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 sixdigit number. It is divisible by
(a) 7 only
(b) 11 only
(c) 13 only
(d) All 7, 11 and 13
Solution:
Sixdigit number 325325 is formed by writing 325 twotimes.
Therefore, required factors are 7, 11 and 13
Answer: (d)
Note: Any sixdigit number is formed by writing a
three digit number twotimes, 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 3digit 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 onedigit) is subtracted from the number, then resulting number is always divisible by 9.
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