Posted by **zamanofdoom** on Thursday, November 15, 2012 at 10:30am.

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.

- counting and probability -
**Anonymous**, Tuesday, July 21, 2015 at 11:30am
hg.kk

- counting and probability -
**ASD**, Friday, October 16, 2015 at 10:49pm
gidae

- counting and probability -
**i need help too**, Saturday, October 17, 2015 at 8:44pm
I need help 2

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- counting and probability -
**BOB**, Thursday, July 28, 2016 at 12:43pm
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 <--- Very suspicious

- counting and probability -
**existing**, Tuesday, August 16, 2016 at 1:18pm
BOB is right.

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