Answer

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**Hint:**

Here, we are given some numbers and we are required to check whether they are divisible by 11 or not. We will use the divisibility rule of 11 and apply it in every number to check which number is divisible by 11 and which one is not.

**Complete step by step solution:**

The divisibility rule of 11 states that the difference between the sum of digits at odd places and even places should be either 0 or divisible by 11.

Sum of digits at odd places \[ - \] Sum of digits at even places \[ = \] 0 or Number divisible by 11.

Hence, this actually means that if we add and subtract the alternate digits then we should get either 0 or a factor of 11 to ensure that the number is divisible by 11.

Hence, applying this rule to all the given parts, we get,

a) 786764

Here the digits 7,6 and 6 are at odd places and 8, 7 and 4 are at even places.

Adding and subtracting the terms, we get

\[\left( {7 + 6 + 6} \right) - \left( {8 + 7 + 4} \right) = 19 - 19 = 0\]

As the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

b) 536393

Here the digits 5, 6 and 9 are at odd places and 3, 3 and 3 are at even places.

Adding and subtracting the terms, we get

\[\left( {5 + 6 + 9} \right) - \left( {3 + 3 + 3} \right) = 20 - 9 = 11\]

We know that 11 is a factor of itself. So, we can say that the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

c) 110011

Here the digits 1, 0 and 1 are at odd places and 1, 0 and 1 are at even places.

Adding and subtracting the terms, we get

\[\left( {1 + 0 + 1} \right) - \left( {1 + 0 + 1} \right) = 2 - 2 = 0\]

As the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

d) 1210121

Here the digits 1, 1, 1 and 1 are at odd places and 2, 0 and 2 are at even places.

Adding and subtracting the terms, we get

\[\left( {1 + 1 + 1 + 1} \right) - \left( {2 + 2 + 0} \right) = 4 - 4 = 0\]

As the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

e) 758043

Here the digits 7, 8 and 4 are at odd places and 5, 0 and 3 are at even places.

Adding and subtracting the terms, we get

\[\left( {7 + 8 + 4} \right) - \left( {5 + 0 + 3} \right) = 19 - 8 = 11\]

As the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

f) 8338472

Here the digits 8, 3, 4 and 2 are at odd places and 3, 8 and 7 are at even places.

Adding and subtracting the terms, we get

\[\left( {8 + 3 + 4 + 2} \right) - \left( {3 + 8 + 7} \right) = 17 - 18 = - 1\]

Since, \[ - 1\] is not a factor of 11. Therefore this number is not divisible by 11.

g) 54678

Here the digits 5, 6 and 8 are at odd places and 4 and 7 are at even places.

Adding and subtracting the terms, we get

\[\left( {5 + 4 + 8} \right) - \left( {4 + 7} \right) = 17 - 11 = 6\]

Since, 6 is not a factor of 11.Therefore this number is not divisible by 11.

h) 13431

Here the digits 1, 4 and 1 are at odd places and 3 and 3 are at even places.

Adding and subtracting the terms, we get

\[\left( {1 + 4 + 1} \right) - \left( {3 + 3} \right) = 6 - 6 = 0\]

As the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

i) 423423

Here the digits 4, 3 and 2 are at odd places and 2, 4 and 3 are at even places.

Adding and subtracting the terms, we get

\[\left( {4 + 3 + 2} \right) - \left( {2 + 4 + 3} \right) = 0\]

As the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

j) 168861

Here the digits 1, 8 and 6 are at odd places and 6, 8 and 1 are at even places.

Adding and subtracting the terms, we get

\[\left( {1 + 8 + 6} \right) - \left( {6 + 8 + 1} \right) = 0\]

As the number satisfies the condition of divisibility of 11, therefore it is divisible by 11.

**Therefore, (a), (b), (c) , (d), (e), (h), (i), (j) are divisible by 11.**

And, (f) and (g) are not divisible by 11.

And, (f) and (g) are not divisible by 11.

**Note:**

We use Divisibility rules as they are shortcut methods which help us to identify that a given number is divisible by another without even dividing them. With the help of different divisibility rules, we can find the prime factorisation of large numbers in less time. Also, in this question, if while applying the alternate method, we get \[ - 11\] as our answer, then also the number is divisible by 11. We look for the factors of 11 and they are not affected by their signs.

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