A fair, standard six faced die is tossed eight times. the sequence of eight results is recorded to form an eight-digit number. What is the probability that the number formed is a multiple of eight. express you answer as a common fraction.
Here we go, so the denominator is 6^8, now we need to find the numerator. The numerator must have values that are multiples of 8. The last three digits must be divisible by 8 for the number to be divisible, so for the first five digits the value is 6*6*6*6*6, now I am just confused on how much 3 digit numbers there are that have their digits add up to something divisible by 8. thank you for help 🙏 . I now that I can use hardcore grinding and listing them out, but if there is a different number other than six, it will be very helpful
+583 If the last three digits of a whole number are divisible by 8, then the entire number is divisible by 8. (By googling)
I don't have an effective way to determine a 3 digit number wehether is multiple of 8 or not.
So I list all 3 digit number consist only numbers 1-6,which are
112,136,144,152,216,232,224,256,264,336,344,312,352,416,456,424,464,432,536,544,512,552,616,624,632,664,656
27 numbers divisible 8 in total (I did that by adding multiple of 40, so it is a bit out of order.)
I also wrote a stupid Java program to check my answer.
public static void main(String []args){
int c=0;
for (int i = 1;i<=6;i++){
for (int j = 1;j<=6;j++)
{
for(int k = 1;k<=6;k++)
{
int n=i*100+j*10+k;
if(n%8==0)
{
c++;
}
}
}
}
System.out.println(c);
}
The first five dight can be any integer in {1,2,3,4,5,6}.Therefore, the probability that the number formed is a multiple of eight is \(\frac{6^5*27}{6^8}=\frac{3^3}{6^3}=\frac{1}{2^3}=\frac{1}{8}\).
PS: I believe there is a way to do this problem in abstract algebra, which I rusted at.
For 8 = 1 permutation
For 16 =6,147 permutations
For 24 =98,813 permutations
For 32 =98,813 permutations
For 40 =6,147 permutations
For 48 =1 permutation
Sum them all up [1 + 6,147 + 98,813 +98,813 +6,147 + 1] =209,922 / 6^8 =34,987 / 279,936
My computer says there are: 54,432 numbers that are mutiples of 8.
Therefore, the probability is: 54,432 / 6^8 =7 / 216
+118609 I get 27 too
P(divisable by 8) = \(\frac{27}{6^3}= \frac{3^3}{3^3*2^3}=\frac{1}{8}\)
So I agree with Fiora
NOTE
27 is not the number that are multiples of 8 and 6^3 is not the total number of numbers.
The fact is that only the last 3 rolls have any relevance to this problem.
So Fiora and I just chose to completely ignore the first 5 rolls.
Hi Melody: 1/8 cannot be right because 1/8 would be multiple of all natural numbers from 0 to 9. But, you are only dealing with 6 numbers from 1 to 6!
The permutations begin like this:
11111112 , 11111136 , 11111144 , 11111152 , 11111176 , 11111184 , 11111192 , 11111216.........and end like this: 66616592 , 66616616 , 66616624 , 66616632 , 66616656 , 66616664 , 66616672 , 66616696, which are ALL multiples of 8. My computer did list all 54,432 permutations which are multiples of 8.
As you can see, you cannot have a number beginning with 5 sixes, or 66666, because you cannot make a multiple of 8 out of 5 sixes in a row! If you want me to, I can print all 54,432 multiples of 8 out of 8 dice.
