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Thread: Sum of digits of numbers between 1 to n

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    Sum of digits of numbers between 1 to n

    Hi, is there any general formula for finding the sum of digits from 1 to n (where n can be upto 10^9) .
    I know it's got to do with some multiple of 45 (sum of digits from 1 to 9) but can't relate that to the required forumula.
    Thanks .
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    The sum of the numbers from \displaystyle 1 to \displaystyle n is \displaystyle \frac{n}{2}(1+n).
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    Thanks but i was interested in sum of digits rather than the sum of numbers
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    Quote Originally Posted by pranay View Post
    Hi, is there any general formula for finding the sum of digits from 1 to n (where n can be upto 10^9) .
    I know it's got to do with some multiple of 45 (sum of digits from 1 to 9) but can't relate that to the required formula.
    I don't think you are going to find a straightforward formula for this, except for some special cases of n.

    For example, you can find the sum of the digits of all the numbers containing k digits, as follows. There are 9*10^{k-1} such numbers (9 possibilities for the first digit, 10 possibilities for each of the remaining k1 digits). Each of these numbers has k digits, so there are 9*k*10^{k-1} digits altogether. Of these, (k-1)*9*10^{k-2} will be zeros (each number stands a 1-in-10 chance of having a 0 in each position except the first). That leaves 9k*10^{k-1} - (k-1)*9*10^{k-2} other digits. Each of the nine other digits (1 to 9) is equally likely to occur, so each digit occurs k*10^{k-1} - (k-1)*10^{k-2} = (9k+1)*10^{k-2} times. The sum of the numbers from 1 to 9 is 45, so the sum of all the digits in all the k-digit numbers is 45*(9k+1)*10^{k-2}.

    If you sum that result for k going from 1 to m, you find that the sum of all the digits in all the numbers from 1 to 10^m-1 is \boxed{45*m*10^{m-1}}.

    But if you want the result for a general value of n (not of the form 10^m-1 ) then you will have a lot more work to do.
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    Thanks a lot for the great explanation , exaclty what i wanted . Thank you
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