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PREP DS 最终仍然不解的问题,等候指教

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楼主
发表于 2008-8-1 06:32:00 | 只看该作者

PREP DS 最终仍然不解的问题,等候指教


    

好不容易把prep数学做了一遍,这里是整理出来的仍然不理解的错题,请达人指教,不少是公倍数公约数问题,疑惑中...谢谢各位 (历史帖我翻了一遍,已解决的问题已经没有写在这里了,感谢CD,不过仍然寄希望与牛牛们解我燃眉之急...

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.


    

 (2) n is
not divisible by 3.


    

 C


    

177.  17595-!-item-!-187;#058&011159


    

If x and
y are positive integers, what is the value of xy ?


    

 (1)  The greatest common factor of x and y is 10.


    

 (2)  The least common multiple of x and y is 180.


    

 C,
        
公倍数公约数问题


    

 


    

202.  19528-!-item-!-187;#058&012279


    

Each
employee of Company Z is an employee of either Division X or Division Y, but
not both.  If each division has some
part-time employees, is the ratio of the number of full-time employees to the
number of part-time employees greater for Division X than for Company Z ?


    

 (1) The
ratio of the number of full-time employees to the number of part-time employees
is less for Division Y than for Company Z.


    

 (2) More
than half of the full-time employees of Company Z are employees of Division X,
and more than half of the part-time employees of Company Z are employees of
Division Y.


    

 D


    

214.  20789-!-item-!-187;#058&013258


    

The cost of a
square slab is proportional to its thickness and
also proportional to the square of its length.  What is the cost of a square slab that is 3
meters long and 0.1 meter thick?


    

 (1) The cost of a
square slab that is 2 meters long and 0.2 meter thick is $160 more than the
cost of a square slab that is 2 meters long and 0.1 meter thick.


    

 (2) The cost of a
square slab that is 3 meters long and 0.1 meter thick is $200 more than the
cost of a square slab that is 2 meters long and 0.1 meter thick.


    

 D, 理解题目以后列式,
但是如何继续求解???


    


            
(1)   
4k + 0.2m
– 160 = 4k + 0.1 m 
9k+0.1m


    

(2)    9k + 0.1m
– 200 = 4k + 0.1m


    

17.    1097-!-item-!-187;#058&000661


    

Store S
sold a total of 90 copies of a certain book during the seven days of last week,
and it sold different numbers of copies on any two of the days.  If for the seven days Store S sold the
greatest number of copies on Saturday and the second greatest number of copies on
Friday, did Store S sell more than 11 copies on Friday?


    

 (1)  Last week Store S sold 8 copies of the book
on Thursday.


    

 (2)  Last week Store S sold 38 copies of the book
on Saturday.


    

 B 不会


    

 
80.    7589-!-item-!-187;#058&005492


    

If x and
y are positive integers such that x = 8y + 12, what is the greatest common
divisor of x and y ?


    

 (1) x = 12u, where u is an integer.


    

 (2) y =
12z, where z is an integer.


    

 B, 公约数问题,这一题是试出来的,想请教个中原理


    

 137.  14447-!-item-!-187;#058&010361


    

If x and
y are positive integers, is xy a multiple of 8 ?


    

 (1)  The greatest common divisor of x and y is 10.


    

 (2)  The least common multiple of x and y is 100.


    

 C, 公倍数公约数问题


    

 
197.  19016-!-item-!-187;#058&012397


    

The
integers m and p are such that 2 < m < p and m is not a factor of p.  If r is the remainder when p is divided by m,
is r > 1 ?


    

 (1)  The greatest common factor of m and p is 2.


    

 2)  The least common multiple of m and p is 30.


    

 A 不会,公倍公约数问题


    

 
200  19224-!-item-!-187;#058&012506


    

If d is a
positive integer and f is the product of the first 30 positive integers, what
is the value of d ?


    

 (1) 10^d
is a factor of f.


    

 (2) d
> 6


    

 C  我觉得是E吧...


推荐
发表于 2008-8-1 10:15:00 | 只看该作者

17

(1)明显不可以
(2)如果星期五卖11本,由于每天卖的都不同,剩下5天最多卖10、9、8、7、6,那么11+10+9+8+7+6< 90-38,所以星期五必须大于11

但我觉得这个方法不算太好,请教大家有没有更好的方法。

沙发
发表于 2008-8-1 09:52:00 | 只看该作者

159、n不是2的倍数,则n=2p+1; n不是3的倍数,则,n=3q+1或n=3q+2, 两者结合起来,n=6p+5或n=6p+7

当n=6p+5时,(n+1)*(n-1)=(6p+6)(6p+4)=12*(p+1)(3p+2),又因为,p+1与3p+2中一定是 一奇一偶,所以他们的乘积一定是2的倍数,结合前面的12,正好是24,所以是24的倍数,余数是0

当n=6p+7时,同理

177、x1=a*m, x2=a*n ;其中,a是x1,x2的最大公约数,显然x1,x2的最大公倍数是,a*m*n

所以 最大公约数*最大公倍数=a*a*m*n=x1*x2

板凳
发表于 2008-8-1 09:58:00 | 只看该作者

214

2*L*0.2T=2L*0.1T+160

是相乘的关系,不是相加

5#
发表于 2008-8-1 15:12:00 | 只看该作者

80.

如果你将条件一代入,得出12U=8Y+12, Y=12U-12/8.

如果你将条件二代入,得出X=8*12z+12 所以X可以被12整除.而根拒条件二本身得知Y也可以被12整除.

6#
 楼主| 发表于 2008-8-1 17:28:00 | 只看该作者
谢谢sarahzhensh和fanshiyou,
sarahzhensh的回复中有一条我没看明白,

159、n不是2的倍数,则n=2p+1; n不是3的倍数,则,n=3q+1或n=3q+2, 两者结合起来,n=6p+5或n=6p+7


没看明白结合两个条件是怎么得出这两个式子的,待指教,谢谢!
7#
 楼主| 发表于 2008-8-1 17:31:00 | 只看该作者

159 (一点疑惑如上), 202, 197, 200 待解, 谢谢!

8#
发表于 2008-8-1 18:02:00 | 只看该作者

197题,相当于是在问p/m 余数有没有可能是1

(1)也就是说两个数是偶数,而两个偶数相除,如果除不尽,余数绝不可能是1,余数绝对应该是2的倍数(这点不太好解释,仔细思考一下)

(2)代入试验:6/5 余1,15/6余3

159的公式 是我把数列出来后自己归纳的。

202其实很简单的,整体与部分的关系。把题看懂了应该没什么问题

9#
发表于 2008-8-1 21:45:00 | 只看该作者

200

考虑25里面有2个5

10#
发表于 2008-8-1 21:51:00 | 只看该作者

159.

(1) n = 2k + 1 或者可以表达为 2k - 1  (2) n = 3m + 1 or n = 3m + 2(或者可以表达为3m - 1)

如果 n = 3m + 1

那么2k + 1 = 3m + 1,那么 n = 6x + 1

如果n = 3m - 1

那么2k - 1 = 3m - 1, 那么 n = 6x - 1

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