best simple answer!If a number is divisible by 3, and you remove any digit and it remains divisible by 3, then the digit that you removed must itself be divisible by 3. If there are no zeros, then the digit must be 3, 6, or 9. And since you can remove *any* digit and the result is still divisible by 3, that means *all* digits must be 3, 6, or 9. And there are 3 * 3 * 3 * 3, or 81, such numbers. However, the original number isn't guaranteed to be divisible by 3 in the first place. So you also have to include the numbers that aren't divisible by 3 but any 3-digit number made by removing one of its digits is. This occurs when each digit is one greater than a multiple of 3 or when each digit is two greater than a multiple of 3. This means how many four-digit numbers exist such that each digit is a 1, 4, or 7, or each digit is a 2, 5, or 8. The answer for each is 3 * 3 * 3 * 3, or 81. So the total number of 4-digit numbers that meet the condition is 81 + 81 + 81, or 243.

说的我兴起了，出一个数学题目一个自然数，她的二分之一能被2整除，她的三分之一能被3整除，她的五分之一能被5整除，她的七分之一能被7整除，请问， 这个自然数最小是几啊？ 赫赫