17d % 3120 = 1
Divident = 17d
Quotient = k
Divisor = 3120
Remainder = 1
Divident = remainder + quotient*divisor
=> 17d = 1 + k*3120
=> d=(1+k*3120)/17
k,d are positive integers
By Brute Force, If I apply k=15, then I get (1+15*3120)/17 = (1+46800)/17 = 46801/17 = 2753
In Exam, depending on Time, u can apply either this or Mk Utkarsh's procedure