Difference between revisions of "User talk:Bobthesmartypants/Solutions"

Latest revision as of 17:36, 24 May 2013

Solutions

Solution 1: For the first number, it will always be $1$ roll. For the second number, there is a $\frac{5}{6}$ chance of getting a new number, so the expected number of rolls is $\frac{6}{5}$ . The expected number of rolls for the third number is $\frac{6}{4}=\frac{3}{2}$ . Continuing the pattern, the expected number of rolls for the 4th, 5th, and final number is $2,3,6$ respectively. So the total expected number of rolls is $1+\frac{6}{5}+\frac{3}{2}+2+3+6=\boxed{\frac{147}{10}, \text{ or }14.7}$ $\Box$

Solution 2: We can imagine the rectangle be refected across the entire plane, so that any time when the point-like ball bounces, it continues in a straight line. The slope of the line with angle measure $60^{\circ}$ is $\sqrt{3}$ , which is an irrational number. The rectangle has integer side lengths, so any line that passes through two lattice points in this grid will have a rational slope. Therefore, the line will never pass through another lattic point, which implies that the point-like ball will $\boxed{\text{Never arrive at a vertex again}}$ . $\Box$

Solution 3: It can only be possible if the number of roads is a positive integer, and that can only be attainable when the number of roads (not inluding double counting) is even. To count the number of roads, we simply add up all the roads per number of cities: for example, to find how many roads are in a country with 4 cities having 3 roads and 3 cities having 6 roads, we simply add $4\times3$ and $3\times6$ to get $30$ roads (with double the counting). To find when it is even, we list the first few Fibonacci numbers' parity (even or odd). It goes like this: $O,O,E,O,O,E\cdots$ We can see a pattern: two odds, then an even, then two odds, then an even, and so on. So when $n=1$ , we have to do $O\times O=O$ , so it is not possible. For $n=2$ , we have to do $O\times O+O\times E=O$ , so it is not possible. For $n=3$ , we have $O\times O+O\times E+E\times O=E$ , so $n=3$ IS possible. We find that if we continue this, then a pattern emerges: every time when $n\equiv0\pmod{3}$ , then it is possible. Otherwise, it isn't. So our answer is $\boxed{\text{Only when }n\text{ is divisible by }3}$ . $\Box$

Solution 4: We first inspect what happens at one "vertex" of the original cube; i.e. the triangle part. If we arrive at the triangle from one edge of the original cube, we can only travel at most $2$ sides of the triangle before exiting at another edge. So now we need to find the most amount of edges possible to travel on the original cube. After some experimentation, we find that the maximum on the cube is $9$ . Every vertex of the original cube contributes to $2$ more edges, and there are $8$ vertices, so their total is $9+2\times8=\boxed{25}$ . $\Box$

Solution 5: This problem seems very tricky, so I will divide it into parts. The first part is to figure out what values of $k$ will allow the ball to bounce on all sides of the $n$ -gon. The only way for the ball to return to the launch spot, Side A, before it has hit all the sides is when there is a common factor of $n$ and $k+1$ other than $1$ . (We do $k+1$ because $k$ is excluding both Side A and Side B, and the $+1$ solves that issue)So that means that for the ball to hit every side, it must have $\gcd(n,k+1)=1$ . Now the problem becomes: "Find the sum of all values $k$ such that $\gcd(n,k+1)=1$ ". How should we do that? Well, let's first inspect some examples. I will use the number $12$ . The numbers that are relatively prime to $12$ are $1,5,7,11$ . Notice how the opposite terms, $1,11$ and $5,7$ , add up to $12$ . This suggest the formula $\frac{n\phi(n)}{2}$ , where $\phi(n)$ is the number of numbers less than $n$ and relatively prime to $n$ ; I came up with the formula because the sum of all the numbers relatively prime to $n$ is equal to half the number of numbers relatively prime to $n$ multiplied by $n$ (I paired up each number with its opposite). It may seem that we are done, but there is one last thing to do. We need to subtract the $+1$ 's we made at the beginning of the solution. To do that we need to subtract the number of $k$ 's there are. There are simply $\phi(n)$ $k$ 's. We subtract, and get the final answer $\frac{n\phi(n)}{2}-\phi(n)=\frac{n\phi(n)}{2}-\frac{2\phi(n)}{2}=\boxed{\frac{(n-2)\phi(n)}{2}}$ . $\Box$

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-test
+==Solutions==
+'''Solution 1:''' For the first number, it will always be <math>1</math> roll. For the second number, there is a <math>\frac{5}{6}</math> chance of getting a new number, so the expected number of rolls is <math>\frac{6}{5}</math>. The expected number of rolls for the third number is <math>\frac{6}{4}=\frac{3}{2}</math>. Continuing the pattern, the expected number of rolls for the 4th, 5th, and final number is <math>2,3,6</math> respectively. So the total expected number of rolls is <math>1+\frac{6}{5}+\frac{3}{2}+2+3+6=\boxed{\frac{147}{10}, \text{ or }14.7}</math> <math>\Box</math>
+'''Solution 2:''' We can imagine the rectangle be refected across the entire plane, so that any time when the point-like ball bounces, it continues in a straight line. The slope of the line with angle measure <math>60^{\circ}</math> is <math>\sqrt{3}</math>, which is an irrational number. The rectangle has integer side lengths, so any line that passes through two lattice points in this grid will have a rational slope. Therefore, the line will never pass through another lattic point, which implies that the point-like ball will <math>\boxed{\text{Never arrive at a vertex again}}</math>. <math>\Box</math>
+'''Solution 3:''' It can only be possible if the number of roads is a positive integer, and that can only be attainable when the number of roads (not inluding double counting) is even. To count the number of roads, we simply add up all the roads per number of cities: for example, to find how many roads are in a country with 4 cities having 3 roads and 3 cities having 6 roads, we simply add <math>4\times3</math> and <math>3\times6</math> to get <math>30</math> roads (with double the counting). To find when it is even, we list the first few Fibonacci numbers' parity (even or odd). It goes like this: <math>O,O,E,O,O,E\cdots</math> We can see a pattern: two odds, then an even, then two odds, then an even, and so on. So when <math>n=1</math>, we have to do <math>O\times O=O</math>, so it is not possible. For <math>n=2</math>, we have to do <math>O\times O+O\times E=O</math>, so it is not possible. For <math>n=3</math>, we have <math>O\times O+O\times E+E\times O=E</math>, so <math>n=3</math> IS possible. We find that if we continue this, then a pattern emerges: every time when <math>n\equiv0\pmod{3}</math>, then it is possible. Otherwise, it isn't. So our answer is <math>\boxed{\text{Only when }n\text{ is divisible by }3}</math>. <math>\Box</math>
+'''Solution 4:''' We first inspect what happens at one "vertex" of the original cube; i.e. the triangle part. If we arrive at the triangle from one edge of the original cube, we can only travel at most <math>2</math> sides of the triangle before exiting at another edge. So now we need to find the most amount of edges possible to travel on the original cube. After some experimentation, we find that the maximum on the cube is <math>9</math>. Every vertex of the original cube contributes to <math>2</math> more edges, and there are <math>8</math> vertices, so their total is <math>9+2\times8=\boxed{25}</math>. <math>\Box</math>
+'''Solution 5:''' This problem seems very tricky, so I will divide it into parts. The first part is to figure out what values of <math>k</math> will allow the ball to bounce on all sides of the <math>n</math>-gon. The only way for the ball to return to the launch spot, Side A, before it has hit all the sides is when there is a common factor of <math>n</math> and <math>k+1</math> other than <math>1</math>. (We do <math>k+1</math> because <math>k</math> is excluding both Side A and Side B, and the <math>+1</math> solves that issue)So that means that for the ball to hit every side, it must have <math>\gcd(n,k+1)=1</math>. Now the problem becomes: "Find the sum of all values <math>k</math> such that <math>\gcd(n,k+1)=1</math>". How should we do that? Well, let's first inspect some examples. I will use the number <math>12</math>. The numbers that are relatively prime to <math>12</math> are <math>1,5,7,11</math>. Notice how the opposite terms, <math>1,11</math> and <math>5,7</math>, add up to <math>12</math>. This suggest the formula <math>\frac{n\phi(n)}{2}</math>, where <math>\phi(n)</math> is the number of numbers less than <math>n</math> and relatively prime to <math>n</math>; I came up with the formula because the sum of all the numbers relatively prime to <math>n</math> is equal to half the number of numbers relatively prime to <math>n</math> multiplied by <math>n</math> (I paired up each number with its opposite). It may seem that we are done, but there is one last thing to do. We need to subtract the <math>+1</math>'s we made at the beginning of the solution. To do that we need to subtract the number of <math>k</math>'s there are. There are simply <math>\phi(n)</math> <math>k</math>'s. We subtract, and get the final answer <math>\frac{n\phi(n)}{2}-\phi(n)=\frac{n\phi(n)}{2}-\frac{2\phi(n)}{2}=\boxed{\frac{(n-2)\phi(n)}{2}}</math>. <math>\Box</math>

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Difference between revisions of "User talk:Bobthesmartypants/Solutions"

Latest revision as of 17:36, 24 May 2013

Solutions