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做真题遇到的几道恶心题目

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匿名  发表于 1970-1-1 08:00:00
9#
发表于 2006-7-10 21:28:00 | 只看该作者
8#
发表于 2006-7-10 17:51:00 | 只看该作者
以下是引用esprit在2005-11-17 16:56:00的发言:

if n is positive integer,R is the remainder of (n^2-1)/24,then R=?

(1)2 is not a factor of n

(2)3 is not a factor of n

ANS:C

(n+1)(n-1)/24         (1)说明n是奇数,所以(n+1)(n-1)是8的倍数              (2)说明(n+1)、(n-1)中有一个是3的倍数。所以选C



One kilogram of a certain coffee blend consists of  x  kilogram of type I coffee and y kilogram of type II coffee ,THe cost of the blend is C dollars per kilogram where C=6.5x+8.5y,is x<0.8

(1)y>0.15

(2)C>=7.30

ANS:B

x+y=1,代入则c=8.5-2x                     (1)得到x<8.5            (2)得到x<=0.6             所以选b



when 15n(positive integer) is divided by 6,remainder is X,then X=?

(1)WHen n is divided by 2 the remainder is 0

(2)When n is divided by 3 ,the remainder is 0

ANS:A

15n/6=3*5n/6                           (1)说明n中有2,所以remainder是0           (2)什么也不能说明。
这种题目都是问能整除的条件。


求解,谢谢

7#
 楼主| 发表于 2005-11-17 21:52:00 | 只看该作者

谢谢 ~~~~~~~~~~~~~~~~~~~~~~~~

6#
发表于 2005-11-17 21:10:00 | 只看该作者
前两句推翻各自陈述,第三行提出不含2,3因子的n的通式
5#
发表于 2005-11-17 20:46:00 | 只看该作者

if n is positive integer,R is the remainder of (n^2-1)/24,then R=?
(1)2 is not a factor of n
(2)3 is not a factor of n
ANS:C

(1) n,R = 1,0 or 3,8 ...
(2) n,R = 1,0 or 2,3 ...
(1)&(2): n = 6k+1 or 6k+5
for n=6k+1, n^2-1=6k(6k+2)=12k(3k+1), k无论奇偶, k(3k+1)总是偶, 故含有24这个因子, R=0
for n=6k+5, n^2-1=6(k+1)(6k+4)=12(k+1)(3k+2), k无论奇偶, (k+1)(3k+2)总是偶, 故含有24这个因子,R=0
therefore ... C

阴影部分是什么道理,请解释,谢谢
地板
发表于 2005-11-17 19:40:00 | 只看该作者

if n is positive integer,R is the remainder of (n^2-1)/24,then R=?
(1)2 is not a factor of n
(2)3 is not a factor of n
ANS:C

(1) n,R = 1,0 or 3,8 ...
(2) n,R = 1,0 or 2,3 ...
(1)&(2): n = 6k+1 or 6k+5
for n=6k+1, n^2-1=6k(6k+2)=12k(3k+1), k无论奇偶, k(3k+1)总是偶, 故含有24这个因子, R=0
for n=6k+5, n^2-1=6(k+1)(6k+4)=12(k+1)(3k+2), k无论奇偶, (k+1)(3k+2)总是偶, 故含有24这个因子,R=0
therefore ... C



One kilogram of a certain coffee blend consists of  x  kilogram of type I coffee and y kilogram of type II coffee ,THe cost of the blend is C dollars per kilogram where C=6.5x+8.5y,is x<0.8
(1)y>0.15
(2)C>=7.30
ANS:B


已知1=x+y, C=6.5x+8.5y
(1) 0.15<y=1-x,则x<0.85, 无法知道x<0.8
(2) 7.3<=C=6.5x+8.5(1-x),则2x<1.2, x<0.6<0.8, 充分!

板凳
发表于 2005-11-17 18:58:00 | 只看该作者
第一题,天山9的吧。欧用2,4,6,8和3,5,7,9分别试,n只可能为5
[此贴子已经被作者于2005-11-17 18:57:50编辑过]
沙发
发表于 2005-11-17 17:42:00 | 只看该作者

when 15n(positive integer) is divided by 6,remainder is X,then X=?


(1)WHen n is divided by 2 the remainder is 0


(2)When n is divided by 3 ,the remainder is 0


ANS:A



n能被2正处,15n is divisable by 6, so remainder is 0



the other two questions need NN help

楼主
发表于 2005-11-17 16:56:00 | 只看该作者

做真题遇到的几道恶心题目

if n is positive integer,R is the remainder of (n^2-1)/24,then R=?


(1)2 is not a factor of n


(2)3 is not a factor of n


ANS:C





One kilogram of a certain coffee blend consists of  x  kilogram of type I coffee and y kilogram of type II coffee ,THe cost of the blend is C dollars per kilogram where C=6.5x+8.5y,is x<0.8


(1)y>0.15


(2)C>=7.30


ANS:B





when 15n(positive integer) is divided by 6,remainder is X,then X=?


(1)WHen n is divided by 2 the remainder is 0


(2)When n is divided by 3 ,the remainder is 0


ANS:A




求解,谢谢

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