We can transform each of the options for 'n' years. (997)214+3≡(p−3)2n−1+3 (1003)215+6≡(p+3)2n+6 (1003)215−3≡(p+3)2n−3 (997)15−3≡(p−3)n−3 As per the condition, in one year, the population ' p′ becomes ′3+2p′ Putting the value of n=1 in each option, and checking to get 3+2p, we have (p−3)2n−1+3≡3≠3+2p