I got the same answer. assume #1 is needed and true, the possible number could be 1,1,3 or 1,2,2. for p,q,r, the smallest prime numbers are 2,3,5. so the min. possible 組合 would be 2^3x 3x 5 = 120, or 2x3^2x5^2=450. so A is enough to prove that the answer is >100. Only B is not enough, so my answer could be A.
am I correct?
V4.【xinnono】问n,p, q 是大于1的不同质数, x, y, z是正整数, (n^x)*(p^y)*(q^z)是否大于100. 条件1: x+y+z=5 条件2: npq=30. 记得最早看JJ时候, npq和xyz说混了, 整理说xyz如果是大于1的不同质数, 最小235, 怎么可能和是5, 这里解答了~ 楼主选了C. 我觉得应该选A吧~npq取最小的质数,2 3 5,x+y+z=5,让他们乘积最小,则 可以为2^3*3^1*5^1=120,大于100,题中就问能不能大于100,最小的情况都能,所以一定能~求NN们讲解~ -- by 会员 奇幻女孩 (2011/12/12 9:28:17)
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