Spoiler: Math ...
We'll define
Part one: showing that the cube is divisible by 125.
(10
(100
(1000
Each term in that final sum is evenly divisible by 125 (since
Part two: showing that the cube is odd.
Any integer multiplied by 1000, 1500, or 750 has to be divisible by two. 125 is odd. even + even + even + odd = odd. Therefore, the entire sum is odd.
n
as an integer. 10n
+ 5 would yield the set of all integers with the final digit as a 5.Part one: showing that the cube is divisible by 125.
(10
n
+ 5) * (10n
+ 5) * (10n
+ 5) = ...(100
n2
+ 100n
+ 25) * (10n
+ 5) = ...(1000
n3
+ 1500n2
+ 750n
+ 125)Each term in that final sum is evenly divisible by 125 (since
n
is an integer); therefore, the entire sum is divisible by 125. Part two: showing that the cube is odd.
Any integer multiplied by 1000, 1500, or 750 has to be divisible by two. 125 is odd. even + even + even + odd = odd. Therefore, the entire sum is odd.