a) Write the smallest digit and the greatest digit in the blank space of each ofthe following numbers so that the number formed is divisible by 6.i. 2567_4ii. 7_614

To determine the smallest and greatest digits that can be placed in the blank spaces to make the numbers divisible by 6, we need to remember that a number is divisible by 6 if and only if it is divisible by both 2 and 3.

1. Divisibility by 2: A number is divisible by 2 if its last digit is even.
2. Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

Let's apply these rules to each number:

i. 2567_4

1. Divisibility by 2: The last digit is 4, which is even, so the number will be divisible by 2 regardless of the blank space digit.

2. Divisibility by 3: We need to find the sum of the digits and ensure it is divisible by 3.

   • Current sum of the digits: 2 + 5 + 6 + 7 + 4 = 24
   • Let the blank space be represented by $x$.
   • New sum: 24 + $x$

   We need $24 + x$ to be divisible by 3. Since 24 is already divisible by 3, $x$ must also be divisible by 3.

   The smallest digit that is divisible by 3 is 0. The greatest digit that is divisible by 3 is 9.

   Therefore, the smallest digit is 0 and the greatest digit is 9.

ii. 7_614

1. Divisibility by 2: The last digit is 4, which is even, so the number will be divisible by 2 regardless of the blank space digit.

2. Divisibility by 3: We need to find the sum of the digits and ensure it is divisible by 3.

   • Current sum of the digits: 7 + 6 + 1 + 4 = 18
   • Let the blank space be represented by $x$.
   • New sum: 18 + $x$

   We need $18 + x$ to be divisible by 3. Since 18 is already divisible by 3, $x$ must also be divisible by 3.

   The smallest digit that is divisible by 3 is 0. The greatest digit that is divisible by 3 is 9.

   Therefore, the smallest digit is 0 and the greatest digit is 9.

Summary:

• For 2567_4, the smallest digit is 0 and the greatest digit is 9.
• For 7_614, the smallest digit is 0 and the greatest digit is 9.