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.
(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.
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Posted by Charlie
on 2015-10-09 08:27:51 |
I did miss something in the previous post.
