n=3599=3600-1 = 60*60 - 1 = 60^2 - 1^2 = (60+1)(60-1) = 61*59 = p*q
p=61
q=59
phi(n) = (p-1)(q-1) = (61-1)(59-1) = 60*58 = 3480
(e*d) % phi(n) = 1
31d % 3480 = 1
Divident = 31d
Quotient = k
Divisor = 3480
Remainder = 1
Divident = remainder + quotient*divisor
=> 31d = 1 + k*3480
=> d=(1+k*3480)/31
k,d are positive integers
By Brute Force, If I apply k=27, then I get (1+27*3480)/31 = (1+93960)/31 = 93961/31 = 3031
In Exam, depending on Time, u can apply either this or AMALJITH 's procedure