Divide 2008 into the sum of n natural numbers, and the digits of these n natural numbers are the same. If you erase all the digits of these n natural numbers, the remaining array will become an arithmetic sequence with tolerance of 6. The maximum of n is______ ．

Divide 2008 into the sum of n natural numbers, and the digits of these n natural numbers are the same. If you erase all the digits of these n natural numbers, the remaining array will become an arithmetic sequence with tolerance of 6. The maximum of n is______ ．

Because 2008 bits are 8, 1 × 8 = 8, 2 × 4 = 8, 6 × 8 = 48, so the possibility is that 8 digits are 1, 2 digits are 4, 4 digits are 2, 6 digits are 8, 8 digits are 6. Because we find the maximum value of N, we consider that 8 digits are 1, if 8 digits