On the number line above, the segment from 0 to 1 has been divided into fifths, as indicated by the large tick marks, and also into sevenths, as indicated by the small tick marks, what is the least possible distance between any of the two tick marks?
A1/70
B1/35
C2/35
D1/12
E1/7
Answer:B
PS:在机经里由于数字小,机主用的方法是死算。但终究不是办法??正确的方法是什么?跪求!!!
1月15日考试在即。。。
谢谢了
就直接把1/5 * 1/7 = 1/35 啦
就这样类推下去 分子都为1 分母为平分的份数
这题PREP里面有详细解释啊:
The small tick marks are placed at 1/7, 2/7, 3/7, 4/7, 5/7, and 6/7, and the large tick marks are at 1/5, 2/5, 3/5 and 4/5. The least common denominator is 35, so the tick marks in ascending order are placed at 5/35, 7/35, 10/35, 14/35, 15/35, 20/35, 21/35, 25/35, 28/35, and 30/35. The least distance between tick marks on this number line is 1/35.
The correct answer is B.
不能直接按照3楼的说法取最小公倍数分之一的,如果分子之间的差值最小不是1,那答案就不对了。
这道题偶的看法是这样:
题目的模型是求单位“1”的4个5的等分点和6个7等分点之间的最小距离。
用数学语言是:a,b为正整数,求│a/5 - b/7│的最小值。隐含的条件是a∈{1,2,3,4},b∈{1,2,3,4,5,6}。
求解,是求│7a-5b│/35 的最小值。
方法只有枚举法(就是挨个试~_~..)。取a=1,最小值为2/35,不够小;a=2,得到b=3时,有最小值1/35;a=3时,得到b=4时,有最小值1/35……
所以,楼上的方法是可行的,应该可以证明│na-mb│总能找到最小值1的。
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