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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Permutation and Combination

Olympiad workbook · IOQM · Permutation and Combination

An ant is at a vertex of a cube. Every 10 minutes it moves to an...

An ant is at a vertex of a cube. Every 10 minutes it moves to an adjacent vertex along an edge. If NN is the number of one hour journeys that end at the starting vertex, find the sum of the squares of the digits of NN.An ant is at a vertex of a cube. Every 10 minutes it moves to an adjacent vertex along an edge. If N is the number of

Correct answer

74

Step-by-step solution

We have divided vertices into four categories An ant is at a vertex of a cube. Every 10 minutes it moves to an adjacent vertex along an edge. If N is the number of
X→X \rightarrow Starting vertex Y→Y \rightarrow Adjacent vertex Z→Z \rightarrow Adjacent to YY but not same as XX W→W \rightarrow Adjacent to ZZ but not same as YY Let an=a_n= number of ways that after nn steps ant is at XX bn=b_n= number of ways that after nn steps ant is at YY cn=c_n= number of ways that after nn steps ant is at ZZ dn=d_n= number of ways that after nn steps ant is at WW We need to find a6a_6 an+1=3bn...(i)bn+1=an+2cn...(ii)cn+1=2bn+dn...(iii) and dn+1=3cn...(iv)\begin{aligned} a_{n+1} & =3 b_n ...(i)\\ b_{n+1} & =a_n+2 c_n ...(ii)\\ c_{n+1} & =2 b_n+d_n ...(iii)\\ \text { and } \quad d_{n+1} & =3 c_n ...(iv) \end{aligned} By eliminating bn,cnb_n, c_n and dnd_n we get an+3=10an+1−9an−1∵a1=0,a2=3,a3=0 and a4=21∴a6=10a4−9a2=210−27=183∴N=183\begin{array}{ll} & a_{n+3}=10 a_{n+1}-9 a_{n-1} \\ \because & a_1=0, a_2=3, a_3=0 \text { and } a_4=21 \\ \therefore & a_6=10 a_4-9 a_2=210-27=183 \\ \therefore & N=183 \end{array} Sum of square of digits of N=74N=74
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