y***n
发帖数: 175 | 1 Find the max integer which satisfies the following requirement:
Starting from left most digit, each digit's position k can be counted as 1,
2, 3, ...., n
The number left to this digit (including current digit), should be divisible
by the current count.
Example:
96005
k=1, value is 9
k=2, value is 96
k=3, value is 960
k=4, value is 9600
k=5, value is 96005
So this int satisfy the requirement, but of course this is just example, so
not the max int satisfies this
requirement.
Any idea how to solve it mathematically or programmatically? | x*****p
发帖数: 1707 | | y***n
发帖数: 175 | | x*****p
发帖数: 1707 | 4 我用程序找到的下面的数可以直接让count到100. 我怀疑没有最大数
98765456405070085200025000001000020802000000002000001000000000000000
00000000000000000040208000000000 | p*******e
发帖数: 37 | 5 至少可以是9876545640546
【在 x*****p 的大作中提到】
: 98765456405
| y***n
发帖数: 175 | 6 Can you paste your code here? | p*******e
发帖数: 37 | 7 最大数是有可能有的，因为在最后一位只有10种可能的情况下，对于大于10的质数
是有可能除不尽的。
【在 x*****p 的大作中提到】
: 我用程序找到的下面的数可以直接让count到100. 我怀疑没有最大数
: 98765456405070085200025000001000020802000000002000001000000000000000
: 00000000000000000040208000000000
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