find the smallest number which is greater than 111,111,000 and divisible by 8 and 9

add 8, add 3

To be divisible by 8, the last 2 digits must be divisible by 8,

to be divisible by 9, the sum of the digits must be divisible by 9
so far they add up to 6, so let's add:
3 ---> no good for the 8 part
12 ---> no good for the 8 part
21 ---> no good for the 8 part
30 ---> no good for the 8 part
39 ---> no good for the 8 part
48 ----- YEAHHH

111,111,048

OR

The LCM of 8 and 9 is 72
so any multiple of 72 is good
111,111,000 ÷ 72 = 1543208 24/72
So if we multiply 72 by the next integer we should have it.
1543209 * 72 = 111,111,048

Reiny is correct, I did not read the problem as divisible by 8 and 9 both, but sequentially, as in two numbers.

To find the smallest number that is greater than 111,111,000 and divisible by both 8 and 9, we need to find the least common multiple (LCM) of these two numbers.

Step 1: Find the LCM of 8 and 9.
To find the LCM, we will prime factorize both numbers.

Prime factorization of 8 = 2^3
Prime factorization of 9 = 3^2

The LCM is the product of the highest powers of all prime factors involved. In this case, the LCM is 2^3 * 3^2 = 72.

Step 2: Divide the number 111,111,000 by the LCM.
111,111,000 ÷ 72 = 1,541,375

Step 3: Add 1 to the quotient.
1,541,375 + 1 = 1,541,376

Therefore, the smallest number greater than 111,111,000 and divisible by both 8 and 9 is 1,541,376.