\( \begin{array}{ll}\begin{array}{ll}\text { a) } 36=2^{2} \times 3^{2} & 90\end{array}=3^{2} \times 2 \times 5 & \\ \text { m.c.m. }(36,90)= & \\ \text { b) } 35=5 \times 7 & =2^{2} \times 3 \times 5 \\ \text { m.c.m. }(35,60)= & \\ \text { c) } 112=2^{4} \times 7 & 72=2^{3} \times 3^{2} \\ \text { m.c.m. }(112,72)= & \\ \text { d) } 24=2^{3} \times 3 & 30=2 \times 3 \times 5\end{array} \quad 45=3^{2} \)

To find the least common multiple (m.c.m.) of numbers, we take the highest power of all prime factors present in any of the numbers. For \( m.c.m.(36,90) \): The prime factorizations are \( 36 = 2^2 \times 3^2 \) and \( 90 = 2^1 \times 3^2 \times 5^1 \). Thus, the m.c.m. is \( 2^2 \times 3^2 \times 5^1 = 180 \). For \( m.c.m.(35,60) \): \( 35 = 5^1 \times 7^1 \) and \( 60 = 2^2 \times 3^1 \times 5^1 \). So, the m.c.m. is \( 2^2 \times 3^1 \times 5^1 \times 7^1 = 420 \). For \( m.c.m.(112,72) \): Here, \( 112 = 2^4 \times 7^1 \) and \( 72 = 2^3 \times 3^2 \). The m.c.m. is \( 2^4 \times 3^2 \times 7^1 = 504 \). For \( m.c.m.(24,30) \): With \( 24 = 2^3 \times 3^1 \) and \( 30 = 2^1 \times 3^1 \times 5^1 \), we find \( 2^3 \times 3^1 \times 5^1 = 120 \). So the final m.c.m. results are: - \( m.c.m.(36,90) = 180 \) - \( m.c.m.(35,60) = 420 \) - \( m.c.m.(112,72) = 504 \) - \( m.c.m.(24,30) = 120 \)