Divisibility challenge

The number must end in 0 if it has to be divisible by 2 and 5, so $B=0.$

The sum of the digits, $15+A,$ must be a multiple of 9 for it to be divisible by 9. Thus, $A=3,$ so $A+B = 3+0=3.$

as the number is divisible by both 2 & 5. so,B must be 0. we know the divisible rule for 9 is that,the sum of the digits is a multiple of 9. if we put 3 in place of A,we can find the sum 18 which is a multiple of 9. so, A+B will be 3+0=3

Lets start with the rules of divisibility of 2, 5, and 9.

2 is divisible to numbers which have an even last digits thus it should be 2, 4, 6, 8, or 0 For example: 23456/2= 11728

5 is divisible to numbers which have the last digits as 5 or 0 For example: 12142380/5= 2428476

9 is divisible to numbers if it's digits are added together which make a sum of a multiple of 9 For example: 18/9= (1+8)= 9/9= 1

Now that's out of the way, we can see the last digit has to be 0 since 0 as the last digit can divisible by 2 and 5 thus B=0.

The next step is to add up the current digits which is 9+2+4+0= 15 NOTE: We still have to find the value of A

The value nearest to 15 but also is a multiple of 9 which is 18 as there is a difference of 3 between 15 and 18, A=3.

The last step is to add the value of A+B which is 3+0=3

Now give yourself a pat on the back or treat yourself with a nice, fresh slice of pizza for reading this much! :)

Divisible by 2,5,9 means, It is divisible lcm ie 90. So the number is 92340, A = 3 & B = 0. A + B = 3+0=3