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Proposed Problem

Discussion in 'Mathematics' started by AVDBA, Sep 18, 2019.

  1. AVDBA

    AVDBA New commenter

    A house has a security system that uses a numeric keyboard as depicted beneath:

    [1] [2] [3]
    [4] [5] [6]
    [7] [8] [9]
    [0]
    A potential intruder, hidden, observes, and perceives that:

    * The first and the last digits are in the same line
    * The second and the third digits are in the adjacently superior line
    * The password being used has four digits

    How many passwords has the potential intruder to test to be sure that
    is going to enter the house?
     
  2. adamcreen

    adamcreen Occasional commenter

  3. gainly

    gainly Lead commenter

    171
     
    BG54 and NoSuchThingAsNormal like this.
  4. AVDBA

    AVDBA New commenter

    Thanks a lot for replying. Hints for working throughout the solution:

    1 - The first and fourth digits cannot be in the first line, since there is not a line that is adjacently superior.

    2 - There exist three disjoint possibilities for the location of the first and fourth digits.

    3 - Using the fundamental counting principle for each of these possibilities, one reaches three disjoint results.

    4 - The disjunction requires that these results be added up.
     
  5. adamcreen

    adamcreen Occasional commenter

    A half.
     
  6. frustum

    frustum Star commenter

    I'm with gainly. That's unless there is additional information you haven't stated, for example that all digits are different.
     
  7. AVDBA

    AVDBA New commenter

    The hints were intended to guide through a solution:

    Being the first and fourth digits in the fourth line, the second and
    third ones must be in the third line, leading to:

    first digit : 1 possibility: digit 0
    second digit: 3 possibilities: digits 7, 8 or 9
    third digit : 3 possibilities: digits 7, 8 or 9
    fourth digit: 1 possibility: digit 0

    The total number of possibilities for this case turns out to be:
    1 x 3 x 3 x 1 = 9 possibilities.

    Being the first and fourth digits in the third line, the second and the
    third ones must be in the second line, leading to:

    first digit : 3 possibility: digits 7, 8 or 9
    second digit: 3 possibilities: digits 4, 5 or 6
    third digit : 3 possibilities: digits 4, 5 or 6
    fourth digit: 3 possibility: digits 7, 8 or 9

    The total number of possibilities for this case turns out to be:
    3 x 3 x 3 x 3 = 81 possibilities.

    Being the first and fourth digits in the second line, the second and the
    third ones must be in the first line, leading to:

    first digit : 3 possibility: digits 4, 5 or 6
    second digit: 3 possibilities: digits 1, 2 or 3
    third digit : 3 possibilities: digits 1, 2 or 3
    fourth digit: 3 possibility: digits 4, 5 or 6

    The total number of possibilities for this case turns out to be:
    3 x 3 x 3 x 3 = 81 possibilities.

    By disjunction, the potential intruder needs to test 9 + 81 + 81 = 171
    passwords to be sure that is going to enter the house.
     
  8. gainly

    gainly Lead commenter

    Do I win a prize?
     
  9. AVDBA

    AVDBA New commenter

    Two prizes.
     
  10. armandine2

    armandine2 Established commenter

    depends what's in the house
     
    blue451 likes this.

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