Art of Problem Solving

MIE 2016/Day 1/Problem 10: Difference between revisions

Anishanne (talk | contribs)
Anishanne (talk | contribs)
 
Line 17: Line 17:


==Solution ==
==Solution ==
==See Also==
{{MIE box|year=2016|num-b=9|num-a=11}}
{{MAA Notice}}

Latest revision as of 21:13, 10 January 2018

Problem 10

A hexagon is divided into 6 equilateral triangles. How many ways can we put the numbers from 1 to 6 in each triangle, without repetition, such that the sum of the numbers of three adjacent triangles is always a multiple of 3? Solutions obtained by rotation or reflection are differents, thus the following figures represent two distinct solutions.


(a) $12$

(b) $24$

(c) $36$

(d) $48$

(e) $96$


Solution