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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've probably miscounted but here goes. | Comment 1 of 3
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.


  Posted by Charlie on 2015-10-08 21:24:23
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