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# 03NS30

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173
6 Can anyone help please

Apr 8, 2019

#1
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The rule for divisibility by 8 is that, if the last three digits are divisible by 8, then the whole number is divisible by 8

The total number of outcomes is  6^8

Notice that  the smallest that the last three digits can be  =  111

And the largest that they can be  = 666

The number of integers between these two numbers that are divisible by 8  =

floor [ (666 - 111) / 8 ]  = 69

So....the number of outcomes that are divisible by 8  =  (6)^5 * 69

So the probability  =   (6)^5 * 69  /  (6)^8   =   69 / 6^3   =   69 / 216   =  23 / 72   Apr 8, 2019
#2
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Wait, won't some of them contain digits larger than 6?

Apr 9, 2019
#3
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No....the integers on the dice cube range from 1 to 6

We will have an 8 digit number.....but no digit > 6   CPhill  Apr 9, 2019
#4
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Six sided die only has digits  1  2  3  4  5  6   on it.

ElectricPavlov  Apr 9, 2019
#5
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But aren't you counting the case where the last three digits are 128?

ROYGBIV  Apr 9, 2019
#6
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The digits aren't ADDED....they are used to form  an eight digit number

First digit   1- 6

Second digit  1-6

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eighth digit 1-6       no digits of the eight will be > 6

Guest Apr 9, 2019