请问大家觉得这道题目是什么答案.

bigwish

Q23. DS: 2m + 3n = 12t, 判斷t與12是否有大於1的公因數? (1)m可被3整除, (2)n可被3整除

无答案:

我觉得是E. 因为12=2*3*2 那么如果它们有共同公因子.T 一定有一个2或者3.即, 2M+3n一定可以被 8或者9整除. 条件1. 式子可以变为: 6X+3N. 但是这个式子/8或者9都不确定是否可以整除. 条件2也推不出来. 结合二者, 原式变成: 6X+9Y. 也无法确定,所以为E.

请问大家觉得答案是什么?

smalllittle

I agree it should be E.

michellemove

2m + 3n = 12t, m = 3x (condition 1)
=> 6x + 3n = 12t
=> 2x + n = 4t, n = 2y (2x is odd, 4t is odd so n must be odd as well)
=> x + y = 2t, y = 3z (condition 2), n = 6z
=> x + 3z = 2t
let t = 1, z = 1 and x = -1
=> m = -3, n = 6 and t = 1 => -6 + 18 = 12
if t, m, n are all postive integer then let t = 5, z = 2 and x = 4
=> m = 12, n = 12, and t = 5 => 24 + 36 = 60