abbcccddddeeeee... When the part with 11 "k‘’s end, the pattern continues with 12 "a‘’s,
13 "b‘’s and so on. What is the 2012th letter in this pattern?

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## Comments

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TopNewest'h'

As I understand the k's always appear in quantities multiple of 11.

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I think it's 'h'.

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It turns out that the answer is just h.

Proof:

define a "cycle" to be going from an a adjacent to a k to the next k adjacent to an a.

The first cycle has 66 letters. The next has 66+11x11=66+121. The next has 66+2x121. Et cetera

We see that the first five cycles have 66x5+121x10=1540 letters. This leaves 472 letters. We see that in the sixth cycle, a occurs 56 times.

This leaves 416 letters after the a's.

This leaves 416-57=359 letters after the b's.

This leaves 359-58=301 letters after the c's.

This leaves 301-59=242 letters after the d's.

182 after e's, 121 after f's, 59 after g's

H.

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Can you please give me a small idea why the digits in the second cycle are 66+121?

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each of the 11 letters appear exactly 11 more times than in the first cycle.

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h

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tttr

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trwrew

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tere

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H , done by geometric sum

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I think L

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There is no 'L'. It goes up to 'k' & starts from 'a' again.

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Sorry , i mean A

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h??

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