Re:Roman Finger Counting
by dTal on Friday May 15, 2020 @07:28AM
(#23190927)
>a base-10 multiplication table has 100 entries
Not quite! You can safely ignore identities (0, 1, and 10 itself) so you only have 8 numbers in your table. And multiplication is commutative so you only need 8+7+6... (= (8+1)(8/2) as per Gauss) = 36 entries.
Base 16 would have (14+1)(14/2) = 105 entries. So proportional to base-10, actually slightly harder than you said.
Not quite! You can safely ignore identities (0, 1, and 10 itself) so you only have 8 numbers in your table. And multiplication is commutative so you only need 8+7+6... (= (8+1)(8/2) as per Gauss) = 36 entries.
Base 16 would have (14+1)(14/2) = 105 entries. So proportional to base-10, actually slightly harder than you said.