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Alphabet Repeat Pattern Poser (Posted on 2010-10-18) Difficulty: 2 of 5
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    
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Solution solution | Comment 1 of 3

The sum of the first 26 whole numbers is 27*26/2 = 351. Dividing 2010 by 351 leaves a remainder of 255, so we seek the 255th letter of the 351-letter sequence. We need the first n for which (n+1)*n/2 equals or exceeds 255.

Solving n^2 + n - 510 = 0 gives a positive value of

      -1 + sqrt(1 + 2040)
      -------------------   ~= 22.0887
               2
              

so that n = 22 fails to meet the mark, but n = 23 does. The letter is w.  As a check 24+25+26 = 75, while 351 - 255 = 96 and the difference between 75 and 96 is 21, which is more than covered by the 23 w's.

Bonus:

This time we need (n+1)*n/2 >= 2010

      -1 + sqrt(1 + 16080)
      --------------------   ~=  62.905     
               2

so we seek the 63rd letter of the repeated alphabet, or the (63-52)th, i.e., the 11th, which is k.              


  Posted by Charlie on 2010-10-18 14:22:19
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