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Difference between revisions of "2019 AMC 12B Problems/Problem 10"

(→‎Problem: Added solution options with good Latex)
(→‎Solution: Added solution.)
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==Solution==
 
==Solution==
 +
Note the following route, which isn't that hard to discover:
 +
 +
!! Someone with good Latex/Asymptote skills please help !!
 +
 +
Look at the two square "loop"s. Each one can be oriented in one of two directions (lower left: either go down or left first; upper right: either go right or up first). Therefore, the answer is 1 route * 2 * 2 = <math>\boxed{\textbf{(E) } 4}</math>. Note that no choice is larger than it.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=9|num-a=11}}
 
{{AMC12 box|year=2019|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:34, 14 February 2019

Problem

The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$, without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.)

!! Someone with good Latex or Asymptote skills please help !!

How many different routes can Paula take?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution

Note the following route, which isn't that hard to discover:

!! Someone with good Latex/Asymptote skills please help !!

Look at the two square "loop"s. Each one can be oriented in one of two directions (lower left: either go down or left first; upper right: either go right or up first). Therefore, the answer is 1 route * 2 * 2 = $\boxed{\textbf{(E) } 4}$. Note that no choice is larger than it.

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

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