Question

说明对于任意正整数n，式子你n(n+5）-（n-3)（n+2）的值都能被6整除

Answer 1

n(n+5)-(n-3)(n+2)
=(n²+5n) - (n²-n-6)
=6n+6
=6(n+1)
所以 [n(n+5)-(n-3)(n+2)] ÷ 6 = 6(n+1)÷6 = n+1
因为n是正整数，所以n+1也是正整数，
所以对于任意正整数n，式子n(n+5)-(n-3)(n+2)的值都能被6整除

Answer 2

原题目:
对于任意正整数n，代数式n(n+5)-(n+2)(n-3)的值是否总能被6整除？请说明理由
证明:
n(n+5)-(n+2)(n-3)
=n^2+5n-(n^2-n-6)
=6n+6
=6(n+1)
所以,对于任意正整数n，代数式n(n+5)-(n+2)(n-3)的值总能被6整除