ 125 \qquad<br />
\textbf{(C)}\ 140 \qquad<br />
\textbf{(D)}\ 165 \qquad<br />
\textbf{(E)}\ 180</math><br />
<br />
<br />
==Solution==<br />
<br />
There are 3 cases: where <math> x^\circ</math> is a base angle with the <math> 70^\circ</math> as the other angle, where <math> x^\circ</math> is a base angle with <math> 70^\circ</math> as the vertex angle, and where <math> x^\circ</math> is the vertex angle with <math> 70^\circ</math> as a base angle.<br />
<br />
Case 1: <math> x^\circ</math> is a base angle with the <math> 70^\circ</math> as the other angle:<br />
Here, <math> x=70</math>, since base angles are congruent.<br />
<br />
Case 2: <math> x^\circ</math> is a base angle with <math> 70^\circ</math> as the vertex angle:<br />
Here, the 2 base angles are both <math> x^\circ</math>, so we can use the equation <math> 2x+70=180</math>, which simplifies to <math> x=55</math>.<br />
<br />
Case 3: <math> x^\circ</math> is the vertex angle with <math> 70^\circ</math> as a base angle:<br />
Here, both base angles are <math> 70^\circ</math>, since base angles are congruent. Thus, we can use the equation <math> x+140=180</math>, which simplifies to <math> x=40</math>.<br />
<br />
Adding up all the cases, we get <math>70+55+40=165</math>, so the answer is <math> \boxed{\textbf{(D)}\ 165}</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/FDgcLW4frg8?t=902<br />
<br />
~ pi_is_3.14<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/iB0dcpVREE8 Soo, DRMS, NM<br />
<br />
==See Also==<br />
{{AMC8 box|year=2009|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_20&diff=1737232013 AMC 8 Problems/Problem 202022-05-02T04:42:24Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
A <math>1\times 2</math> rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?<br />
<br />
<math>\textbf{(A)}\ \frac\pi2 \qquad \textbf{(B)}\ \frac{2\pi}3 \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}3 \qquad \textbf{(E)}\ \frac{5\pi}3</math><br />
<br />
<br />
==Solution==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br />
import graph; usepackage("amsmath");<br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = 2.392515856236789, xmax = 4.844947145877386, ymin = 6.070697674418619, ymax = 8.062241014799170; /* image dimensions */<br />
pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); <br />
<br />
draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)--cycle, zzttqq); <br />
/* draw figures */<br />
draw((2.912600422832983,6.903678646934476)--(4.326813985206080,6.903678646934476)); <br />
draw(shift((3.619707204019532,6.903678646934476))*xscale(0.7071067811865487)*yscale(0.7071067811865487)*arc((0,0),1,0.000000000000000,180.0000000000000)); <br />
draw((3.619707204019532,6.903678646934476)--(4.119707204019532,6.903678646934476)); <br />
draw((3.619707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476)); <br />
draw((3.119707204019531,7.403678646934482)--(4.119707204019532,7.403678646934476), zzttqq); <br />
draw((4.119707204019532,7.403678646934476)--(4.119707204019532,6.903678646934476), zzttqq); <br />
draw((4.119707204019532,6.903678646934476)--(3.119707204019531,6.903678646934476), zzttqq); <br />
draw((3.119707204019531,6.903678646934476)--(3.119707204019531,7.403678646934482), zzttqq); <br />
label("$1$",(3.847061310782247,6.924820295983102),SE*labelscalefactor); <br />
label("$1$",(4.155729386892184,7.208118393234687),SE*labelscalefactor); <br />
draw((3.619707204019532,6.903678646934476)--(4.119707204019532,7.403678646934476)); <br />
label("$\sqrt{2}$",(3.711754756871041,7.288456659619466),SE*labelscalefactor); <br />
label("$2$",(3.563763213530660,7.563298097251601),SE*labelscalefactor); <br />
/* dots and labels */<br />
dot((2.912600422832983,6.903678646934476)); <br />
dot((4.326813985206080,6.903678646934476)); <br />
dot((3.619707204019532,6.903678646934476)); <br />
dot((4.119707204019532,6.903678646934476),blue); <br />
dot((3.619707204019532,6.903678646934476)); <br />
dot((3.119707204019531,6.903678646934476),blue); <br />
dot((3.119707204019531,7.403678646934482),blue); <br />
dot((4.119707204019532,7.403678646934476),blue); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. The area is <math>\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}</math>.<br />
<br />
==Solution 2==<br />
Double the figure to get a square with side length <math>2</math>. The circle inscribed around the square has a diameter equal to the diagonal of this square. The diagonal of this square is <math>\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}</math>. The circle’s radius ,therefore, is <math>\sqrt{2}</math><br />
<br />
The area of the circle is <math>\left ( \sqrt{2} \right ) ^2 \pi = 2\pi</math><br />
<br />
Finally, the area of the semicircle is <math>\pi</math>, so the answer is <math>\boxed{C}</math>.<br />
<br />
<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=6WPBluEpmMA<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/0g14IJJ2Z-8 Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2013|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
Thank You for reading these answers by the followers of AoPS.</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_18&diff=1737222005 AMC 8 Problems/Problem 182022-05-02T04:33:58Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
How many three-digit numbers are divisible by 13?<br />
<br />
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77</math><br />
<br />
<br />
==Solution==<br />
Let <math>k</math> be any positive integer so that <math>13k</math> is a multiple of <math>13</math>. For the smallest three-digit number, <math>13k>100</math> and <math>k>\frac{100}{13} \approx 7.7</math>. For the greatest three-digit number, <math>13k<999</math> and <math>k<\frac{999}{13} \approx 76.8</math>. The number <math>k</math> can range from <math>8</math> to <math>76</math> so there are <math>\boxed{\textbf{(C)}\ 69}</math> three-digit numbers.<br />
<br />
<br />
==Video Solution==<br />
https://youtu.be/7an5wU9Q5hk?t=393<br />
<br />
<br />
==Video Solution 2==<br />
https://youtu.be/101Rgutx1R0 Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2005|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Sootommyleehttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_17&diff=1737212008 AMC 8 Problems/Problem 172022-05-02T04:26:22Z<p>Sootommylee: </p>
<hr />
<div>==Problem==<br />
Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of <math>50</math> units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?<br />
<br />
<math>\textbf{(A)}\ 76\qquad<br />
\textbf{(B)}\ 120\qquad<br />
\textbf{(C)}\ 128\qquad<br />
\textbf{(D)}\ 132\qquad<br />
\textbf{(E)}\ 136</math><br />
<br />
==Solution==<br />
A rectangle's area is maximized when its length and width are equivalent, or the two side lengths are closest together in this case with integer lengths. This occurs with the sides <math>12 \times 13 = 156</math>. Likewise, the area is smallest when the side lengths have the greatest difference, which is <math>1 \times 24 = 24</math>. The difference in area is <math>156-24=\boxed{\textbf{(D)}\ 132}</math>.<br />
<br />
<br />
==Video Solution==<br />
https://youtu.be/9bVwSsWa8IY Soo, DRMS, NM<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2008|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Sootommylee