Question

Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by $11$: $\left(a\right)92\_\_\_\_\_\_389\left(b\right)8\_\_\_\_\_\_9484$

Answer 1

Determine the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by $11$

The concept of divisibly by $11$implies When a number has an even number of digits, add the first digit and remove the final digit from of the remainder of the number.

Calculate the value for (A): $92\_\_389$

Let the blank value is $x$

Calculate the sum of its digits at odd places:

$9+3+2=14$

Calculate the sum of its digits at even places:

$8+x+9=17+x$

Calculate the difference:

$17+x–14=3+x$

Calculate the difference that is $0$ or a multiple of $11$, the integer is divisible by $11$.

$$\begin{gathered} 3+x=0 \\ \Rightarrow x=-3 \end{gathered}$$

Similarly, it cannot be negative. Using the nearest multiple of $11$, which is close to $3$, It is $11$ and close to $3$.

Now,

$$\begin{gathered} 3+x=11 \\ \Rightarrow x=11–3 \\ \Rightarrow x=8 \end{gathered}$$

So, the required digit is 8.

Calculate the value for (B): $8\_\_9484$

Let the blank value is $x$

Calculate the sum of its digits at odd places:

$4+4+x=8+x$

Calculate the sum of its digits at even places:

$8+9+8=25$

Calculate the difference:

$25–(8+x)=17–x$

Calculate the difference that is $0$ or a multiple of $11$, the integer is divisible by $11$.

$$\begin{gathered} 17–x=0 \\ \Rightarrow x=17(notpossible) \end{gathered}$$

Now consider a multiple of $11$.

given that$11$

$$\begin{gathered} 17–x=11 \\ \Rightarrow x=17–11 \\ \Rightarrow x=6 \end{gathered}$$

So, the required digit is $6$.

Hence, the digit in the blank space of each of the following numbers so that the number formed is divisible by $11$ are $8$ and $6$ respectively.