Difference between revisions of "2006 AMC 12B Problems/Problem 18"
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== Solution == | == Solution == | ||
| − | Let the starting point be <math>(0,0)</math>. After <math>10</math> steps we can only be in locations <math>(x,y)</math> where <math>|x|+|y|\leq 10</math>. Additionally, each step changes the parity of exactly one coordinate. Hence after <math>10</math> steps we can only be in locations <math>(x,y)</math> where <math>x+y</math> is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable. | + | Let the starting point be <math>(0,0)</math>. After <math>10</math> steps we can only be in locations <math>(x,y)</math> where <math>|x|+|y|\leq 10</math>. Additionally, each step changes the [parity] of exactly one coordinate. Hence after <math>10</math> steps we can only be in locations <math>(x,y)</math> where <math>x+y</math> is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable. |
Once we pick <math>x\in\{-10,\dots,10\}</math>, we have <math>11-|x|</math> valid choices for <math>y</math>, giving a total of <math>\boxed{121}</math> possible positions. | Once we pick <math>x\in\{-10,\dots,10\}</math>, we have <math>11-|x|</math> valid choices for <math>y</math>, giving a total of <math>\boxed{121}</math> possible positions. | ||
| − | |||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2006|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:24, 8 December 2013
Problem
An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
Solution
Let the starting point be 
. After 
steps we can only be in locations 
where 
. Additionally, each step changes the [parity] of exactly one coordinate. Hence after steps we can only be in locations where 
is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable.
Once we pick 
, we have 
valid choices for 
, giving a total of 
possible positions.
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
