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Difference between revisions of "2020 CIME I Problems/Problem 1"
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==Solution==
 
==Solution==
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Let <math>A</math> denote a move of <math>2</math> units north and <math>1</math> unit east, and let <math>B</math> denote a move of <math>1</math> unit north and <math>2</math> units east. To get to the point <math>(15,15)</math> using only these moves, say <math>x</math> moves in direction <math>A</math> and <math>y</math> moves in direction <math>B</math>, we must have <math>2x+1y=1x+2y=15</math> because both the <math>x</math> and <math>y</math>-coordinates have increased by <math>15</math> since the knight started. Solving this system of equations gives us <math>x=y=5</math>. This means we need the knight to make <math>10</math> moves, <math>5</math> of which are headed in direction <math>A</math>, and the remaining <math>5</math> are headed in direction <math>B</math>. Any combination of these moves work, so the answer is <math>\binom{10}{5}=\boxed{252}.</math>
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Let <math>A</math> denote a move of <math>2</math> units north and <math>1</math> unit east, and let <math>B</math> denote a move of <math>1</math> unit north and <math>2</math> units east. To get to the point <math>(15,15)</math> using only these moves, say <math>a</math> moves in direction <math>A</math> and <math>b</math> moves in direction <math>B</math>, we must have <math>2a+1b=1a+2b=15</math> because both the <math>x</math> and <math>y</math>-coordinates have increased by <math>15</math> since the knight started. Solving this system of equations gives us <math>a=b=5</math>. This means we need the knight to make <math>10</math> moves, <math>5</math> of which are headed in direction <math>A</math>, and the remaining <math>5</math> are headed in direction <math>B</math>. Any combination of these moves work, so the answer is <math>\binom{10}{5}=\boxed{252}.</math>
  
 
==See also==
 
==See also==

Revision as of 10:36, 1 September 2020

Problem 1

A knight begins on the point $(0,0)$ in the coordinate plane. From any point $(x,y)$ the knight moves to either $(x+2,y+1)$ or $(x+1,y+2)$. Find the number of ways the knight can reach the point $(15,15)$.

Solution

Let $A$ denote a move of $2$ units north and $1$ unit east, and let $B$ denote a move of $1$ unit north and $2$ units east. To get to the point $(15,15)$ using only these moves, say $a$ moves in direction $A$ and $b$ moves in direction $B$, we must have $2a+1b=1a+2b=15$ because both the $x$ and $y$-coordinates have increased by $15$ since the knight started. Solving this system of equations gives us $a=b=5$. This means we need the knight to make $10$ moves, $5$ of which are headed in direction $A$, and the remaining $5$ are headed in direction $B$. Any combination of these moves work, so the answer is $\binom{10}{5}=\boxed{252}.$

See also

2020 CIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions

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