▁it ▁would ▁be ▁amongst ▁the ▁numbers ▁ 1 , ▁ 2 , ▁ 4 , ▁ 8 ▁or ▁ 1 6 . ▁Therefore , ▁ 3 1 ▁cannot ▁divide ▁. ▁And ▁since ▁ 3 1 ▁does ▁not ▁divide ▁ ▁and ▁ ▁measures ▁ 4 9 6 , ▁the ▁fundamental ▁theorem ▁of ▁arithmetic ▁implies ▁that ▁ ▁must ▁divide ▁ 1 6 ▁and ▁be ▁amongst ▁the ▁numbers ▁ 1 , ▁ 2 , ▁ 4 , ▁ 8 ▁or ▁ 1 6 . ▁Let ▁ ▁be ▁ 4 , ▁then ▁ ▁must ▁be ▁ 1 2 4 , ▁which ▁is ▁impossible ▁since ▁by ▁hypothesis ▁ ▁is ▁not ▁amongst ▁the ▁numbers ▁ 1 , ▁ 2 , ▁ 4 , ▁ 8 , ▁ 1 6 , ▁ 3 1 , ▁ 6 2 , ▁ 1 2 4 ▁or ▁ 2 4 8 . ▁ ▁The ▁ 0 th ▁through ▁ 9 5 th ▁powers ▁of ▁two ▁▁ ▁Starting ▁with   2 ▁the ▁last ▁digit ▁is ▁periodic ▁with ▁period   4 , ▁with ▁the ▁cycle ▁ 2 – 4 – 8 – 6 – , ▁and ▁starting ▁with   4 ▁the ▁last ▁two ▁digits ▁are ▁periodic ▁with ▁period   2 0 . ▁These ▁patterns ▁are ▁generally ▁true ▁of ▁any ▁power , ▁with ▁respect ▁to ▁any ▁base . ▁The ▁pattern ▁continues ▁where ▁each ▁pattern ▁has ▁starting ▁point ▁, ▁and ▁the ▁period ▁is ▁the ▁multip lic ative ▁order ▁of ▁ 2   mod ulo   , ▁which ▁is ▁ ▁( see ▁Mult ip lic ative ▁group ▁of