2个数字的问题这是我数学老师说的,最好有过程.甲想

1个回答

这是一个逻辑题首先我们来说一下两个数字的组合可能, 10,11,12,13,14,15,16,17,18,19,20,22,23,24,25,26,27,28,29,30,33,34,35,36,37,38,39,40,44,45,46,47,48,49,50,55,56,57,58,59,60,66,67,68,69,70,77,78,79,80,88,89,99 这些数的两位相加是从1,2,~~~~17,18,其中为1,和17,19的只有一个数,其它都有两个或两个以上,如果只有一个,那就直接知道了,所以排除了10,89,99 剩下 11,12,13,14,15,16,17,18,19,20,22,23,24,25,26,27,28,29,30,33,34,35,36,37,38,39,40,44,45,46,47,48,49,50,55,56,57,58,59,60,66,67,68,69,70,77,78,79,80,88,90. 两个数字相乘依次是1,2,3,4,5,6,7,8,9,0,4,6,8,10,12,14,16,18,0,9,12,15,18,21,24,27,0,16,20,24,28,36,0,25,30,35,40,45,0,36,42,48,54,0,49,56,63,0,64,0 因为乙也不知道,所以排除乘积属只出现一次对应的数字 11,12,13,15,17,25,27,35,37,39,45,47,55,56,57,58,59,67,68,69,77,78,79,88 所以剩下的数字为 14,16,18,19,20,22,23,24,26,28,29,30,33,34,36,38,40,44,46,48,49,50,60,66,70,80,90. 这时对应的和为 5,7,9,10,2,4,5,6,8,10,11,3,6,7,9,11,4,8,10,12,13,5,6,12,7,8,9 因为乙还是不知道,所以再排除和只出现一次对应的数字 20,30,48,49 所以数字还剩下 14,16,18,19,22,23,24,26,28,29,33,34,36,38,40,44,46,50,60,66,70,80,90. 这时,数字积依次 4,6,8,9,4,6,8,12,16,18,9,12,18,24,0,16,24,0,0,36,0,0 而此时乙知道了数字,所以在这里出现的次数是唯一的,只有24,对应的是38 所以答案是38或者83