 # Divisibility Models

A rule that is in short and the best way to determine whether the given integer is divisible by a divisor that is fixed without carrying out the method of division is called the divisibility rule. It is usually done by inspecting the digits.

Divisibility Model:

1] A number is divisible by 3: If the sum of the digits is divisible by 3.

2] A number is divisible by 4: If its last 2 digits are divisible by 4.

3] A number is divisible by 6: If it is divisible by 2 and 3.

4] A number is divisible by 7: [(2 * units-place digits) – (number formed by other digits)].

5] A number is divisible by 8: If its last 3 digits are divisible by 8.

6] A number is divisible by 9: If the sum of its digits is divisible by 9.

7] A number is divisible by 11: (sum of digits in the odd places) – (sum of digits in even place) is divisible by 11.

Example 1: For all positive integral values of n, 32n − 2n + 1 is divisible by

A) 2

B) 4

C) 8

D) 12

Solution: A

Putting n = 2 in 32n − 2n + 1 then, 32×2 − 2 × 2 + 1 = 81 − 4 + 1 = 78, which is divisible by 2.

Example 2: If n ∈ N, then 72n + 23n−3 . 3n−1 is always divisible by

A) 25

B) 35

C) 45

D) None of these

Solution: A

Putting n=1 in 72n + 23n−3 . 3n−1 then,

72×1 + 23×1 − 3 . 31−1

= 72 + 20 . 30

= 49 + 1 = 50 ……(i)

Also, n = 2,

72×2 + 23×2 − 3 . 32−1

= 2401 + 24

= 2425 ……(ii)

From (i) and (ii) it is always divisible by 25.

Example 3: For every natural number n, n (n2 − 1) is divisible by

A) 4

B) 6

C) 10

D) None of these

Solution: B

n (n2 − 1) = (n − 1) (n) (n + 1)

It is the product of three consecutive natural numbers, so according to Lagrange’s theorem, it is divisible by 3 that is 6.

Example 4: If p is a prime number, then np − n is divisible by p when n is a

A) Natural number greater than 1

B) Irrational number

C) Complex number

D) Odd number

Solution: A

np − n is divisible by p for any natural number greater than 1.

It is Fermat’s theorem.

Trick: Let n = 4 and p = 2

(4)2 − 4 = 16 − 4 = 12, it is divisible by 2.

So, it is true for any natural number greater than 1.

Example 5: 10n + 3 (4n+2) + 5 is divisible by (n ∈ N)

A) 7

B) 5

C) 9

D) 17

E) 13

Solution: C

10n + 3 (4n+2) + 5

Taking n = 2;

102 + 3 × 44 + 5

= 100 + 768 + 5

= 873

Therefore, this is divisible by 9.

Example 6: A 5-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4, 5 without repetitions. The total number of ways that it can be done is

Solution:

The given digits are 0, 1, 2, 3, 4, 5.

A number is divisible by 3 if the sum of its digits is a multiple of 3.

The sum of given 6 digits = 0 + 1 + 2 + 3 + 4 + 5 = 15

Possible sum of 5 digits such that the number formed is divisible by 3 = 12 or 15

Case (i):

The given digits are 0, 1, 2, 3, 4, 5. (sum of digits = 12)

Number of ways = 5P5 = 5! – 4! = 120 – 24

= 96

Case (ii):

The given digits are 0, 1, 2, 3, 4, 5. (sum of digits = 15)

Number of ways = 5P5 = 5! = 120

Total number of ways = 120 + 96 = 216

#### Divisibility Models – Video Lesson 