https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Fhdsaukfaioifk&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T15:03:18ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_9&diff=1158582018 AMC 10B Problems/Problem 92020-01-30T01:11:53Z<p>Fhdsaukfaioifk: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
<br />
The faces of each of <math>7</math> standard dice are labeled with the integers from <math>1</math> to <math>6</math>. Let <math>p</math> be the probabilities that when all <math>7</math> dice are rolled, the sum of the numbers on the top faces is <math>10</math>. What other sum occurs with the same probability as <math>p</math>?<br />
<br />
<math>\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}</math><br />
<br />
==Solution 1==<br />
It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.<br />
<br />
So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows:<br />
<br />
(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...<br />
<br />
However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math>, and we are done.<br />
<br />
Written By: Archimedes15<br />
<br />
==Solution 2==<br />
<br />
Let's call the unknown value <math>x</math>. By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and <math>x</math>. So, <br />
<br />
<math>10 - 7 = 42- x </math><br />
<br />
<math> x = 39 </math> and our answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
By: Soccer_JAMS<br />
<br />
==Solution 3==<br />
<br />
For the sums to have equal probability, the average sum of both sets of <math>7</math> dies has to be <math>(6+1)\cdot 7 = 49</math>. Since having <math>10</math> is similar to not having <math>10</math>, you just subtract 10 from the expected total sum. <math>49 - 10 = 39</math> so the answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
<br />
By: epicmonster<br />
<br />
==Solution 4==<br />
The expected value of the sums of the die rolls is <math>3.5*7=24.5</math>, and since the probabilities should be distributed symmetrically on both sides of <math>24.5</math>, the answer is <math>24.5+24.5-10=39</math>, which is <math>\boxed{\textbf{(D)} \text{ 39}}</math>.<br />
<br />
By: dajeff<br />
<br />
==Solution 5==<br />
Another faster and easier way of doing this, without using almost any math at all, is realizing that the possible sums are <math>{7,8,9,10,...,39,40,41,42}</math>. By symmetry, (and doing a few similar problems in the past), you can realize that the probability of obtaining <math>7</math> is the same as the probability of obtaining <math>42</math>, <math>P(8)=P(41)</math> and on and on and on. This means that <math>P(10)=P(39)</math>, and thus the correct answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math>.<br />
<br />
By: fhdsaukfaioifk<br />
<br />
=== Note ===<br />
<br />
Calculating the probability of getting a sum of <math>10</math> is also easy. There are <math>3</math> cases:<br />
<br />
<br />
Case <math>1</math>: <math>\{1,1,1,1,1,1,4\}</math><br />
<br />
<br />
<math>{7 \choose 6}=7</math> cases<br />
<br />
<br />
Case <math>2</math>: <math>\{1,1,1,1,1,2,3\}</math><br />
<br />
<br />
<math>{7 \choose 5}=6 \cdot 7=42</math> cases<br />
<br />
<br />
Case <math>3</math>: <math>\{1,1,1,1,2,2,2\}</math><br />
<br />
<br />
<math>{7 \choose 4}=\frac {7 \cdot 6 \cdot 5}{3 \cdot 2}=35</math> cases<br />
<br />
<br />
The probability is <math>{84 \over 6^7} = \frac{14}{6^6}</math>. <br />
<br />
Calculating <math>6^6</math>: <br />
<br />
<math>6^6=(6^3)^2=216^2=46656</math> <br />
<br />
Therefore, the probability is <math>{14 \over 46656} = \boxed{{7 \over 23328}}</math><br />
<br />
~Zeric Hang (Main writer) and fhdsaukfaioifk (Editor)<br />
<br />
==Related Problems==<br />
There is a similar to problem 11 of the AMC 10A in the same year, which is almost an replica of the problem mentioned by Zeric Hang in the Note section:<br />
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 <br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Fhdsaukfaioifkhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_9&diff=1158572018 AMC 10B Problems/Problem 92020-01-30T01:04:23Z<p>Fhdsaukfaioifk: /* Note */</p>
<hr />
<div>==Problem==<br />
<br />
The faces of each of <math>7</math> standard dice are labeled with the integers from <math>1</math> to <math>6</math>. Let <math>p</math> be the probabilities that when all <math>7</math> dice are rolled, the sum of the numbers on the top faces is <math>10</math>. What other sum occurs with the same probability as <math>p</math>?<br />
<br />
<math>\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}</math><br />
<br />
==Solution 1==<br />
It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.<br />
<br />
So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows:<br />
<br />
(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...<br />
<br />
However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math>, and we are done.<br />
<br />
Written By: Archimedes15<br />
<br />
==Solution 2==<br />
<br />
Let's call the unknown value <math>x</math>. By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and <math>x</math>. So, <br />
<br />
<math>10 - 7 = 42- x </math><br />
<br />
<math> x = 39 </math> and our answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
By: Soccer_JAMS<br />
<br />
==Solution 3==<br />
<br />
For the sums to have equal probability, the average sum of both sets of <math>7</math> dies has to be <math>(6+1)\cdot 7 = 49</math>. Since having <math>10</math> is similar to not having <math>10</math>, you just subtract 10 from the expected total sum. <math>49 - 10 = 39</math> so the answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
<br />
By: epicmonster<br />
<br />
==Solution 4==<br />
The expected value of the sums of the die rolls is <math>3.5*7=24.5</math>, and since the probabilities should be distributed symmetrically on both sides of <math>24.5</math>, the answer is <math>24.5+24.5-10=39</math>, which is <math>\boxed{\textbf{(D)} \text{ 39}}</math>.<br />
<br />
By: dajeff<br />
<br />
<br />
<br />
=== Note ===<br />
<br />
Calculating the probability of getting a sum of <math>10</math> is also easy. There are <math>3</math> cases:<br />
<br />
<br />
Case <math>1</math>: <math>\{1,1,1,1,1,1,4\}</math><br />
<br />
<br />
<math>{7 \choose 6}=7</math> cases<br />
<br />
<br />
Case <math>2</math>: <math>\{1,1,1,1,1,2,3\}</math><br />
<br />
<br />
<math>{7 \choose 5}=6 \cdot 7=42</math> cases<br />
<br />
<br />
Case <math>3</math>: <math>\{1,1,1,1,2,2,2\}</math><br />
<br />
<br />
<math>{7 \choose 4}=\frac {7 \cdot 6 \cdot 5}{3 \cdot 2}=35</math> cases<br />
<br />
<br />
The probability is <math>{84 \over 6^7} = \frac{14}{6^6}</math>. <br />
<br />
Calculating <math>6^6</math>: <br />
<br />
<math>6^6=(6^3)^2=216^2=46656</math> <br />
<br />
Therefore, the probability is <math>{14 \over 46656} = \boxed{{7 \over 23328}}</math><br />
<br />
~Zeric Hang (Main writer) and fhdsaukfaioifk (Editor)<br />
<br />
==Related Problems==<br />
There is a similar to problem 11 of the AMC 10A in the same year, which is almost an replica of the problem mentioned by Zeric Hang in the Note section:<br />
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 <br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Fhdsaukfaioifkhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_17&diff=1157982017 AMC 10A Problems/Problem 172020-01-29T02:22:54Z<p>Fhdsaukfaioifk: /* Solution */</p>
<hr />
<div>==Problem==<br />
Distinct points <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> lie on the circle <math>x^{2}+y^{2}=25</math> and have integer coordinates. The distances <math>PQ</math> and <math>RS</math> are irrational numbers. What is the greatest possible value of the ratio <math>\frac{PQ}{RS}</math>?<br />
<br />
<math>\mathrm{\textbf{(A)}}\ 3\qquad\mathrm{\textbf{(B)}}\ 5\qquad\mathrm{\textbf{(C)}}\ 3\sqrt{5}\qquad\mathrm{\textbf{(D)}}\ 7\qquad\mathrm{\textbf{(E)}}\ 5\sqrt{2}</math><br />
<br />
==Solution 1==<br />
<br />
Because <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are <math>(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),</math> and <math>(\pm 5,0).</math> We want to maximize <math>PQ</math> and minimize <math>RS.</math> They also have to be the square root of something, because they are both irrational. The greatest value of <math>PQ</math> happens when <math>P</math> and <math>Q</math> are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be <math>(-4,3)</math> and <math>(3,-4)</math> because the two points are almost across from each other. Another possible pair could be <math>(-4,3)</math> and <math>(5,0)</math>. To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from <math>(-4,3)</math> to <math>(4,-3)</math>. The distance between <math>(3,-4)</math> and <math>(4,-3)</math> is shorter than the distance between <math>(5,0)</math> and <math>(4,-3)</math>. Therefore, the segment from <math>(-4,3)</math> to <math>(3,-4)</math> is the longest attainable. The least value of <math>RS</math> is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, <math>R</math> is <math>(3,4)</math> and <math>S</math> is <math>(4,3).</math> They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point <math>(3,4)</math> than <math>(4,3).</math> Using the distance formula, we get that <math>PQ</math> is <math>\sqrt{98}</math> and that <math>RS</math> is <math>\sqrt{2}.</math> <math>\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{\mathrm{\textbf{(D)}}\ 7}</math><br />
<br />
==Solution 2==<br />
So what we can do is look at the option choices. Since we are aiming for the highest possible ratio, let's try using <math>7</math>. Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let <math>RS</math> be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of <math>RS</math> is <math>\sqrt{1^{2}+1^{2}} = \sqrt{2}</math>. Assuming that <math>\frac{PQ}{RS} = 7</math>, we plug in <math>RS = \sqrt{2}</math> and solve for PQ: <math>PQ=7\sqrt{2}</math>. Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of <math>x</math> and <math>y</math> does <math>\sqrt{x^{2}+y^{2}}=7\sqrt2</math>? Aha! We see that this can easily be made into a <math>45-45-90</math> triangle. But, instead of substituting <math>y=x</math> into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a <math>45-45-90</math> triangle, the two <math>45</math> degree sides have side length <math>s</math>, then the hypotenuse is <math>s\sqrt2</math>. Using this, we can see that <math>s=7</math>, and since our equation does in fact yield a sensible solution, we can be assured that our answer is <math>\boxed{\mathrm{\textbf{(D)}}\ 7}</math>. As an extra bonus, <math>7</math> is the largest of the options available, so we can be double reassured<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Fhdsaukfaioifkhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_8&diff=1157732017 AMC 10A Problems/Problem 82020-01-28T23:47:54Z<p>Fhdsaukfaioifk: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
At a gathering of <math>30</math> people, there are <math>20</math> people who all know each other and <math>10</math> people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?<br />
<br />
<math>\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490</math><br />
<br />
==Solution 1==<br />
Each one of the ten people has to shake hands with all the <math>20</math> other people they don’t know. So <math>10\cdot20 = 200</math>. From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands, or <math>\binom{10}{2} = 45</math>. Thus the answer is <math>200 + 45 = \boxed{\textbf{(B)}\ 245}</math>.<br />
<br />
==Solution 2==<br />
We can also use complementary counting. First of all, <math>\dbinom{30}{2}=435</math> handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from <math>435</math> to find the handshakes. Hugs only happen between the <math>20</math> people who know each other, so there are <math>\dbinom{20}{2}=190</math> hugs. <math>435-190= \boxed{\textbf{(B)}\ 245}</math>.<br />
<br />
==Solution 3==<br />
We can focus on how many handshakes the <math>10</math> people who don't know anybody get. <br />
<br />
The first person gets <math>29</math> handshakes with other people not him/herself, the second person gets <math>28</math> handshakes with other people not him/herself and not the first person, ..., and the tenth receives <math>20</math> handshakes with other people not him/herself and not the first, second, ..., ninth person. We can write this as the sum of an arithmetic sequence:<br />
<br />
<math>\frac{10(20+29)}{2}\implies 5(49)\implies 245.</math><br />
Therefore, the answer is <math>\boxed{\textbf{(B)}\ 245}</math><br />
<br />
<br />
==Solution 4==<br />
First, we can find out the number of handshakes that the <math>10</math> people who don't know anybody share with the <math>20</math> other people. This is simply <math>10 \cdot 20 = 200</math>. Next, we need to find out the number of handshakes that are shared within the <math>10</math> people who don't know anybody. Here, we can use the formula <math>\frac{n(n-1)}{2}</math>, where <math>n</math> is the number of people being counted. The reason we divide by <math>2</math> is because <math>n(n-1)</math> counts the case where the <math>1^{st}</math> person shakes hands with the <math>2^{nd}</math> person <math>and</math> the case where the <math>2^{nd}</math> shakes hands with the <math>1^{st}</math> (and these 2 cases are the same). Thus, plugging <math>n=10</math> gives us <math>\frac{10 \cdot 9}{2} \implies 45</math>. Adding up the 2 cases gives us <math>200+45=\boxed{\textbf{(B)}\ 245}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/pxg7CroAt20<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Fhdsaukfaioifk