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Suppose $N = 1+11+101+1001+10001+\dots+1\underbrace{00\dots00}_{50\text{ zeroes}}1$.

When $N$ is written as a single integer, i.e. all terms added, what is the sum of the digits of $N$?

I tried subtracting $1$ from each term to get:

$$0+10+100+\dots\;,$$

therefore ending with a sum of $\underbrace{111\dots111}_{50\text{ ones}}$.

Then I added $50$ to the end two numbers of $N$ ($50$ is the total number of ones I minused); there the end two numbers would be $6$ and $0$; therefore add all the ones, ($49$ I think) and the $6$ to obtain $55$.

Not sure if this is the way to do it though?

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  • $\begingroup$ I may have miscounted my ones $\endgroup$
    – Daniel
    Commented May 15, 2012 at 7:45
  • $\begingroup$ The method seems all right. Be careful though in counting things; for instance your sum has $52$ terms to begin with. $\endgroup$
    – Marc van Leeuwen
    Commented May 15, 2012 at 7:46

1 Answer 1

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You might try again to count the numbers of terms. If you had started with $1+11+101 =113$ your method seems to give an answer of $2$ when the answer should be $5$.

I think you have $52$ terms in the original sum, giving $51$ ones in $111\ldots1110$ to which you add $52$, giving giving $50$ ones in $111\ldots1162$ a digit sum of $58$.

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  • $\begingroup$ do you add 52 though? because we subtracted the ones, we must add them at the end of the number not the actual number 52 $\endgroup$
    – Daniel
    Commented May 15, 2012 at 7:47
  • $\begingroup$ @Daniel: yes - as in "to which you add $52$" $\endgroup$
    – Henry
    Commented May 15, 2012 at 7:48
  • $\begingroup$ on the end we have 1110 i propose we add the 52 to the 10 at the end to obtain 62 at the end $\endgroup$
    – Daniel
    Commented May 15, 2012 at 7:48
  • $\begingroup$ no you add the sum of the digits not the digit itself $\endgroup$
    – Daniel
    Commented May 15, 2012 at 7:48
  • $\begingroup$ @Daniel: I have adjusted my reply $\endgroup$
    – Henry
    Commented May 15, 2012 at 7:51

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