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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? |