请问~~ GWD-3-16???

111cara

不知为何我下载的GWD31, 数学题号跟大家的版本不对应~~

查不到, 只好在这里开新帖提问了, 斑斑请原谅!

GWD-3-Q16:

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

(1)     2 is not a factor of n.

(2)     3 is not a factor of n.

                  

答案C.

 

谢谢!

 

 

        

zlily424

from (1), we can get numbers such as 1,3,5,7,9,11,13,17,19,and etc.

from (2), we can get numbers such as 1,2,4,5,7,8,10,11,13,14,16,20, and etc.

combine (1) and (2), we can get numbers 1,5,7,13,and etc.

1-1=0, 1+1=2->0*2=0->0/24=0, remaining=0

5-1=4, 5+1=6->4*5=24->24/24=1,remaining=0

7-1=6, 7+1=8->6*8=48->48/24=2, remaining=0

11-1=10, 11+1=12->10*12=120->120/24=5, remaining=0

13-1=12,13+1=14->12*14=168->168/24=7, remaining=0

Therefore, C is the correction answer.

flyinglamb

n=6k+5/6k+1

这样简单些

111cara

好聪明,谢谢两位MM了```

rouslinj

若 n = 6k+5，那么 (n-1)(n+1) = (6k+4)(6k+6) = 12(3k+2)(k+1)，无法判定是否能被 24 整除？

同理，若 n = 6k+1，也无法判定？

graceguo98

 n = 6k+1, 和n=6k+5是怎么得出来的？为什么会有这个因式？

不好意思，