, ▁then ▁as ▁the ▁excess ▁of ▁the ▁second ▁is ▁to ▁the ▁first — so ▁is ▁the ▁excess ▁of ▁the ▁last ▁to ▁all ▁those ▁before ▁it . ▁( This ▁is ▁a ▁rest at ement ▁of ▁our ▁formula ▁for ▁geometric ▁series ▁from ▁above .) ▁App lying ▁this ▁to ▁the ▁geometric ▁pro gression ▁ 3 1 , ▁ 6 2 , ▁ 1 2 4 , ▁ 2 4 8 , ▁ 4 9 6 ▁( which ▁results ▁from ▁ 1 , ▁ 2 , ▁ 4 , ▁ 8 , ▁ 1 6 ▁by ▁multip lying ▁all ▁terms ▁by ▁ 3 1 ), ▁we ▁see ▁that ▁ 6 2 ▁minus ▁ 3 1 ▁is ▁to ▁ 3 1 ▁as ▁ 4 9 6 ▁minus ▁ 3 1 ▁is ▁to ▁the ▁sum ▁of ▁ 3 1 , ▁ 6 2 , ▁ 1 2 4 , ▁ 2 4 8 . ▁Therefore , ▁the ▁numbers ▁ 1 , ▁ 2 , ▁ 4 , ▁ 8 , ▁ 1 6 , ▁ 3 1 , ▁ 6 2 , ▁ 1 2 4 ▁and ▁ 2 4 8 ▁add ▁up ▁to ▁ 4 9 6 ▁and ▁further ▁these ▁are ▁all ▁the ▁numbers ▁that ▁divide ▁ 4 9 6 . ▁For ▁suppose ▁that ▁ ▁div ides ▁ 4 9 6 ▁and ▁it ▁is ▁not ▁amongst ▁these ▁numbers . ▁Assume ▁ ▁is ▁equal ▁to ▁, ▁or ▁ 3 1 ▁is ▁to ▁ ▁as ▁ ▁is ▁to ▁ 1 6 . ▁Now ▁ ▁cannot ▁divide ▁ 1 6 ▁or