Difference between revisions of "User talk:Bobthesmartypants/Solutions"
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'''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 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> | ||
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| + | '''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> verteces, so ther total is <math>9+2\times8=\boxed{25}</math>. <math>\Box</math> | ||
Revision as of 16:13, 21 May 2013
Solutions
Solution 1: For the first number, it will always be 
roll. For the second number, there is a 
chance of getting a new number, so the expected number of rolls is 
. The expected number of rolls for the third number is 
. Continuing the pattern, the expected number of rolls for the 4th, 5th, and final number is 
respectively. So the total expected number of rolls is 
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 
is 
, 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 
.
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 
and 
to get 
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: 
We can see a pattern: two odds, then an even, then two odds, then an even, and so on. So when 
, we have to do 
, so it is not possible. For 
, we have to do 
, so it is not possible. For 
, we have 
, so IS possible. We find that if we continue this, then a pattern emerges: every time when 
, then it is possible. Otherwise, it isn't. So our answer is 
.
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 
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 
. Every vertex of the original cube contributes to more edges, and there are 
verteces, so ther total is 
.
