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Triangular Trial (Posted on 2010-07-20) Difficulty: 2 of 5
All the positive triangular numbers are written successively without commas or spaces resulting in this infinite string.

13610152128364555667891105120136153171..............

Determine the 2010th digit in the above pattern.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts one possible approach. | Comment 1 of 2

1. The formula for the nth triangular number is 1/2n(n+1).

2. for small n it's probably easiest to set out the terms:

 - 3 less than 10, for 3 digits

 - another 10 less than 100, for 20  digits

 - another 31 less than 1000, for 93 digits

3. Then 1/2n(n+1) = 10000 gives n about 141, and n = 141 is in fact 10011. So we get n = 45 to 140 inclusive for an additional 96 numbers of 4 digits each: 384 digits.

4. Using the same method, 1/2n(n+1) = 100000 gives n about 446, and n=447 is in fact 100128. So we get n = 141 to 446 inclusive for an additional 306 numbers of 5 digits each: 1530 digits and the running total is over 2010; to be exact, it is 2030.

5. So the sought digit is the last digit of the fourth number   before n = 446, namely n = 442, which is 97903: the last digit of which is a 3.

 

 


  Posted by broll on 2010-07-20 12:46:36
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