帮我看看：prep-ds1-159这么分析对吗？

lisony · 发表于 2007-10-18 16:55:00

159. 15957-!-item-!-187;#058&010660

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) n is not divisible by 2.—>n肯定是奇数；n^2-1肯定是偶数可以被2整除；

(2) n is not divisible by 3.-->n^2-1肯定能被3整除. 这两个数能被6整除，但不能被24整除

[此贴子已经被作者于2007-10-18 17:14:31编辑过]

sydneyang · 发表于 2007-10-18 17:04:00

The answer on this one is C, correct?

Plug in some numbers, you quickly find that neither is sufficient.

Combine. N must be a prime greater 3. So sufficient.

lisony · 发表于 2007-10-18 17:12:00

I choose C,but E is correct answer

sydneyang · 发表于 2007-10-18 17:51:00

Then your PREP is different from mine. Just kidding.

I went back a minute ago and checked the answer on PREP. It says C!

lisony · 发表于 2007-10-18 22:59:00

看了一天书真是看晕了，是C，我看串了，谢谢！

leenny · 发表于 2007-10-19 15:07:00

是不是可以为样理解？
（1）n不能被2整除则n＝2X-1，X为正整数。则n^2-1=4X(X-1),因为X和（X-1)是两个相邻的整数，所以n^2-1＝4x(x-1)能被8整数.
(2)n不能被3整除，则n^2-1=(n-1)(n+1)。因n-1,n and n=1是三个相邻的整数而n又不能被3整除，则n-1,n+1其中一个必能被3整数，即(n-1)(n+1) is divisible by 3.
so, (1) and (2) together, n^2-1 is divisible by 24. C is correct answer.

Mayanist · 发表于 2009-3-8 11:32:00

以下是引用leenny在2007-10-19 15:07:00的发言：
是不是可以为样理解？
（1）n不能被2整除则n＝2X-1，X为正整数。则n^2-1=4X(X-1),因为X和（X-1)是两个相邻的整数，所以n^2-1＝4x(x-1)能被8整数.
(2)n不能被3整除，则n^2-1=(n-1)(n+1)。因n-1,n and n=1是三个相邻的整数而n又不能被3整除，则n-1,n+1其中一个必能被3整数，即(n-1)(n+1) is divisible by 3.
so, (1) and (2) together, n^2-1 is divisible by 24. C is correct answer.

fantastic explanation, thank you .

帮我看看：prep-ds1-159这么分析对吗？

帮我看看：prep-ds1-159这么分析对吗？

回复：（lisony）帮我看看：prep-ds1-159这么分析对吗...

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