1. Someone forms an integer by writing the integers from 1 to 82 in ascending order, i.e. 1234567891011 ...808182. Find the sum of the digits of this integer.
某人把 1 到 82 之間的整數順序窞出來, 從而得到另一個整數 1234567891011...808182。求這個整數的數字之和。
1. 667
1. We don't have to care about zeros as far as sum of digits is concerned, so we simply count the number of occurrences of the non-zero digits. The digit ' 3 ' occurs 18 times ( 10 as tens digits in $30,31, \ldots, 39$ and 8 as unit digits in $3,13, \ldots, 73$ ). The same is true for the digits ' 4 ', ' 5 ', ' 6 ' and ' 7 '. A little modification shows that the digits ' 1 ' and ' 2 ' each occurs 19 times. Finally the digit ' 8 ' occurs 11 times $(8,18,28, \ldots, 78,80,81,82)$ and the digit ' 9 ' occurs 8 times $(9,19$, $29, \ldots, 79)$. Hence the answer is
$$
(3+4+5+6+7) \times 18+(1+2) \times 19+8 \times 11+9 \times 8=667 \text {. }
$$