Question

Number of ordered triplets (a, b, c) of positive integers less than 10, for which the product abc is divisible by 20, are –

• A. 48
• B. 102
• C. 54
• D. 51

Answer 1

Answer: (B)

102

Answer 2

Sol. Case-I

a, b, c having exactly one 5 with

(i) 2 even digits (different or same)

$4C_{2}.\begin{matrix} \begin{matrix} 3 \end{matrix} + \end{matrix}^{4}C_{1}.\begin{matrix} \frac{\begin{matrix} 3 \end{matrix}}{\begin{matrix} 2 \end{matrix}} = 48 \end{matrix}$

(ii) one digit divisible by 4 and one odd

\begin{matrix} 3 \end{matrix} = 48 \end{matrix}$$ **Case-II** a, b, c having exactly two 5 with : one digit divisible by 4, 2 .$\begin{matrix} \frac{\begin{matrix} 3 \end{matrix}}{\begin{matrix} 2 \end{matrix}} \end{matrix}$= 6 Hence total number of ways = 102