扫描二维码，直接登录
题库 / Prep2007E2-DS-166
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p2t ?
(1) m has more than 9 positive factors.
(2) m is a multiple of p3.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
还没有题目讲解（毕出老师会陆续发布对官方考题的解读，请保持关注）。
If the prime numbers p and t are the only prime factors of integer m
=> m=(p^a)(t^b) 且a>=1, b>=1
is m a multiple of p^2t? => 即要求证a>=2
1) m has more than 9 positive factors => (a+1)(b+1)>9 不能保证a>=2 比如a为1，b为4就不行
2) m is a multiple of p^3 ＝> 直接给出 a>=3
任何一个自然数的因子个数 为质因数分解中 每个质因子的指数加1的乘积
i.e. 20=(2^2)(5^1) 所以20的因子个数为（2＋1）（1＋1）＝6个，分别是1，2，4，5，10，20
所以B.
是否为tp的倍数
任何一个自然数的因子个数 为质因数分解中 每个质因子的指数加1的乘积
i.e. 20=(2^2)(5^1) 所以20的因子个数为（2＋1）（1＋1）＝6个，分别是1，2，4，5，10，20
A选项的情况，如果有9个因素，先假设一个P, 一个t， 剩余7可能都是合数，或者含有或全部都是P和t，如果都是合数或者t更多，那么不成立。
B选项，用以上相同思维，是成立的，
since m is also a multiple of p3, then m must be a multiple of p2. (e,g. p=2, m can be divided by 8, then m can also be divided by 4.)
so m is a multiple of p2t
次方和t的4次方也符合1
INSUFFICIENT: it doesnt tell exponent/powers of prime factors p & t. We dont know whether m is multiple of p^2.
2) m is a multiple of p3p3
SUFFICIENT: If m is a multiple of p3p3, then m must be multiple of p2p2. As 't' is also a prime factor of m, then m must be multiple of p2∗tp2∗t
e.g. say m=24, p=2, t=3. As 24 is multiple of p3=23=8p3=23=8, 24 must be multiple of p2=22=4p2=22=4, and therefore 24 is also multiple of p2∗t=22∗3=6
(1) M有9个因子，不能判断M是否能被P^2T整除
(2) M=BP^3，由条件可知M=APT联立的P^2=AT/B为整数，两边同乘以T得到P^2T=AT/B，为整数，所以M可以被P^T整除，条件充分
为什么是整数呀登录 或 注册 后可以参加讨论