Answer :
Alright! Let's solve each part step-by-step to find the smallest digit that can replace the asterisk () to make the given number divisible by the mentioned divisor.
### (i) [tex]\( 65 5 \)[/tex]: Make divisible by [tex]\( 5 \)[/tex]
For a number to be divisible by 5, its last digit must be either 0 or 5.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 6505 \)[/tex] - Not divisible by 5.
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 6515 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 6525 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 3 \)[/tex]: [tex]\( 6535 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 6545 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 6500 \)[/tex] - Divisible by 5.
The smallest digit replacing `` in [tex]\( 65 5 \)[/tex] is [tex]\( \mathbf{0} \)[/tex].
### (ii) [tex]\( 2 135 \)[/tex]: Make divisible by [tex]\( 2 \)[/tex]
For a number to be divisible by 2, its last digit must be even.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 2035 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 2235 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 2435 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 6 \)[/tex]: [tex]\( 2635 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 8 \)[/tex]: [tex]\( 2835 \)[/tex] - Not divisible by 2.
None of the replacements provide a divisible number by 2. The closest answer here is an incorrect number unless there's a typo.
### (iii) [tex]\( 6702 \)[/tex]: Make divisible by [tex]\( 3 \)[/tex]
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
- Sum of known digits: [tex]\( 6 + 7 + 0 + 2 = 15 \)[/tex]
- Sum must still be divisible by 3; possible values for replacing [tex]\( \)[/tex] could be:
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 67020 \)[/tex] - Sum is 15, Divisible by 3.
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 67021 \)[/tex], sum is 16 - Not divisible by 3.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 67022 \)[/tex], sum is 17 - Not divisible by 3.
Thus, smallest digit is [tex]\( \mathbf{0} \)[/tex].
### (iv) [tex]\( 91 67 \)[/tex]: Make divisible by [tex]\( 7 \)[/tex]
Using digit replacement:
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 91067 \)[/tex] - Not divisible by 7 (q = 13009, r = 4)
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 91167 \)[/tex] - Not divisible by 7 (q = 13023, r = 6)
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 91267 \)[/tex] - Not divisible by 7 (q = 13037, r = 2)
- If `` is replaced by [tex]\( 3 \)[/tex]: [tex]\( 91367 \)[/tex] - Divisible by 7 (q = 13052, r = 0)
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 91467 \)[/tex] - Not divisible by 7 (q = 13066, r = 3)
So, the smallest digit is [tex]\( \mathbf{3} \)[/tex].
### (v) [tex]\( 6678 1 \)[/tex]: Make divisible by [tex]\( 11 \)[/tex]
For a number to be divisible by 11, the difference between the sum of digits in odd positions and the even positions should be a multiple of 11.
- Known digits [tex]\( 6,6,7,8 \)[/tex] and 1
- [tex]\(6+2+0+1=9-8=1\)[/tex]
- [tex]\(6-6+-10\)[/tex]
The sum is already a multiple of 11.
Thus the answer is [tex]\( \mathbf{0} \)[/tex].
### (vi) [tex]\( 835 86 \)[/tex]: Make divisible by [tex]\( 2 \)[/tex]
For a number to be divisible by 2, its last digit must be even.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 835086 \)[/tex] - Divisible by 2.
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 835186 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 835286 \)[/tex] - Divisible by 2.
- If `` is replaced by [tex]\( 3 \)[/tex]: [tex]\( 835386 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 835486 \)[/tex] - Divisible by 2.
The smallest value here smallest digit is [tex]\( \mathbf{0} \)[/tex].
Therefore, the answers to the parts are:
(i) [tex]\( 0 \)[/tex]
(ii) Not possible.
(iii) [tex]\( 0 \)[/tex]
(iv) [tex]\( 3 \)[/tex]
(v) [tex]\( 0 \)[/tex]
(vi) [tex]\( 0 \)[/tex]
### (i) [tex]\( 65 5 \)[/tex]: Make divisible by [tex]\( 5 \)[/tex]
For a number to be divisible by 5, its last digit must be either 0 or 5.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 6505 \)[/tex] - Not divisible by 5.
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 6515 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 6525 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 3 \)[/tex]: [tex]\( 6535 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 6545 \)[/tex] - Divisible by 5.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 6500 \)[/tex] - Divisible by 5.
The smallest digit replacing `` in [tex]\( 65 5 \)[/tex] is [tex]\( \mathbf{0} \)[/tex].
### (ii) [tex]\( 2 135 \)[/tex]: Make divisible by [tex]\( 2 \)[/tex]
For a number to be divisible by 2, its last digit must be even.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 2035 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 2235 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 2435 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 6 \)[/tex]: [tex]\( 2635 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 8 \)[/tex]: [tex]\( 2835 \)[/tex] - Not divisible by 2.
None of the replacements provide a divisible number by 2. The closest answer here is an incorrect number unless there's a typo.
### (iii) [tex]\( 6702 \)[/tex]: Make divisible by [tex]\( 3 \)[/tex]
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
- Sum of known digits: [tex]\( 6 + 7 + 0 + 2 = 15 \)[/tex]
- Sum must still be divisible by 3; possible values for replacing [tex]\( \)[/tex] could be:
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 67020 \)[/tex] - Sum is 15, Divisible by 3.
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 67021 \)[/tex], sum is 16 - Not divisible by 3.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 67022 \)[/tex], sum is 17 - Not divisible by 3.
Thus, smallest digit is [tex]\( \mathbf{0} \)[/tex].
### (iv) [tex]\( 91 67 \)[/tex]: Make divisible by [tex]\( 7 \)[/tex]
Using digit replacement:
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 91067 \)[/tex] - Not divisible by 7 (q = 13009, r = 4)
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 91167 \)[/tex] - Not divisible by 7 (q = 13023, r = 6)
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 91267 \)[/tex] - Not divisible by 7 (q = 13037, r = 2)
- If `` is replaced by [tex]\( 3 \)[/tex]: [tex]\( 91367 \)[/tex] - Divisible by 7 (q = 13052, r = 0)
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 91467 \)[/tex] - Not divisible by 7 (q = 13066, r = 3)
So, the smallest digit is [tex]\( \mathbf{3} \)[/tex].
### (v) [tex]\( 6678 1 \)[/tex]: Make divisible by [tex]\( 11 \)[/tex]
For a number to be divisible by 11, the difference between the sum of digits in odd positions and the even positions should be a multiple of 11.
- Known digits [tex]\( 6,6,7,8 \)[/tex] and 1
- [tex]\(6+2+0+1=9-8=1\)[/tex]
- [tex]\(6-6+-10\)[/tex]
The sum is already a multiple of 11.
Thus the answer is [tex]\( \mathbf{0} \)[/tex].
### (vi) [tex]\( 835 86 \)[/tex]: Make divisible by [tex]\( 2 \)[/tex]
For a number to be divisible by 2, its last digit must be even.
- If `` is replaced by [tex]\( 0 \)[/tex]: [tex]\( 835086 \)[/tex] - Divisible by 2.
- If `` is replaced by [tex]\( 1 \)[/tex]: [tex]\( 835186 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 2 \)[/tex]: [tex]\( 835286 \)[/tex] - Divisible by 2.
- If `` is replaced by [tex]\( 3 \)[/tex]: [tex]\( 835386 \)[/tex] - Not divisible by 2.
- If `` is replaced by [tex]\( 4 \)[/tex]: [tex]\( 835486 \)[/tex] - Divisible by 2.
The smallest value here smallest digit is [tex]\( \mathbf{0} \)[/tex].
Therefore, the answers to the parts are:
(i) [tex]\( 0 \)[/tex]
(ii) Not possible.
(iii) [tex]\( 0 \)[/tex]
(iv) [tex]\( 3 \)[/tex]
(v) [tex]\( 0 \)[/tex]
(vi) [tex]\( 0 \)[/tex]
