One counterfeit coin problem which is either heavy/light: Using 3 weighs, I found that we can solve for up to "13" coins where as my professor says we can solve up to "14". I DO understand the concept of (3^n)/2 but I believe, its always [(3^n)/2]-1. I even tried with "2" weighs and could solve for only up to "4" coins instead of "5". I also tried with "13" coins and was able to succeed but failing with "14" coins.

Question

One counterfeit coin problem which is either heavy/light: Using 3 weighs, I found that we can solve for up to "13" coins where as my professor says we can solve up to "14". I DO understand the concept of (3^n)/2 but I believe, its always [(3^n)/2]-1.  I even tried with "2" weighs and could solve for only up to "4" coins instead of "5". I also tried with "13" coins and was able to succeed but failing with "14" coins.