Question

find smallest number that must be added to 128 so that it becomes exactly divisible by 5 ​

Answer 1

Answer : 125 is the first number divided by 5 and comes after 123.

so to get 125 from 123 , we must need to add the number 2 in it

so. 123+2=125

that is divisible by 5.

125/5=25

Answer 2

Given:

A number 128.

To Find:

The smallest number to be added with 128 such that the number is exactly divisible by 5 will be?

Solution:

The given problem can be solved using the concepts of divisibility rules.

1. The given number is 128.

2. For a number to be divisible by 5, the units digit of the number must be either 0 or 5 at all the cases irrespective of the digits at the other places.

3. The number 128 when divided by 5 gives a remainder 3,

=> 128 = 5 x 25 + 3. ( Where 25 is the quotient and 3 is the remainder).

4. The smallest number that can be added to the number 128 to be divisible by 5 is 2. Hence, after adding 2 the number is 128 + 2 = 130.

=> 130 = 26 x 5 + 0, ( quotient - 26 and remainder is 0).

5. Hence, the smallest number is 2.

Therefore, the smallest number that must be added to 128 so that it can be divisible by 5 is 2.