We will discuss here about the rules of divisibility tests by 3 and 6 with the help of different types of problems.

**1.** 325325 is a six-digit number. It is divisible by

(a) 7 only

(b) 11 only

(c) 13 only

(d) All 7, 11 and 13

Solution:

Six-digit number 325325 is formed by writing 325 two-times.

Therefore, required factors are 7, 11 and 13

Answer: (d)

**Note:** Any six-digit number is formed by writing a
three digit number two-times, that number is always divisible by 1001 and its
prime factors 7, 11 and 13.

**2.** The sum of
three consecutive odd numbers is always divisible by

(a) 2

(b) 3

(c) 5

(d) 6

Solution:

Solution:

Sum of any three consecutive odd numbers divisible by 3

Answer: (b)

**Note:** Sum of any three consecutive numbers is
divisible by 3, but four numbers divisible by 2.

Sum of any three consecutive odd numbers divisible by 3 but even numbers divisible by 6

**3.** The largest
natural number which exactly divides the product of any four consecutive
natural numbers is:

(a) 6

(b) 12

(c) 24

(d) 120

Solution: Product of any four consecutive natural numbers is always divisible by 1 × 2 × 3 × 4 = 24

Answer: (c)

**Note:** Product of any three consecutive natural
numbers is divisible by 6 and four numbers divisible by 24.

The first natural number is 1.

**4.** The largest
natural number by which the product of three consecutive even natural numbers
is always divisible is:

(a) 16

(b) 24

(c) 48

(d) 96

Solution:

Product of any three consecutive even numbers is divisible by {2^(3 + 1) × 3} = {2^4 × 3} = 16 × 3 = 48

Answer: (c)

**Note:** Product of any three consecutive odd natural
numbers is divisible by 3. But even numbers is divisible by 48.

** **

**5.** The difference
between the squares of two consecutive odd integers is always divisible by:

(a) 3

(b) 6

(c) 7

(d) 8

Solution:

Required number is 8.

Answer: (d)

**Note:** The difference of squares of two consecutive
odd integers is divisible by 8 but even integers are divisible by 4.

** **

**6.** The sum of the
digits of a 3-digit number is subtracted from the number. The resulting number
is

(a) divisible by 6

(b) divisible by 9

(c) divisible neither by 6 nor by 9

(d) divisible by both 6 and 9

Solution:

The resulting number is divisible by 9

Answer: (b)

**Note:** If sum of digits of any number (more than
one-digit) is subtracted from the number, then resulting number is always
divisible by 9.

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