Given: 112 = q × 6r

To find: The possible values of r

Solution:

We are given the equation 112 = q × 6r.

To find the possible values of r, we need to consider the factors of 112 and see which ones satisfy the given equation when multiplied by 6.

Step 1: Find the factors of 112:

The factors of 112 are:
1, 2, 4, 7, 8, 14, 16, 28, 56, 112

Step 2: Substitute the factors into the equation:

Let's substitute the factors into the equation and check which ones satisfy the equation:

For r = 1:
112 = q × 6 × 1
112 = 6q
q = 112/6 = 18.67 (not an integer)

For r = 2:
112 = q × 6 × 2
112 = 12q
q = 112/12 = 9.33 (not an integer)

For r = 4:
112 = q × 6 × 4
112 = 24q
q = 112/24 = 4.67 (not an integer)

For r = 7:
112 = q × 6 × 7
112 = 42q
q = 112/42 = 2.67 (not an integer)

For r = 8:
112 = q × 6 × 8
112 = 48q
q = 112/48 = 2.33 (not an integer)

For r = 14:
112 = q × 6 × 14
112 = 84q
q = 112/84 = 1.33 (not an integer)

For r = 16:
112 = q × 6 × 16
112 = 96q
q = 112/96 = 1.17 (not an integer)

For r = 28:
112 = q × 6 × 28
112 = 168q
q = 112/168 = 0.67 (not an integer)

For r = 56:
112 = q × 6 × 56
112 = 336q
q = 112/336 = 0.33 (not an integer)

For r = 112:
112 = q × 6 × 112
112 = 672q
q = 112/672 = 0.17 (not an integer)

Step 3: Identify the possible values of r:

From the above calculations, we can see that none of the factors of 112 satisfy the equation when multiplied by 6 to give an integer value for q.

Therefore, there are no possible values of r for the given equation.

Answer: The possible values of r are none.

This can be answered by using Euclid's division lemma: Let 'a' be a +ve integer and 'b' = 6, such that "a=bq+r" where 0