Question

In each of the following, fill in the blanks with the smallest digit to make it divisible by 11 (a)1083_2 (b)620_53

Answer 1

SOLUTION

TO FILL IN THE BLANK

The smallest digit to make it divisible by 11

(a) 1083_2

(b) 620_53

CONCEPT TO BE IMPLEMENTED

DIVISIBILITY RULE FOR 11

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely

EVALUATION

(a) Here the given number = 1083_2

Let a be the number in blank place

So the number = 1083a2

Now the numbers in odd places 1 , 8 , a

The number in even places 0 , 3 , 2

Difference of the sum of alternative digits

= ( 1 + 8 + a ) - ( 0 + 3 + 2 )

= 4 + a

The smallest value of a for which 4 + a is divisible by 11 is 7

Hence the required number 108372

(b) Here the given number = 620_53

Let b be the number in blank place

So the number = 620b53

Now the numbers in odd places 6 , 0 , 5

The number in even places 2 , b , 3

Difference of the sum of alternative digits

= ( 6 + 0 + 5 ) - ( 2 + b + 3 )

= 6 - b

The smallest value of a for which 6 - b is divisible by 11 is 6

Hence the required number 620653

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ there are the same remainder three when 135 111 87 divided by the certain number find the largest possible value of this...

https://brainly.in/question/33996172

2. what number should be added to each of the numbers 8,15,18,30, such that resulting number are in proportion

https://brainly.in/question/13256357

Answer 2

Answer:

add in alternative =1+8+a=4a

0+3+2=5

4+a = 7 is divisible by 11

so 1083

add in alternative =6+0+5= 11

2+a+3=5a

5+a= 6 is divisible by 11

so 620653

please mark