|
发表于 5-7-2014 15:22:46|来自:新加坡
|
显示全部楼层
为啥是8呢? 不是 7 吗?请教老师怎么算得?
5*8*11*14* .... * 95*98*101
a) There are 33 terms [(101-5)/3 +1]. These 33 terms can be divided into 3 groups of 10 numbers plus the 3 remaining numbers ie 95, 98, 101
b) 1st group of numbers are 5, 8, 11, 14, 17, 20, 23, 26, 29, 32 should result in A*10^2 since only 5*8*20 will result in 2 zeros for the last 2 digits. Note that A must be an integer. We also know that the last digit for A must be a 2 since the remaining one digits 1*4*7*3*6*9*2 will end with a 2.
c) For the 2nd and 3rd group of numbers, since the digit in the one place are the same, they will result in B*10^2 and C*10^2 (same logic as (b)). For B and C, it must end with a 2 too.
d) For the last 3 numbers: 95*98*101, it must be D*10^1 with the last digit of D is a 1.
e) So (A*10^2) * (B*10^2) * (C*10^2) * (D*10^1) = E * 10^7 (note that the last digit of A, B and C is a 2 and when one multiplies them, one does not get a zero).
So I think that the answer should be a 7 instead of 8.
|
|