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PP- DS- T1Q70

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PP- DS- T1Q70

文章小花 » 2008-01-07 20:25

70. 6067-!-item-!-187;#058&004849
Is the integer x divisible by 6 ?

(1) x + 3 is divisible by 3.

(2) x + 3 is an odd number.


ANS:C

各位有快速解法嗎
我用土法煉鋼 一個個數字慢慢代
又慢又容易錯
小花
高級會員
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文章: 392
註冊時間: 2007-08-23 14:59

Re: PP- DS- T1Q70

文章chris8888 » 2008-01-07 22:36

小花 \$m[1]:70. 6067-!-item-!-187;#058&004849
Is the integer x divisible by 6 ?

(1) x + 3 is divisible by 3.

x + 3 = 3a
x = 3a -3
x = 3(a-1)
if a = 1, x = 0 (is divisible by 6)
if a = 2, x = 3 (is not divisible by 6)
if a = 3, x = 6 (is divisible by 6)
if a = 4, x = 9 (is not divisible by 6)

(2) x + 3 is an odd number.
if x = 2, x+3=5 odd number, 2 is not divisible by 6

(1)+(2),
x = 3(a-1) and x+3 is odd number
a = 1, x = 0, 0+3=3 (x+3 is odd but x is not divisible by 6)
a = 2, x = 3, 3+3=6 (x+3 is not odd)
a = 3, x = 6, 6+3=9(x+3 is odd and x is divisible by 6)
a = 4, x = 9, 9+3=12(x+3 is not odd)
a = -1, x = -6, -6+3=-3(x+3 is odd and x is divisible by 6)

Answer 應該是 E, 除非限定x > 0, 那就會是C


ANS:C

各位有快速解法嗎
我用土法煉鋼 一個個數字慢慢代
又慢又容易錯
頭像
chris8888
高級會員
高級會員
 
文章: 444
註冊時間: 2007-07-31 22:47

文章小花 » 2008-01-08 09:07

(2) x + 3 is an odd number
我認為他給的條件是告訴我們X是偶數
(偶+奇)=奇
(奇+奇)=偶
(偶+偶)=偶

(1)(2)合起來看
(1)x = 3(a-1) =>3是X的因數
(2)X是偶數
故C
最後由 小花 於 2008-01-09 18:27 編輯,總共編輯了 1 次。
小花
高級會員
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文章: 392
註冊時間: 2007-08-23 14:59

文章puja » 2008-01-09 10:16

小花 \$m[1]:(2) x + 3 is an odd number
問認為他給的條件是告訴我們X是偶數
(偶+奇)=奇
(奇+奇)=偶
(偶+偶)=偶

(1)(2)合起來看
(1)x = 3(a-1) =>3是X的因數
(2)X是偶數
故C


但若x=0 則 x+3=3不能為6整除,故還是 E比較合理~~~ 投 E一票!!
puja
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初級會員
 
文章: 46
註冊時間: 2007-12-24 16:37

文章mupeane » 2008-01-09 16:20

由(1),x+3是3的倍數
由(2),條件不足

so由(1)+(2)
when x+3=3, 不可被6整除
when x+3=9,不可被6整除
以此類推
mupeane
初級會員
初級會員
 
文章: 25
註冊時間: 2007-05-20 10:21

文章小花 » 2008-01-09 18:26

puja \$m[1]:
小花 \$m[1]:(2) x + 3 is an odd number
問認為他給的條件是告訴我們X是偶數
(偶+奇)=奇
(奇+奇)=偶
(偶+偶)=偶

(1)(2)合起來看
(1)x = 3(a-1) =>3是X的因數
(2)X是偶數
故C


但若x=0 則 x+3=3不能為6整除,故還是 E比較合理~~~ 投 E一票!!


謝謝puja加入討論的行列
我剛剛檢查PP給的答案 答案的確是C
0好像不是偶數
小花
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高級會員
 
文章: 392
註冊時間: 2007-08-23 14:59


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