Suppose that three consecutive natural numbers are of the form: n – 1, n, n + 1.
Then their sum will be: n – 1 + n + n + 1 = 3 * n, that is, this number is a multiple of 3.
The sum of the digits of a multiple of 3 must be divisible by 3, we get:
* 1234: * + 1 + 2 + 3 + 4 = * + 10. In order for the indicated amount to be divisible by 3 and have the smallest value, instead of *, you must put 2.
2 + 10 = 12.
Thus, the smallest sum of three consecutive numbers that meets the conditions of the problem is 21234.
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