We will discuss here about the rules of divisibility tests by 9 and 11 with the help of different types of problems.

**1.** What least positive integral value must be given to * so that the number 7654*21 is divisible by 9?

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

Sum of known digits of 7654*21 is 25. The no. Just greater than 25 which is divisible by 9 is 27.

Now, 25 + (*) = 27

Therefore, * = 2

Answer: (b)

**Note:** Sum of digits when divisible by 9, then the
no. is divisible by 9.

**2. **Which of the
following numbers is exactly divisible by ninety-nine?

(a) 114345

(b) 3572404

(c) 135792

(d) 913464

Solution:

Co-prime factors of 99 are 9 and 11.

114345 is divisible by 99 because sum of digits is 18 and difference of (5 + 3 + 1) - (4 + 4 + 1) = 0

Therefore, required number is 114345.

Answer: (a)

**Note:** The difference of sums of the digits in odd
and even places is zero or multiple of 11, then the no. is divisible by 11.

**3.** 4\(^{91}\) + 4\(^{92}\) +
4\(^{93}\) + 4\(^{94}\) is divisible by

(a) 17

(b) 13

(c) 11

(d) 3

Solution:

4\(^{91}\) + 4\(^{92}\) + 4\(^{93}\) + 4\(^{94}\)

= 4\(^{91}\)(4\(^{0}\) + 4\(^{1}\) + 4\(^{2}\) + 4\(^{3}\))

= 4\(^{91}\)(1 + 4 + 16 + 64)

= 4\(^{91}\) × 85

= 4\(^{91}\) × 5 × 17, which is divisible by 17

Therefore, the required number is 17

Answer: (a)

**4.** The digits
indicated by ⨂ in 3422213⨂⨂ so that this number is divisible by ninety-nine, are:

(a) 1, 9

(b) 3, 7

(c) 4, 6

(d) 5, 5

Solution:

Co-prime factors of 99 are 9 and 11. Sum of the digits of 3422213xy is (17 + x + y)

According to the given options,

x + y = 10

And, (3 + 2 + 2 + 3 + y) - (4 + 2 + 1 + x) = 11

Or, 10 + y - 7 - x = 11

Or, y - x = 8

Now, x + y = 10 and y - x = 8

Therefore, x = 1 and y =9

Thus, the required numbers are 1, 9

Answer: (a)

**5.** The number (10\(^{25}\) - 7) is divisible by

(a) 3

(b) 7

(c) 11

(d) 13

Solution:

The number (10\(^{25}\) - 7) is divisible by 3.

Answer: (a)

**Note:** (10\(^{n}\) - 7) is always divisible by 3, for all
values of n

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