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attempt:

numbers you can use (0,2,3,4,6,7,8,9) = 8 numbers

first position: (2,3,4,6,,8, 7) = 6 numbers

last position: (3,7,9) = 3 numbers

Rest of the two positions = 6 and 5 numbers left

answer:

6 x 5 x 3 x 6

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  • $\begingroup$ The second and third digits can be any of 0 through 9 (except 5 and 1) $\endgroup$ – David P Jan 30 '14 at 23:02
  • $\begingroup$ You can also use $8$ in 1st position right? $\endgroup$ – voldemort Jan 30 '14 at 23:03
  • $\begingroup$ you two are too fast, edited it right after pressing comment haha $\endgroup$ – David P Jan 30 '14 at 23:03
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    $\begingroup$ @voldemort right you are..my bad $\endgroup$ – Jessica Jan 30 '14 at 23:05
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    $\begingroup$ @Jessica: numbers can be repeated, unless the problem states that all digits have to be different. $\endgroup$ – voldemort Jan 30 '14 at 23:07
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If I read the question correctly, there are 3 rules:

  1. The number must be odd
  2. The number cannot contain the digits 1 or 5
  3. The number must be between 1000 and 9000
  4. Digits may be repeated within the number.

If these are the only rules, then the first digit can be 2, 3, 4, 6, 7, or 8. (6 numbers) 1 and 5 are not allowed, and this digit cannot be 9 because that would make the number over 9000. It also can't be 0 because that would make the number less than 1000.

The second digit can be 0, 2, 3, 4, 6, 7, 8, or 9. (8 numbers)

The third digit can also be 0, 2, 3, 4, 6, 7, 8, or 9. (8 numbers)

The final digit can only be 3, 7, or 9. (3 numbers) These are the only allowed odd numbers.

Therefor, there are 1152 possible numbers (6 * 8 * 8 * 3).

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  • $\begingroup$ this is correct,thank you $\endgroup$ – Jessica Jan 30 '14 at 23:17

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