Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:
(a) 92 _ 389 (b) 8 _ 9484

AcademicMathematicsNCERTClass 6

To do :

We have to find the digits in the blank spaces so that the numbers formed are divisible by 11.

Solution :

Divisibility rule of 11:

A positive integer N is divisible by $11$ if the difference of the alternating sum of digits of N is a multiple of $11$.

(a) Let the digit in the blank space be $x$ where $0 \leq x \leq 9$.

Sum of the odd digits $= 9 + 3 + 2$

$= 14$

Sum of the even digits $= 8 + x + 9$

$= x+17$

Difference $=x+17-14$

$=x+3$

Now, $x+3$ is divisible by 11.

If $x=9, x+3=9+3=12$ which is not divisible by 11.

If $x=8, x+3=8+3=11$ which is divisible by 11.

If $x=7, x+3 = 7+3 = 10$ which is not divisible by 11 and the same is the case for any value of $x<7$.

Hence, the required digit is 8.

(b) Let the digit in the blank space be $x$ where $0 \leq x \leq 9$.

Sum of the odd digits $= 4 + 4 + x$

$= x+8$

Sum of the even digits $= 8 + 9 + 8$

$= 25$

Difference $=25-(x+8)$

$=25-8-x$

$=17-x$

Now, $17-x$ is divisible by 11.

If $x=9, 17-x=17-9=8$ which is not divisible by 11.

If $x=8, 17-x=17-8=9$ which is not divisible by 11.

If $x=7, 17-x=17-7=10$ which is not divisible by 11.

If $x=6, 17-x=17-6=11$ which is divisible by 11.

If $x=5, 17-x = 17-5 = 12$ which is not divisible by 11 and the same is the case for any value of $x<5$.

Hence, the required digit is 6.

raja
Updated on 10-Oct-2022 13:30:33

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