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sum of digits and 100th term (Posted on 2015-10-08) Difficulty: 3 of 5
s(x) denotes the sum of digits of x.
Consider all the positive integer values of n satisfying s(n) = 100 arranged in strictly ascending order of magnitude.
Determine the 100th term of this sequence.

No Solution Yet Submitted by K Sengupta    
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Solution I did miss something in the previous post. Comment 3 of 3 |
(In reply to I've probably miscounted but here goes. by Charlie)


                                                                       total
                                                        count         so far
1, then 11 9's                                              1              1
2, then an 8 instead of one of the 9's                     11             12
3, then a 7 instead of one of the 9's                      11             23
3, then 8's replacing two of the 9's            C(11,2) =  55             78
46, then 10 9's  (we're not getting to 49...)               1             79
47, then an 8 and nine 9's                      C(10,1) =  10             89
487, then nine 9's                                          1             90 
48, then two 8's with 8 9's                     C(10,2) =  55       more than 100

It's among the last category

#91    488899999999
#92    488989999999
#93    488998999999
#94    488999899999
#95    488999989999
#96    488999998999
#97    488999999899
#98    488999999989
#99    488999999998
#100   489889999999

However, pondering this last one, which I previously gave as the answer, I see that 489799999999  (formatted 489,799,999,999) would be the next in line, as #100, and the categorization was flawed.

The answer is 489,799,999,999.

  Posted by Charlie on 2015-10-09 08:27:51
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