PROBLEMS ON DIVISIBILITY RULES

Problem 1 :

Test the following numbers for divisibility by 2, 3, 4, 5 and 9.

a)  250     b)  3609       c)  12345      d) 14641

Solution :

a) 250

Divisible by 2 :

Since the given number ends with 0, it is even number. Then 250 is divisible by 2. 

Divisible by 3 :

The sum of the digits,

= 2 + 5 + 0

= 7

Since the sum of the digits is not divisible by 3, then 250 is not divisible by 3.

Divisible by 4 :

Number formed by last two digits is 50, it is not divisible by 4. Then 250 is not divisible by 4.

Divisible by 5 :

250 ends with 0, then it must be divisible by 5.

Divisible by 9 :

The sum of digits in 250 is 2 + 5 = 7.

Since it is not multiple of 9, It is not divisible by 9.

Problem 2 :

A four digit number form, a5b1. If the number is divisible by 3, what are the possible values of a + b ?

Solution :

Given number a5b1 is divisible by 3.

Then the sum of the digits should be the multiple of 3. Finding the sum of the digits, we get

= a + 5 + b + 1

= a + b + 6

When a + b = 3, the value of a + b + 6 will be 9. It is divisible by 3.

When a + b = 6, the value of a + b + 6 will be 12. It is divisible by 3.

When a + b = 9, the value of a + b + 6 will be 15. It is divisible by 3.

When a + b = 12, the value of a + b + 6 will be 18. It is divisible by 3.

When a + b = 15, the value of a + b + 6 will be 21. It is divisible by 3.

When a + b = 18, the value of a + b + 6 will be 24. It is divisible by 3.

Then possible values of a + b are 3, 6, 9, 12, 15 and 18.

Problem 3 :

Consider the five digit number 8251_ .What digit could replace __, so that the number is divisible by

a)  3       b)  4       c)  5      d)  6      e) 9     f)  11 ?

Solution :

8251_

Let x be the unknown.

a)  Divisible by 3 :

Sum of the digits of the given number

= 8 + 2 + 5 + 1 + x

= 16 + x

When x = 2, the sum will become 18. Then it is divisible by 3. So, the missing digit is 2.

b)  Divisible by 4 :

Considering the last two digits 1 x

If x is 2, then the number created by last two digits will be 12. It is divisible by 4. So, the required digit is 2.

c)  Divisible by 5 :

If the number ends with 0 or 5, it will be divisible by 5.  So, the required digit is 0 or 5.

d)  Divisible by 6 :

If the number is divisible by 2 and 3, it is also divisible by 6. So, the required digit is 2.

e)  Divisible by 9 :

Sum of the digits of the given number

= 8 + 2 + 5 + 1 + x

= 16 + x

When x = 2, then the sum will become 18. It will be divisible by 9. So, the required digit is 2.

f)  Divisible by 11 :

Sum of the digits in odd place :

= 8 + 5 + x

= 13 + x

Sum of the digits in the even place :

= 2 + 1

= 3

Difference between them = 13 + x - 3

= 10 + x

If x = 1, then we get the value of 10 + x as 11. It is divisible by 11. Then the required digit is 1.

Problem 4 :

Rearrange the digits 1, 4, 5 and 8 to form a number which is divisible by :

i)  5    ii)  4

Solution :

i) To create a number which is divisible by 5, we have to create a number which ends with 0 or 5.

So, the possible numbers are 

1485, 1845, 8145, 8415, 4185, 4815

ii) To create a number which is divisible by 4, we have to look at the last two digits. By considering the given four digits, 48 is the number which is divisible by 4.