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 Alphabet Repeat Pattern Poser (Posted on 2010-10-18)
The string abbcccddddeeeee.... continuously repeats - with a repeating once, b repeating twice and so on, such that after the final z, the letters abbcccddddeeeee.... begin again.

What will be the 2010th letter in the above pattern?

As a bonus, what will be the 2010th letter in the above pattern, if instead - after the final z, the next a repeats 27 times, then b repeats 28 times and so forth?

 See The Solution Submitted by K Sengupta No Rating

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The cycle's length is   ( 1+26)*26/2=351

2010 mod 351= 255

lowest INTEGER n to satisfy n*(n+1)>2*255    IS 23

23th letter is w.

For bonus question -same reasoning:

lowest INTEGER n to satisfy n*(n+1)>2*2010    IS 63

63 mod 26=63-52=11

11th letter is k.

wHO  kNOWS?

 Posted by Ady TZIDON on 2010-10-18 14:53:27

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