Question 12: Given a natural number consisting of consecutive natural numbers from 1 to 1999 written in consecutive order as follows: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.. ..1994, 1995, 1996, 1997, 1998, 1999. Sum all those digits.

Comment:

– Two numbers 0 and 1999 have the sum of digits: 0 + 1 + 9 + 9 + 9 = 28

– Two numbers 1 and 1998 have the sum of digits: 1 + 1 + 9 + 9 + 8 = 28

– Two numbers 9 and 1990 have the sum of digits: 9 + 1 + 9 + 9 + 0 = 28

– Two numbers 10 and 1989 have the sum of digits: 1 + 0 + 1 + 9 + 8 + 9 = 28

– Two numbers 99 and 1900 have the sum of digits: 9 + 9 + 1 + 9 + 0 + 0 = 28

– Two numbers 100 and 1899 have the sum of digits: 1 + 0 + 0 + 1 + 8 + 9 + 9 = 28

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– Two numbers 998 and 1001 have the sum of digits: 9 + 9 + 8 +1 + 0 + 0 + 1 = 28

– Two numbers 999 and 1000 have the sum of the digits: 9 + 9 + 9 +1 + 0 + 0 + 0 = 28

Thus, in the sequence of numbers 0,1,2,3,4….,1997,1998, 1999, the sum of the digits that make up each pair of numbers counting from both ends upwards (such as 0 and 1999 , 1 and 1998, 2 and 1997, ….) are equal and equal to 28.

There are 1000 such pairs. So the sum of all the digits that make up the above sequence is: 28 x 1000 = 28000

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