Suppose N is any positive integer. Add the digits of N to obtain a smaller integer. Repeat this process of digit-addition till you get a single digit number n. Find the number of positive integers N ≤ 1000, such that the final single-digit number n is equal to 5.


Let $s(n)$ be the sum of digits function. Then $n\equiv s(n)\pmod{9}$, and by induction $$ n\equiv s(\cdots(s(n))\cdots)\pmod{9}. $$ Since the last obtained number is $5$ then it equivalent to find the number of positive integers which have remainder $5$ modulo $9$, i.e., $\lfloor 1000/9\rfloor=111$.

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  • $\begingroup$ why "positive integers which have remainder 5 modulo 9 will be [1000/9]"? please explain. $\endgroup$ – mnulb Feb 1 '17 at 17:19
  • $\begingroup$ You just pick one integer every nine consecutive ones... $\endgroup$ – Paolo Leonetti Feb 1 '17 at 17:20
  • $\begingroup$ first will be 5 then 14 then 23.... last will be 999? $\endgroup$ – mnulb Feb 1 '17 at 17:21
  • $\begingroup$ $999$ has not remainder $5$ when it is divided by $9$; the last one is $995$.. $\endgroup$ – Paolo Leonetti Feb 1 '17 at 17:38

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