Infinite String and Letter Puzzle

  • Thread starter K Sengupta
  • Start date
  • #1
111
0
The string abbcccddddeeeee… continuously repeats such that after the final z, the letters abbcccddddeeeee… begin again.

What will be the 3000th letter in the pattern?
 
Last edited:

Answers and Replies

  • #2
328
0
T
I think
 
  • #3
LAF
31
0
I Think it´s:
R
 
  • #4
DavidSnider
Gold Member
502
143
Here's my guess:

Code:
var alpha = "abcdefghijklmnopqrstuvwxyz".split("");
var a = [];
var i=0;
var j=0;
var k=0;

while(i<=3000) {
    for(j=0;j<k;j++) {
        if(i<3000) {
            a.push(alpha[k-1]);
        }
        i++;
    }
    k++;
    if(k > alpha.length) k = 0;
}


a[a.length-1];

t
 
Last edited:
  • #5
668
3
Code:
#!perl
for(A..Z) {$s.=$_ x++$i;}
$t=$s.($s x int(3000/length $s));
print substr($t,2999,1)."\n";

Or, more appropriately:

There are 26 letters in the alphabet, hence the string from the first A to the final Z will be ((1+26)/2)*26 characters long (351 characters). 3000 modulo 351 is 192. Hence, the 3000th character will be the same as the 192nd character. You can find this by finding when N choose 2 is equal to or greater than 192, which happens to be at N=20 (20 choose 2 being 210). And the 20th letter of the alphabet is T.

And the next question-- what happens if instead, after the final Z, the next A repeats 27 times, then B repeats 28 times and so forth?

DaveE
 
Last edited:
  • #6
12
0
Code:
#!perl
for(A..Z) {$s.=$_ x++$i;}
$t=$s.($s x int(3000/length $s));
print substr($t,2999,1)."\n";

Or, more appropriately:

There are 26 letters in the alphabet, hence the string from the first A to the final Z will be ((1+26)/2)*26 characters long (351 characters). 3000 modulo 351 is 192. Hence, the 3000th character will be the same as the 192nd character. You can find this by finding when N choose 2 is equal to or greater than 192, which happens to be at N=20 (20 choose 2 being 210). And the 20th letter of the alphabet is T.

And the next question-- what happens if instead, after the final Z, the next A repeats 27 times, then B repeats 28 times and so forth?

DaveE

I used basic math for this.
x(x+1) = 6000 and x = int -> x^2 < 6000 < x^2 + x ; x= 77. My answer is Y.
 
  • #7
328
0
I used basic math for this.
x(x+1) = 6000 and x = int -> x^2 < 6000 < x^2 + x ; x= 77. My answer is Y.

There appears to be something wrong with your basic math.

26*(26+1)/2 = 351
3000 / 351 = 8.547...
3000-351(8) = 192

192-1-2-3-4...-19 = 2

So T is the 3000th letter.
 

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