Total number of numbers using 1-6
= 6^8 = 1679616
For any number to be divisible by 8, its last 3 digits must be divisible by 8
but these last two digits must contain only the digits from 1 to 6, with no zeros.
possible cases are:
112 136 144 152
216 224 232 256 264
312 336 344 352
416 424 432 456 464
512 536 544 552
616 624 632 656 664
Hoping that I didn't miss any, I count 27
so the front 5 numbers could be anything
there are 6^5 or 7776 different front numbers, each of those could have 27 different last three numbers
so there are 7776x27 or 209952 which are divisible by 8
prob that a number is a multiple of 8
= 209952/1679616
26244/209952
= 6561/52488
= 1/8
mmmmh, very interesting
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.
3 answers
That is right
That is indeed correct.
A number is a multiple of 8 if and only if the number formed by its last three digits is a multiple of 8.
We want to find the three-digit multiples of 8, where all the digits are between 1 and 6. (More precisely, we want to find how many there are.) First, we find those where the first digit is 1. These are 112, 136, 144, and 152.
Next, we find those where the first digit is 2. These are 216, 224, 232, 256, and 264.
Next, we find those where the first digit is 3. Since 200 is a multiple of 8, we can find these numbers by adding 200 to all the numbers where the first digit is 1. This gives us 312, 336, 344, and 352.
Next, we find those where the first digit is 4. Since 200 is a multiple of 8, we can find these numbers by adding 200 to all the numbers where the first digit is 2. This gives us 416, 424, 432, 456, and 464.
We can do the same calculation in the cases where the first digit is 5 and 6. This gives us $4 + 5 + 4 + 5 + 4 + 5 = 27$ three-digit multiples of 8.
There are a total of $6^3 = 216$ possible outcomes for the last three digits, all of which are equally probable, so the probability that the number is divisible by 8 is $27/216 = \boxed{1/8}$.
A number is a multiple of 8 if and only if the number formed by its last three digits is a multiple of 8.
We want to find the three-digit multiples of 8, where all the digits are between 1 and 6. (More precisely, we want to find how many there are.) First, we find those where the first digit is 1. These are 112, 136, 144, and 152.
Next, we find those where the first digit is 2. These are 216, 224, 232, 256, and 264.
Next, we find those where the first digit is 3. Since 200 is a multiple of 8, we can find these numbers by adding 200 to all the numbers where the first digit is 1. This gives us 312, 336, 344, and 352.
Next, we find those where the first digit is 4. Since 200 is a multiple of 8, we can find these numbers by adding 200 to all the numbers where the first digit is 2. This gives us 416, 424, 432, 456, and 464.
We can do the same calculation in the cases where the first digit is 5 and 6. This gives us $4 + 5 + 4 + 5 + 4 + 5 = 27$ three-digit multiples of 8.
There are a total of $6^3 = 216$ possible outcomes for the last three digits, all of which are equally probable, so the probability that the number is divisible by 8 is $27/216 = \boxed{1/8}$.
1 answer