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 one hour journeys that end at the starting vertex, find the sum of the squares of the digits of N.

Step 1: We have divided vertices into four categories

Step 2: X  Starting vertex

Step 3: Y  Adjacent vertex

Step 4: Z  Adjacent to Y but not same as X

Step 5: W  Adjacent to Z but not same as Y

Step 6: Let an = number of ways that after $n$ steps ant is at X

Step 7: bn = number of ways that after $n$ steps ant is at Y

Step 8: cn = number of ways that after $n$ steps ant is at Z

Step 9: dn = number of ways that after $n$ steps ant is at W

Step 10: We need to find a6

Step 11: an+1 = 3bn …(i)

Step 12: bn+1 = an + 2cn …(ii)

Step 13: cn+1 = 2bn + dn …(iii)

Step 14: and dn+1 = 3cn …(iv)

Step 15: By eliminating bn, cn and dn we get

Step 16: an+3 = 10an+1 – 9an–1

Step 17:  a1 = 0, a2 = 3, a3 = 0 and a4 = 21

Step 18: => a6 = 10a4 – 9a2 = 210 – 27 = 183

Step 19: => N = 183

Step 20: Sum of square of digits of N = 74