Once you attempt the question then PrepInsta explanation will be displayed.

Condition for divisibility by 3--

The sum of digits of number must be divisible by 3.

i.e a+b+6+4+a+b = 0 mod 3

=> 10 + 2(a+b) = 0 mod 3

=> 1 + 2(a+b) = 0 mod 3 (10 = 1 mod 3)

=> 2(a+b) = -1 mod 3

=> 2(a+b) = 2 mod 3

=> a+b = 1 mod 3

So, a+b must be of the form 3n + 1 for some positive integer n.

Max value of (a+b) is 18 as a and b can take values upto 9 since they are digits.

The possible values are 1,4,7,10,13,16.

2. Condition for divisibility by 4--

For divisibility by 4, Last two digits must be divisible by 4.

i.e 10a+b = 0 mod 4

=> 2a + b = 0 mod 4 (Since, 8a = 0 mod 4)

So, 2a + b can be 4,8,12,16,20,24,28.

Combining these,we get

a + b = 3n +1

2a + b = 4k

The smallest values which satisfies these

two equation is a = 4 ,b = 0.

So, the minimum sum of a + b is 4

**Video solution:**

Login/Signup to comment