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[求助]还有几道GWD数学向nn们请教~~

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楼主
发表于 2009-8-23 09:12:00 | 只看该作者

[求助]还有几道GWD数学向nn们请教~~

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

正确答案是D。
貌似三个连续正整数的积能被3整除,而这题问的是连续三个正整数的积能被8整除。是不是要考虑连续三个正整数积能被24整除的数?

呵呵,具体该怎么做捏~~

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数据充分题(选项省略)

If n and k are positive integers, is n divisible by 6?

(1) n = k(k +1)(k - 1)
(2) k - 1 is a multiple of 3

我选的是C,即两个联合才够充分;正确答案是A,条件1就足够充分。

后来想想 ,是不是正如上题所说,三个连续正整数乘积一定能被3整除,就像连续两个正整数乘积一定能被2整除一样。那么k(k + 1)(k - 1)中, 既有2的倍数,也有3的倍数,因此就一定能被6整除?

我这样想 对吗?

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If an automobile averaged 22.5 miles per gallon of gasoline, approximately how many kilometers per liter of gasoline did the automobile average? (1 mile = 1.6 kilometers and 1 gallon = 3.8 liters, both rounded to the nearest tenth)

A. 3.7
B. 9.5
C. 31.4
D. 53.4
E. 136.8

我选的是E,正确答案是B
这题怎么酱紫麻烦捏~~

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沙发
发表于 2009-8-23 10:17:00 | 只看该作者

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

正确答案是D。
貌似三个连续正整数的积能被3整除,而这题问的是连续三个正整数的积能被8整除。是不是要考虑连续三个正整数积能被24整除的数?

呵呵,具体该怎么做捏~~

All the even # within the range(1..96) will be divisible by n*(n+1)*(n+2) (because "2 ,4 6 ...96" divided by 2 is 1,2,3.... within which 2 4 6 ... are factor of 2, so n & n+2 will be a factor of 2 & 4 if n is even)

If n is odd n(n-1)(n-2) will be divisible by 8 iif n+1 = 8, if you count all those "n"s u should find around 8*12 of them, then u should get the answer.

BTW, if I see this in exam and need to solve it quickly, I will just take some random samples and guess the probability.

2) statement 2 alone is never sufficient. statment 1 is sufficient because it tells n=1 2 3 ; 2 3 4.....  all of which have factors 2, 3

" averaged 22.5 miles per gallon of gasoline" what is that suppose to mean?

板凳
发表于 2009-8-23 10:19:00 | 只看该作者

第一题:

第一种情况 n为偶数,n(n+2)能被8整除,如果两个连续偶数相乘,肯定能被8整除。第二种情况n为奇数,这时n(n+2)为奇数,要指望n(n+1)(n+2)被8整除,只能是N+1是8的倍数。96/8=12 有12种 1-96中奇数的个数为48

所以总的概率=1/2+(1/2)*(12/48)=5/8



地板
发表于 2009-8-23 10:32:00 | 只看该作者

第二题你想法是对的

第三题 只要用22.5*1.6/3.8就行了

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