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    How many 5 digit numbers that have 5 different digits are divisible by 5?

    The answer says
    (9×8×7×6×1)+(8×8×7×6×1) = 5712
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    Last digit has to be 0 or 5 First digit must be between 1 and 9 and can't be the last digit. Because "last digit = 5" removes one of the options for the first digit, but "last digit = 0" does not, you need to consider these separately (the first bracket in the answer is the case "last digit = 5" the 2nd bracket the case "last digit = 0").
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    (Original post by Paing Thet)
    How many 5 digit numbers that have 5 different digits are divisible by 5?

    The answer says
    (9×8×7×6×1)+(8×8×7×6×1) = 5712
    A number divisible by 5 must end in either 5 or 0. You need to consider the two separate cases since the first digit can never be 0.

    So I recommend writing out the number like this _ _ _ _ _

    For the case where last digit equals 0 you have _ _ _ _ 0

    Go through the digits left to right and consider how many choices you have. So the leftmost digit has 9 choices since it can't be a 0...

    EDIT: Too late!
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    (Original post by DFranklin)
    the first bracket in the answer is the case "last digit = 5" the 2nd bracket the case "last digit = 0".

    Other way around, I believe.
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    (Original post by ghostwalker)
    Other way around, I believe.
    Yes, thanks.
 
 
 
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