求教两道数学题

brendaqian

1.for every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. if p is the smallest prime factor of h(100)+1, then p is?
A. 2 to 10
B.10 to 20
C.20 to 30
D. 30 to 40
E. >40
2.a certain library assesses fines for overdue books as follows: on the first day that a book is overdue, the total fine is 0.1, for each additional day that the book is overdue, the total fine is either increase by 0.3 or doubled,whichever results in the lessen amount.what's the total fine on fourth day a book overdued?
(答案给出的是0.7)

WesternCder

This one has been explained before.

Basically,
1) These consecutive natural numbers h(100) and [h(100) + 1]  are co-prime, meaning that they do not share any prime numbers as their cofactors.  
2) h(100) contains all the prime numbers between 2 (from 2) and 47 (from 94).
3) Then prime numbers among factors of [h(100) + 1] would not include any prime numbers between 2 and 47.
4) Then the smallest prime facotr of [h(100) + 1] is bigger than 47.

bah-torontoer

第一题还是没弄明白,请讲解下,谢谢!

momohe

第一题，
有一个小技巧，就是相邻的自然数的最小公倍数是1.
这个怎么想到的。。。。

brendaqian

嗯嗯，谢谢~当时题目意思都没怎么理解

haileywang

LZ啊，偶前两天做Prep这两个题目也错了...第一题的曼哈顿解释很精彩：
Guest, this is definitely a difficult number properties question. Let's first consider the prime factors of h(100). According to the given function,
h(100) = 2*4*6*8*...*100


By factoring a 2 from each term of our function, h(100) can be rewritten as
2^50*(1*2*3*...*50).


Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).


Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.


Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.

第二题：
a1=0.1
a2=0.2
a3=0.4(double because 0.2+0.3>0.4)
a4=0.7(add 0.3 because 0.4*2>0.7)

ozymendias

第一题不会算，第二题还好只有四天的要算，都写出来就好了，题目的意思，是超过一天以后每多一天则罚款为
min(n+0.3,2n),n为前一天的罚款。