Hard Probability

The 3-digit no. formed by the arrangement of the last 3 digits should be divisible by 8 since 1000 is divisible by 8.

Thus the form should be "XXXXXYY2", "XXXXXYY4", "XXXXXYY6" or "XXXXXYY8" since 8 is even.

For "XXXXXYY2", YY ranges from 11, 15, 19, ... to 87 with a total of 20 possibilities.

However, YY cannot contain any duplicate digits and any of 0, 2 or 9. Therefore only 15, 31, 35, 43, 47, 51, 63, 67, 71, 75, 83 and 87 are possible, total = 12

Hence no. of possible permutations with last digit = 2 is 5! x 12

For "XXXXXYY4", YY ranges from 10, 14, 18, ... to 86 with a total of 20 possibilities.

However, YY cannot contain any duplicate digits and any of 0, 4 or 9. Therefore only 18, 26, 38, 58, 62, 78, 82 and 86 are possible, total = 8

Hence no. of possible permutations with last digit = 4 is 5! x 8 For "XXXXXYY6", YY ranges from 13, 17, 21, ... to 85 with a total of 19 possibilities.

However, YY cannot contain any duplicate digits and any of 0, 6 or 9. Therefore only 13, 17, 21, 25, 37, 41, 45, 53, 57, 73, 81 and 85 are possible, total = 12

Hence no. of possible permutations with last digit = 6 is 5! x 12 For "XXXXXYY8", YY ranges from 12, 16, 20, ... to 76 with a total of 17 possibilities. However, YY cannot contain any duplicate digits and any of 0, 8 or 9. Therefore only 12, 16, 24, 32, 36, 52, 56, 64, 72 and 76 are possible, total = 10Hence no. of possible permutations with last digit = 8 is 5! x 10Therefore total no. of possible permutations that the resulting 8-digit no. is divisible by 8 is 5! x (18 + 8 + 12 + 10) = 5! x 38So the probability is 5! x 38/8! = 38/(6 x 7 x 8) = 19/168

2011-05-28 20:28:31 補充：
Correction to last 3 lines:

Therefore total no. of possible permutations that the resulting 8-digit no. is divisible by 8 is 5! x (12 + 8 + 12 + 10) = 5! x 42

2011-05-28 20:28:53 補充：
So the probability is 5! x 42/8! = 42/(6 x 7 x 8) = 1/8