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anonymous

  • one year ago

how do i find the 7000th registration plate of a series of plates using numbers 0,1,2,3,5,7 and 6 OR 9

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  1. anonymous
    • one year ago
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    This is an interesting question.

  2. anonymous
    • one year ago
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    What do you think the question actually means?

  3. anonymous
    • one year ago
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    Yep any ideas?? the reg. number must be 7 digits long and cannot start with a zero. The numbers go in ascending order so the first three numbers are 1023567, 1023576, 1023579

  4. anonymous
    • one year ago
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    I believe there is a bit of confusion with the way question is worded.

  5. anonymous
    • one year ago
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    I think you use factorial numbers to figure it out. so number of combinations using 9 would be 5040 subtract combinations which start with zero (720) then add the combinations using 6 instead of nine and total number of combinations for reg. number is 8640 which would be the largest number possible with the combination of numbers: 9753210

  6. anonymous
    • one year ago
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    This is very curious... I am trying my best.

  7. anonymous
    • one year ago
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    Do you mind precisely showming me the original question you seem to be reiterating over and over..? It's a little confusing and I am not on the same page as you.

  8. anonymous
    • one year ago
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    A harbour has boat registration number of 7 digits. None can start with zero. the harbour has been able to get numbers 0,1,2,3,5,7,9. No digit is used more than once in a registration number. You can use the 9 as a six upside down. Both numbers 6 and 9 cannot be in the same registration number. Numbers are issued in ascending order. What is the 7000th number?

  9. IrishBoy123
    • one year ago
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    .

  10. mathmate
    • one year ago
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    @jimmy10 Thank you for posting the original question. It is much appreciated.

  11. mathmate
    • one year ago
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    There are 7 digits, and each number has 7 digits, so there are 7! permutations. Removing zero at the beginning, leaves us with 6(6!) permutations. Taking mutation of the 6 & 9 into account, there are 2*6(6!)=8640 permutations, which is greater than 7000, so there is a solution. Here's an idea of what I did. Check it out to make sure there are no errors. Hope someone finds a faster way of doing it (apart from using a computer program) Numbers of the form 1XXXXXX, 2XXXXXX, 3XXXXXX, 5XXXXXX take up 2*4(6!)=5760 numbers. (note: the multiplier "2" is for the mutation of the 6/9). Numbers of the form 6XXXXXX take up 6!=720 numbers (no more duplicates with 6 & 9) for a total of 6480 numbers. Numbers of the form 70XXXXX, 71XXXXX take up 2*2(5!)=480 numbers for a total of 6960 numbers Numbers of the form 720XXXX take up 2(4!) = 48 numbers, for at total of 7008 numbers. So the 7000th number must be of this form. Numbers of the form 7201XXX, 7203XXX, 7205XXX take up 2*3(3!)=36 numbers for a total of 6996 numbers The next numbers are: 7206135 7206153 7206315 7206351 (7000th number) ...

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