# Proposed Problem

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

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1. ### AVDBANew 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?

6

3. ### gainlyStar commenter

171

BG54 and NoSuchThingAsNormal like this.
4. ### AVDBANew 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.

A half.

6. ### frustumStar commenter

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

7. ### AVDBANew 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. ### gainlyStar commenter

Do I win a prize?

Two prizes.

10. ### armandine2Established commenter

depends what's in the house

blue451 likes this.