|
|
| Line 7: |
Line 7: |
| | | | |
| | [[User talk:Bobthesmartypants/Solutions]] | | [[User talk:Bobthesmartypants/Solutions]] |
| − |
| |
| − | ==Answers==
| |
| − | '''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>
| |
Revision as of 16:05, 21 May 2013
-sided die. Every time he rolls a number that he has already rolled before, he rolls again. He stops when he has rolled all the numbers. What is the expected number of rolls it will take Bob?
and 
, where 
. We launch a point-like ball from one of the vertices with an angular degree of 
. The ball bounces off the sides of the box. Pretend there is no friction, drag, or anything else to slow down the ball. Prove or disprove that the ball won't ever hit a vertex again.
cities have 
roads leading out of it. What values of 
are there such that this setup is possible?
