- UID
- 436964
- 在线时间
- 小时
- 注册时间
- 2009-4-22
- 最后登录
- 1970-1-1
- 主题
- 帖子
- 性别
- 保密
|
2 翻译过来就是“如果2n的因子是n的因子的2倍”
如果n是偶数，2n的因子个数由于重叠则肯定少于n因子个数的两倍（n的因子包含2）
如果n为奇数，2n的因子个数由于没有重叠则肯定等于n因子个数的两倍（n的因子中不包含2）
精确解释如下：
Statement (1) - insufficient as n can be both even and odd.
Statement (2) - sufficient - Here n will have a unique prime factorisation as a product of
p1 ^ q1 * p2 ^ q2 * ....... pk ^qk
where p1 is less than p2 , p2 is less than p3 and so on.
Now p1 is either equal to 2 or it is an odd integer greater than 2.
Thus if n is odd then 2n has a unique prime factorisation of
2 * p1^q1 * p2^q2 * .... * pk ^ qk
but if n is even then 2n can be factorised as
p1^(q1+1) * p2^q2 * ... where p1=2, q1 >= 1
Hence , now the total number of factors is based on choice of prime powers for each prime in term.
Thus if n is odd, then 2n must have twice as many factors as n.
Alternatively in simpler language
If n is even --- 2n will not have twice as many divisors as n
e.g . For n=2; 2n or 4 has 1,2,4 as divisors ,
For n = 4 divisors are 1,2,4; 2n=8 has 1,2,4,8
But if n is odd
then if n=3 divisors are 1,3 and 2n=6 divisors are 1,2,3,6
n=5 divisors are 1,5 and 2n=10 divisors are 1,2,5,10
Thus if n is odd the 2nd condition is satisfied. Hence ans is B |
|
京公网安备11010202008513号 京ICP证101109号 京ICP备12012021号
发表于 2009-11-9 10:58:29
收藏
收藏
