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.
The first on the list would presumably be 199,999,999,999. What next?
I'd think 289,999,999,999 as the second number on the list. Then the 8 would slide to the right, until 299,999,999,998 would be the 12th number on the list.
Then 379,999,999,999 as #13. Next is 388,999,999,999 as # 14, and now the second 8 can slide to the right until 389,999,999,998 is #23
Then 397,999,999,999 as #24, followed by 398,899,999,999 through 398,999,999,998, accounting for #25 through #33. Then 399,799,999,999 for #34 and 399,889,999,999 through 399,899,999,998 as #35 through #42.
The next set of a 7-type and multiple double-8's takes us to #50 (399,989,999,998). The next, 57; next, 63; next 68; next 72; next 75; next 77 and next 78. That would be 399,999,999,988.
Then 469,999,999,999 is #79 and 478,999,999,999 through 479,999,999,998 account for #80 through #89.
487,999,999,999 is #90.
Then 488,899,999,999 through 488,999,999,998 account for #91 through #99. We're getting close.
489,889,999,999 is then the 100th.
But I've probably miscounted somewhere. I know I caught a couple of miscounts along the way, but probably missed some.
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Posted by Charlie
on 2015-10-08 21:24:23 |
I've probably miscounted but here goes.
