diff --git a/docs/Chap30/Problems/30-1.md b/docs/Chap30/Problems/30-1.md
index f431482b62..8054f72ad5 100755
--- a/docs/Chap30/Problems/30-1.md
+++ b/docs/Chap30/Problems/30-1.md
@@ -4,4 +4,134 @@
 >
 > **c.** Show how to multiply two $n$-bit integers in $O(n^{\lg 3})$ steps, where each step operates on at most a constant number of $1$-bit values.
 
-(Omit!)
+**a.** Since $(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd$
+
+We can compute it using three multiplications:
+
+$$
+\begin{aligned}
+M_1 &= ac \\\\
+M_2 &= bd \\\\
+M_3 &= (a + b)(c + d)
+\end{aligned}
+$$
+
+Then:
+$$
+ad + bc = (a + b)(c + d) - ac - bd = M_3 - M_1 - M_2
+$$
+
+Thus:
+$$
+(ax + b)(cx + d) = M_1x^2 + (M_3 - M_1 - M_2)x + M_2
+$$
+
+**b.**
+
+Let's denote:
+
+$$
+A(x) = \sum_{i=0}^{n-1} a_i x^i, \quad
+B(x) = \sum_{i=0}^{n-1} b_i x^i
+$$
+
+1. Split into high and low halves
+
+Let $m = \lceil n/2 \rceil$.
+Split each polynomial into two parts:
+
+$$
+\begin{aligned}
+A(x) &= A_{\text{low}}(x) + x^m A_{\text{high}}(x) \\\\
+B(x) &= B_{\text{low}}(x) + x^m B_{\text{high}}(x)
+\end{aligned}
+$$
+
+Compute three recursive products:
+$$
+\begin{aligned}
+P_1 &= A_{\text{low}} \cdot B_{\text{low}} \\\\
+P_2 &= A_{\text{high}} \cdot B_{\text{high}} \\\\
+P_3 &= (A_{\text{low}} + A_{\text{high}})(B_{\text{low}} + B_{\text{high}})
+\end{aligned}
+$$
+
+Combine results:
+$$
+A(x)B(x) = P_1 + x^m (P_3 - P_1 - P_2) + x^{2m} P_2
+$$
+
+Each recursive call is on size at most $\lceil n/2 \rceil$, and the combining step takes $\Theta(n)$ time. The recurrence is:
+$$
+T(n) = 3T(\lceil n/2 \rceil) + \Theta(n) \implies T(n) = \Theta(n^{\log_2 3})
+$$
+
+2. Split by even and odd indices
+
+Instead of splitting by halves, we split by index parity:
+$$
+\begin{aligned}
+A(x) &= A_{\text{even}}(x^2) + xA_{\text{odd}}(x^2) \\\\
+B(x) &= B_{\text{even}}(x^2) + xB_{\text{odd}}(x^2)
+\end{aligned}
+$$
+
+where:
+
+$$
+A_{\text{even}}(x) = a_0 + a_2 x + a_4 x^2 + ..., \quad
+A_{\text{odd}}(x) = a_1 + a_3 x + a_5 x^2 + ...
+$$
+
+Apply the same trick:
+$$
+\begin{aligned}
+Q_1 &= A_{\text{even}} \cdot B_{\text{even}} \\\\
+Q_2 &= A_{\text{odd}} \cdot B_{\text{odd}} \\\\
+Q_3 &= (A_{\text{even}} + A_{\text{odd}})(B_{\text{even}} + B_{\text{odd}})
+\end{aligned}
+$$
+
+Then:
+$$
+A(x)B(x) = Q_1(x^2) + x(Q_3(x^2) - Q_1(x^2) - Q_2(x^2)) + x^2 Q_2(x^2)
+$$
+
+Again,
+$$
+T(n) = 3T(n/2) + \Theta(n) = \Theta(n^{\log_2 3})
+$$
+
+**c.** We apply the same structure to integers, interpreting them as polynomials in base 2.
+Let two $n$-bit integers be:
+
+$$
+U = u_{\text{high}} 2^m + u_{\text{low}}, \quad
+V = v_{\text{high}} 2^m + v_{\text{low}}, \quad
+m = \lceil n/2 \rceil
+$$
+
+Then:
+$$
+UV = u_{\text{high}}v_{\text{high}} 2^{2m} + (u_{\text{high}}v_{\text{low}} + u_{\text{low}}v_{\text{high}})2^m + u_{\text{low}}v_{\text{low}}
+$$
+
+Compute using three multiplications:
+
+$$
+\begin{aligned}
+A &= u_{\text{low}} \cdot v_{\text{low}}, \
+C &= u_{\text{high}} \cdot v_{\text{high}}, \
+B &= (u_{\text{low}} + u_{\text{high}})(v_{\text{low}} + v_{\text{high}})
+\end{aligned}
+$$
+
+Then:
+$$
+UV = C \cdot 2^{2m} + (B - A - C) \cdot 2^m + A
+$$
+
+The recurrence:
+$$
+T(n) = 3T(n/2) + \Theta(n) \implies T(n) = \Theta(n^{\log_2 3})
+$$