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Math Help - Is there a better way?

  1. #1
    Member billym's Avatar
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    Is there a better way?

    (i) Find the number of integers between 500 and 5000 that do not contain 0 as a digit.

    (ii) How many integers between 500 and 5000 contain the digit 0 exactly once?

    (iii) ...at least twice?

    ---

    Assuming that 500 and 5000 are not included,

    (i) (4999-500)-18-4*19-4(100+9*19)=3321

    (ii) 18*5+4*9*18=738

    (iii) 4+4(19+9)=116


    I assume these are right(ish), but is there a simpler method?
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  2. #2
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    Hello, billym!

    I'll assume that 500 and 5000 are included.


    (1) Find the number of integers between 500 and 5000 that do not contain 0 as a digit.
    Three-digit numbers: _ _ _

    They can begin with: {5, 6, 7, 8, 9} . . . 5 choices.
    . . There are 9 choices for the second digit.
    . . There are 9 choices for the third digit.
    There are: . 5\cdot9^2 \:=\:405 three-digit numbers with no 0s.


    Four-digit numbers: _ _ _ _

    They can begin with: {1, 2, 3, 4} . . . 4 choices.
    . . There are 9 choices for each of the other 3 digtits.
    There are: . 4\cdot9^3 \:=\:2916 four-digit numbers with no 0s.


    Therefore, there are: . 405 + 2916 \:=\:\boxed{3321} numbers with no 0s.




    (2) How many integers between 500 and 5000 contain the digit 0 exactly once?
    Three-digit numbers: _ _ _

    They can begin with: {5, 6, 7, 8, 9} . . . 5 choices.
    . . There are 2 choices for the position of the 0.
    . . There arte 9 choices for the third digit.
    There are: 5\cdot2\cdot9 \:=\:90 three-digit numbers with one 0.


    Four-digit number: _ _ _ _

    They can begin with: {1, 2, 3, 4} . . . 4 choices.
    . . There are 3 choices for the position of the 0.
    . . There are: 9^2 choices for the other two digits.
    There are: 4\cdot3\cdot 9^2 = 972 four-digit numbers with one 0.


    Therefore, there are: . 90 + 972 \:=\:\boxed{1062} numbers with exactly one 0.




    (3) ...at least twice?
    From 500 to 5000, there are: . 4501 numbers.

    3321 of them have no 0s.
    1062 of them have one 0.

    Therefore: . 4501 - 3321 - 1062 \:=\:\boxed{118} numbers with two or more 0s.


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  3. #3
    Member billym's Avatar
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    Revised:

    Assuming 500 and 5000 not included,

    Forms of 3-digit numbers containing 0:

    AA0
    A00
    A0A

    5*9*1*2 + 4*1*1 = 94

    Forms of 4-digit numbers containing 0:

    AAA0
    AA0A
    A0AA

    AA00
    A00A
    A0A0

    A000

    4*9*9*3 = 972

    4*9*3 = 108

    4 = 4

    (i) 4999 - 500 - 94 - 108 - 972 - 4 = 3321

    (ii) 5*9*2 + 4*9*9*3 = 1062

    (iii) 4 + 4*9*3 + 4 = 116

    Correct?
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  4. #4
    Member billym's Avatar
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    Okay so you beat my revision. I guess our answers are the same except you have +2 in the last question because you included 500 and 5000. Great.
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