數學知識交流---整數部份(3)

Solve: n^3 + n^2 + n^1 + n^0 = 1000
n^2 (n+1) + n^0 (n+1) = 1000
(n^2 + 1) (n+1) = 1000
Let (n^2 + 1) (n+1) = a
When the value of n increases, the value of a will also increase.
When n = 10
101×11 = 1111
a = 1111 > 1000
so n<10
When n = 9
82×10 = 820
a = 820 < 1000
so n>9
so 9<n<10
so the 整數部分 of n is 9.