The integers m and p are 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 greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.
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我不是你说的NN,但是抛砖引玉,等大家批判:
我选C,因为
(1)--> assume m=2a and p=2b, a and b are prime numbers:2,3,5... --> p/m=2b/2a=b/a
-->r=1(e.g. b=3,a=2) or r>1 (e.g. b=5,a=3)
(2) -->30=2*3*5 and m>2 --> r=1(e.g. p=10,m=3) or r>1(e.g. p=10,m=6)
(1)+(2) --> LCM of m&p = 2ab=30 -->a=3,b=5 -->p/m=10/6 -->r>1
我觉得楼上的给的答案不对,我直接举反例:
条件(1)最大公约数2,说明m和p都是偶数,偶数最小也差2因此r一定大于1,比如m=4,p=6,或者m=8,p=10。因此条件1充分
条件(2)最小公倍数30,m=5,p=6则r=1,m=6,p=10,则r=4。因此条件2不充分
所以我认为本题答案应该选A
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