Difference between revisions of "1986 AJHSME Problems/Problem 9"
5849206328x (talk | contribs) (New page: ==Problem== Using only the paths and the directions shown, how many different routes are there from <math>\text{M}</math> to <math>\text{N}</math>? <asy> draw((0,0)--(3,0),MidArrow); dra...) |
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==Solution== | ==Solution== | ||
| − | + | There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN. Since A can only go to C or D, which each only have 1 way to get to N each, there are 2 ways to get from A to N. Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are 4 ways to get from B to N. | |
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| + | M can only go to either B or A, A has 2 ways and B has 4 ways, so M has 6 ways to get to N. | ||
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| + | 6 is E. | ||
==See Also== | ==See Also== | ||
[[1986 AJHSME Problems]] | [[1986 AJHSME Problems]] | ||
Revision as of 18:02, 24 January 2009
Problem
Using only the paths and the directions shown, how many different routes are there from 
to 
?
![[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(6,0),MidArrow); draw(6*dir(60)--5.1961524227066318805823390245176*dir(30),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--3*dir(60),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(3,0),MidArrow); label("M",6*dir(60),N); label("N",(6,0),SE); label("A",3*dir(60),NW); label("B",5.1961524227066318805823390245176*dir(30),NE); label("C",(3,0),S); label("D",(0,0),SW); [/asy]](http://latex.artofproblemsolving.com/d/7/b/d7b6d3b20583aeac57d4bae5e7e0367f8b123720.png)
Solution
There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN. Since A can only go to C or D, which each only have 1 way to get to N each, there are 2 ways to get from A to N. Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are 4 ways to get from B to N.
M can only go to either B or A, A has 2 ways and B has 4 ways, so M has 6 ways to get to N.
6 is E.
