(4)

$$\begin{gathered} HCF (26, 169) \times LCM (26, 169) = 26 \times 169 (\because LCM (a, b) \times HCF (a, b) = a \times b) \\ \Rightarrow 13 \times LCM (26, 169) = 26 \times 169 \\ \Rightarrow LCM (26, 169) = (26 \times 169) ÷ 13 = 338 \end{gathered}$$

(4)

$$\begin{gathered} We have, 5 = 1 \times 5; 15 = 3 \times 5 and 20 = 2 \times 2 \times 5 \\ \therefore LCM (5, 15, 20) = 3 \times 4 \times 5 = 60 \\ and HCF (5, 15, 20) = 5 \\ Therefore, ratio between the LCM and HCF = 60 ÷ 5 = 12 : 1 \end{gathered}$$

(4)

$$\begin{gathered} We have, a \times b = 1152 and HCF (a, b) = 12 \\ We know that LCM (a, b) \times HCF (a, b) = a \times b \\ \Rightarrow LCM (a, b) \times 12 = 1152 \\ \Rightarrow LCM (a, b) = 1152 ÷ 12 = 96 \end{gathered}$$

(3)  

For any number to be divisible by all the numbers from 1 to 10, we need LCM of the numbers from 1 to 10. So, it must have maximum powers of each prime number that can be obtained in prime factorisation of numbers from 1 to 10.

$$\begin{gathered} \therefore 1 = 1 \\ 2 = 2¹ \\ 3 = 3¹ \\ 4 = 2² \\ 5 = 5¹ \\ 6 = 2¹ \times 3¹ \\ 7 = 7¹ \\ 8 = 2³ \\ 9 = 3² \\ 10 = 2¹ \times 5¹ \\ So, required least number = 2³ \times 3² \times 5¹ \times 7¹ = 8 \times 9 \times 5 \times 7 = 2520 \end{gathered}$$