Puzzle by IBM: PONDER!

Replies

• 

  shalini_goel14

  This one is too difficult for me. I do not have enough time to try so many permutations and combinations . I give up for it 😔.
• 

  Kaustubh Katdare

  Here we go -
  
  Instead of computing the thousands of digits of ponderthis, we can use the following inequality to quickly verify that the first six digits are, indeed, “ponder” (frac stands for the fractional part):
  
  frac(Log10(ponder)) ≤ frac(this*Log10 (ponder)) < frac(Log10 (ponder+1))
  
  Out of the all permutations, only three match the condition:
  ponder=504271 this=6983
  ponder=813760 this=2459
  ponder=835402 this=7619
  
  Here comes the interesting part – although the expression gives three different results, in all three cases, the first four digits, translated back to letters, are “oded”.
  
  Source: #-Link-Snipped-#