Answer: $81$ Solution: Lemma 1: $100|11^k-10k-1$ for every integer $k$ and $0 \leq k \leq 9$ Proof: This is true for $k=0$ since $11^k-10k-1=0$ Assume that this is true for $k=n$, we will prove that this is true for $k=n+1$, meaning: Prove that: If $100|11^n-10n-1$ then $100|11^{n+1}-10(n+1)-1$ $11^{n+1}-10(n+1)-1=11 \cdot 11^n-10n-11=11(11^n-10n-1)+100n$ which is a multiple of 100. Lemma 1 proven. Using Lemma 1, we obtain: $11^{10} =11^9 \cdot 11 \equiv 91 \cdot 11 \equiv (-9) \cdot 11 = -99 \equiv 01 \mod 100$ Lemma 2: $11^{10n+k} \equiv 10k+1 \mod 100$ for every integer $k$ and $0 \leq k \leq 9$ and for every positive integer $n$ Proof: $11^{10n+k} = (11^{10})^n \cdot 11^k \equiv 01 \cdot 11^k=11^k \cdot 10k+1 \mod 100$ Using Lemma 2 for $n=1990$ and $k=8$, we obtain: $11^{1998} \equiv 10 \cdot 8+1=81 \mod 100$. So the last 2 digits of $11^{1998}$ is $81$.