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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
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.
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