Directions Read the given passage carefully and find the answers of following questions. LCM of two numbers A and B is 60. B is a 2-digit number which leaves remainder as 2 when divided by 6. It is also given B>A.Read the instructions carefullyWhat is remainder when B is divided by 2?

Sure, let's break down the problem step by step to find the remainder when B is divided by 2.

1. Given Information:

   • The Least Common Multiple (LCM) of two numbers A and B is 60.
   • B is a 2-digit number.
   • B leaves a remainder of 2 when divided by 6.
   • B is greater than A (B > A).

2. Finding B:

   • Since B leaves a remainder of 2 when divided by 6, we can write B in the form: $B = 6k + 2$ where $k$ is an integer.

3. Identifying Possible Values of B:

   • B must be a 2-digit number, so we need to find values of $k$ such that $10 \leq 6k + 2 \leq 99$.
   • Solving for $k$: $10 \leq 6k + 2 \leq 99$ Subtract 2 from all parts of the inequality: $8 \leq 6k \leq 97$ Divide by 6: $\frac{8}{6} \leq k \leq \frac{97}{6}$ Simplify: $1.33 \leq k \leq 16.17$ Since $k$ must be an integer, the possible values of $k$ are 2 through 16.

4. Calculating Possible Values of B:

   • For $k = 2$: $B = 6(2) + 2 = 14$
   • For $k = 3$: $B = 6(3) + 2 = 20$
   • Continue this process up to $k = 16$: $B = 6(16) + 2 = 98$
   • The possible values of B are: 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 80, 86, 92, 98.

5. Finding the Remainder When B is Divided by 2:

   • All the possible values of B are even numbers (since they are in the form $6k + 2$).
   • When an even number is divided by 2, the remainder is always 0.

Conclusion: The remainder when B is divided by 2 is 0.