16
$\begingroup$

Using the clues provided, determine the 'password'.

501 — Two correct numbers in wrong places
135 — One correct number in the right place
483 — All numbers are wrong
167 — One correct number in wrong place
430 — One correct number in the wrong place

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ +1 for reminding me that I haven't played Mastermind in a while...such a fun game. $\endgroup$
    – Brandon_J
    Jan 29, 2019 at 19:06
  • 1
    $\begingroup$ While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :) $\endgroup$
    – Ian MacDonald
    Jan 29, 2019 at 21:10

9 Answers 9

Reset to default
12
$\begingroup$ 
[5, 0, 1]: 2  
[1, 3, 5]: 1

So either 5 or 1 is correct, but not both.

[4, 8, 3]: none.

[1, 6, 7]: 1

Because 1 is in the same place as it was before, but the "one correct" is in the wrong place, it cannot be a right number. This means that 5 (from earlier) is definitely the last number, and also one of either 6 or 7 is correct, but in the wrong place.

[4, 3, 0]:

We know from above that 4 and 3 are both wrong, and we deduced from the last step that 1 was not a number, meaning the first line shows us that 5 and 0 are both numbers, so this actually gives us no new information on which numbers, but does tell us that 0 is still in the wrong place.


This means that we have "0 x 5" with x being either 6 or 7. From the second-last line, we know that the correct number is in the wrong place, so it can't be 6 (which was in the middle).


The final answer is: 0 7 5.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.) $\endgroup$
    – Kevin
    Jan 29, 2019 at 22:20
4
$\begingroup$

The following are my musings to find the answer:

Starting with 430 (Rule 5):

4?? Eliminated by rule 5
?4? Eliminated by rule 3
??4 Eliminated by rule 3
3?? Eliminated by rule 3
?3? Eliminated by rule 5
??3 Eliminated by rule 3
0?? Remaining - Now Rule 6
?0? Remaining - Now Rule 6
??0 Eliminated by rule 5

Starting with 501 (Rule 1):

5?1 Eliminated by rule 1
51? Eliminated by rule 1
15? Eliminated by rule 6
1?5 Eliminated by rule 2
?15 Eliminated by rule 6
?51 Eliminated by rule 1
50? Eliminated by rule 1
5?0 Eliminated by rule 1
?50 Eliminated by rule 2
?05 Eliminated by rule 1
0?5 Remaining - Now Rule 7
05? Eliminated by rule 2
?01 Eliminated by rule 1
?10 Eliminated by rule 2
1?0 Eliminated by rule 6
10? Eliminated by rule 1
01? Eliminated by rule 2
0?1 Eliminated by rule 1

Starting with 167 (Rule 4):

1?? Elimated by rule 4
?1? Elimated by rule 1
??1 Elimated by rule 7
6?? Elimated by rule 7
?6? Elimated by rule 4
??6 Elimated by rule 7
7?? Elimated by rule 7
?7? Remaining - Now Rule 8
??7 Elimated by rule 4

Combining the two new rules, 7 and 8 gives:

Result 075

Improve this answer
$\endgroup$
4
$\begingroup$

I've found the answers too complicate, so I share mine. Only three rules are needed:

501 — Two correct numbers in wrong places

So, only one of those numbers should be removed. Which one?

135 — One correct number in the right place

Since this shares 2 numbers with the first one, the 3 cannot be the correct one. (If the 3 comes into play, the rule would give info about 2 numbers instead of one). So, from the previous rule, we should remove only the 1 or the 5. Now we know the number has a 0 in one side, and starts with 1 or end with 5. Only 2 chances: 1_0 or 0_5.

167 — One correct number in wrong place

This number start with 1 as the previous one, and the 1 can't be in right and wrong place at the same time. So, we have only one possibility now: 0_5. We are only missing the middle number and can't be the 6 because it's already in the middle and not in the right place. So the only option left is the 7. -> 075.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more? $\endgroup$
    – Prince North Læraðr
    Jan 30, 2019 at 3:06
  • $\begingroup$ Excellent observation about the 3 to get it down to three rules! $\endgroup$
    – par
    Jan 30, 2019 at 20:31
3
$\begingroup$

Answer Should be

075

because

0 Is right because there are two numbers right in first condition. And other one is wrong because of second condition in which 5 or 1 Has to be wrong. and From last condition 0 is selected and other 2 values got discard because of the 3rd condition. We got the position of the 0 from these two upper conditions. as 0 is wrong at second place in first condition and 0 is also wrong at last place in last condition.
5 Is selected from first condition if we select 1 then we get only 2 numbers as we need 3 to make the key. 5 is also right in condition second in which we also got the right position of the 5.
7 is right because of 4th condition as the other two number which is 1 is wrong due to first and second condition because we selected 5 and 6 is also wrong due to the position of 6. we need number that fits in position between 5 and 0. because 7 is at wrong place that fits at second position.
As it satisfy all conditions
501 — Two correct numbers in wrong places(5,0, As 1 or 5 get discard in second condition, so i took 5)
135 — One correct number in the right place(5, As we can select only one that should be from first condition.)
483 — All numbers are wrong(If presented above shall be discard.)
167 — One correct number in wrong place(7, Both the last and first place is booked thats why only chance 7)
430 — One correct number in the wrong place(0, as we took in the first step.)

Improve this answer
$\endgroup$
0
2
$\begingroup$

Only the first four rules are necessary to find the answer.

a. Via rules 2 & 3 the answer contains a 1 or a 5 but not both.
b. Via (a), rule 1, and rule 2 the 1 or 5 in the answer is not in the middle.
c. Rules 2 and 4 contradict if 1 is in the number, so 1 is not in the number, therefore 5 is at the end.
d. Via (b), (c), and rule 1 the answer contains a 0 which is not in the middle, therefore 0 is at the front.
e. Via (c), (d), and rule 4, 7 is in the middle.

The answer is 075

Improve this answer
$\endgroup$
2
$\begingroup$

I will share my process:

  • Notes of format !5 | !0 | !1 means that the first position can't be 5, second can't be 0, etc.
  • Possible is short for values which are possible and also have been previously discussed
  • Impossible represents values which have been ruled out
  • Definite represents values which have been confirmed to exist, though makes no statement on order
  • Conditions represent restrictions on password. For example: !(1+4) means the code cannot contain both one and 4

Step one:

!5 | !0 | !1

Possible: 0,1,5
Conditions: !(0+1+5)

Step Two

!3!5 | !0!1!5 | !1!3
Possible: 0,1,3,5
Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)

Step Three

!5 | !0!1!5 | !1
Possible: 0,1,5
Impossible: 3,4,8 -> as defined
Conditions: !(0+1+5)

Step Four

!1!5 | !0!1!5!6 | !1!7 -> 1 is eliminated
Possible: 0,5,6,7
Impossible: 1,3,4,8
Conditions: !(1+6+7)!(1+6)!(1+7)!(6+7)

Step Five

!5 | !0!5!6 | !7 -> !5 | 7 | !7 -> 6 is eliminated here, as explained in below:
Possible: 0,5,6,7
Impossible: 1,3,4,8
Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed

Conclusion

Possible: 0,5,7
First Term: !5 -> 0 (since 7 is second term)
Second Term: 7
Third Term: 5
.
0 7 5

Improve this answer
$\endgroup$
1
$\begingroup$

1) 501
2) 1X5 — 3 gets eliminated by the next clue
3) XXX — All eliminated
4) X67 — 1 gets eliminated for being in the "wrong" place here but the "right place" on Clue 2
5) XX0 — 4 and 3 gets eliminated by clue 3, leaving 0

So 0 is in the 1st position, because of the first clue that says it can't be in 2nd

2) XX5 — The 0 took the 1's place, so only 5 is left

So we have 0X5 at this point.

4) XX7 — 1 gets eliminated by clues 1 and 2, and 6 would be in the "right" place if it was one of the numbers

Answer: 075

Improve this answer
$\endgroup$
1
$\begingroup$

Rule 1) 501 — Two correct numbers in wrong places
Rule 2) 135 — One correct number in the right place
Rule 3) 483 — All numbers are wrong
Rule 4) 167 — One correct number in wrong place
Rule 5) 430 — One correct number in the wrong place

Starting from the most elimination Rule 3

Because of Rule 3, Rule 5 proves that 0 is at position 1 or 2.
But 0 can't be at position 2 because of Rule 1.

So 0 is at position 1.

Finding the 2nd digit

Because of Rule 1, and we know about 0, it has to contain a 1 or a 5, but not both.
And because of Rule 2, it can't be 1 because that position is already taken by 0.

So 5 is at position 3.

The remaining digit

Then there's only the 2nd position to figure out.
And because of Rule 4, it can't be 6.
The 1 had already been discarded when we discovered 5 is at position 3.

So 7 is at position 2.

Conclusion:

0 7 5

Improve this answer
$\endgroup$
1
$\begingroup$

The required password is

075 .

Improve this answer
$\endgroup$
1
  • $\begingroup$ please explain your approach to this answer. thanks and happy puzzling ;) $\endgroup$
    – Omega Krypton
    Feb 1, 2019 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.