The billion period has three different even digits whose sum is 12. each of the other places is occupied by 8. what is the number?
The 3 digits must be 2,4,6 in any order.
So the sum is 12 right, so it might be 246. It says the remaining digits is occupied by 8. So i figure that the answer is 246, 888, 888, 888
To solve this problem, we need to determine the three different even digits whose sum is 12 to fill the three different positions in the billion period.
Let's start by listing the even digits: {0, 2, 4, 6, 8}.
We can now try different combinations of these digits to find the ones that sum up to 12.
Let's start with the highest digit, 8. We can subtract 8 from 12 to see if there are any combinations of two digits that sum up to the remaining 4.
- If we subtract 8, we have 12 - 8 = 4. We can find the combination (2, 2) that sums up to 4.
Now we have identified two digits, 8 and 2, that sum up to 12. To fill the third position, we can place 8 in the remaining empty place.
So, the number we obtain is: 828882888.