All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Like CABACB (Posted on 2017-10-08)
Let q(ltr) represent a numerical value of the position in English alphabet of the letter ltr (ltr being one of the letters in ABC}.
So q(A)=1 , q(B)=2 , …q(K)=11 , …q(Z)=26.

Let's take 2k letters, 2 of each of the first k letters of the English ABC.
Now create a string in which between each of the identical letters say each of ltr there are q(ltr) other letters. The title string and its reversal - qualify as such concatenations for k=3.

Not for every k such solutions exist (try e.g. for k=5...).

It is impossible to create such a string for k=26 (either trust me or explain why), but adding one extra symbol, say #,- which will be counted as a letter with q(#)=0 - it can be done (try for k=5 as well)

Solve for k=26 adding "#" to 2*"ABC string".

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( You must be logged in to post comments.)
 Subject Author Date A solution for k=26 Jer 2017-10-09 13:25:55 possible k's. k=26 not solved. Jer 2017-10-09 11:34:59

 Search: Search body:
Forums (4)