P2-DS-122

[bibbyshiu]

122. 12685-!-item-!-187;#058&009461 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.

先把24分解=3 x 8
(1) 2不是n的因數, n=2k+1, (n-1)(n+1) = (2k)(2k+2) ==> 連續偶數乘積可被8整除, 連續偶數一定某個被2整除,某個被4整除
(2) 3不是n的因數 n=3y+1, (n-1)(n+1)=(3y)(3y+2) 或是 n=3y+2, (n-1)(n+1)=(3y+1(3y+3) ==> 可被3整除.

兩個條件合起來, 可同時被8與3整除的數..除以24..餘數為0.
~_~...這只是商學院的考試有必要考這麼難嗎?

[chris8888]

122. 12685-!-item-!-187;#058&009461
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.
n = 2k + 1 insufficient

(2) 3 is not a factor of n.
n = 3k + 1 insufficient
n = 3k + 2 insufficient

1+2.
(2k+1-1)(2k+1+1) => (2k+0) * (2k+2) <== 可以看出是連續的偶整數,
連續偶整數最小是2 and 4, 若相承至少會有8.
P.S. (0是整數, 但不是偶數, 也不是正數)
(3k+0)(3k+2) => 3k(3k+2) 有3的factor
(3k+1)(3k+3) => 3(3k+1)(k+1) 有3的factor
因此, 兩個條件合在一起, 要符合不是3的倍數又不是2的倍數, 那麼(n-1)(n+2) = 24的倍數(8*3)

若考試沒想到, 還有個更快的作法
1. 不是3又不是2的倍數, 依序是1, 5, 7 ....
代入1, (1-1)(1+1)/24 = 餘0
代入5, (5-1)(5+1)/24 = 餘0
.....

Ans: (C)