prep一道题，好像答案不对？

The integers m and p such that 2<m<p and m is not a factor of p.   If r is the remainder when p is divided by m, is r>1?

1)the greated common factor of m and p is 2

2)the least common multiple of m and p is 30

The answer of this question is A, but I choose D, because 2) is sufficient:

Under the condition of 2) the possible solution is m=3,p=10 or m=5,p=6 so r=1

Who can explain the ets answer to me?

if m=10, p=15,then r=5>1

I don't know why you choose D,it seems your explaination proves that D is wrong...

The integers m and p such that 2<m<p and m is not a factor of p.   If r is the remainder when p is divided by m, is r>1?

1)the greated common factor of m and p is 2

第一个条件怎么知道是对的?

（１）is sufficent ,for r>1 must be true!

(2) is not sufficent ,for r>1 or r=1.

这样理解DS题,对路吗?

我只能推出(1)但是对于(2) 我没能推出是不是充分的:

假设2<2a<2b

当然a不是b的约数

余数要么大于一,要么就等于一,因为不为约数所以余数不为零.

假设余数等于一,则2b可以表示成: 2ak+1=2b 其中k为2b除以2a的商,余数为一

则,2ak必为偶数,而加一为奇数,可是2b也为偶数,所以不能成立.

即,余数要么没有,要么就大于一.