A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? Express your answer as a common fraction.
I think the last 3 digits have to be divisible by 8 to be a multiple of 8...
1/8
You're correct! In order for an eight-digit number to be divisible by 8, the last three digits must form a multiple of 8.
To find the probability, we'll need to count the number of possible outcomes that satisfy this condition, and then divide by the total number of possible outcomes.
Let's start by determining the total number of possible outcomes for each digit. Since we have a fair, six-sided die, there are 6 possible outcomes for each roll. Since there are 8 rolls, the total number of possible outcomes is 6^8.
Now, let's count the number of possible outcomes for the last three digits, keeping in mind that they need to form a multiple of 8. To do this, we'll consider all the different combinations of the last three digits that are multiples of 8.
A number is divisible by 8 if its last three digits together form a multiple of 8. Let's list all the possible combinations for the last three digits:
- 008, 016, 024, 032, 040, 048, 056
- 064, 072, 080, 088, 096
- 104, 112, 120, 128, 136, 144, 152
- 160, 168, 176, 184, 192
- ...
We can see a pattern here. The last two digits will keep increasing by 8 until they reach 96, then they will restart at 00 with the third digit increasing by 1.
From this pattern, we can observe that there are 24 different combinations of the last three digits that form a multiple of 8.
Now, let's calculate the probability. The number of favorable outcomes (combinations that form a multiple of 8) is 24, and the total number of possible outcomes is 6^8 = 1679616.
Therefore, the probability that the resulting number is a multiple of 8 is 24/1679616, which simplifies to 1/69984.
So, the probability is 1/69984.