crack this code #1965
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Using the clues provided, determine the 'password'.
501 — Two correct numbers in wrong places135 — One correct number in the right place483 — All numbers are wrong167 — One correct number in wrong place430 — One correct number in the wrong place
mastermind
edited Feb 1 at 9:46
JonMark Perry
20k64096
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
$endgroup$
add a comment |
$begingroup$
Using the clues provided, determine the 'password'.
501 — Two correct numbers in wrong places135 — One correct number in the right place483 — All numbers are wrong167 — One correct number in wrong place430 — One correct number in the wrong place
mastermind
edited Feb 1 at 9:46
JonMark Perry
20k64096
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
$endgroup$
1
$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 at 19:06
1
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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 at 21:10
add a comment |
$begingroup$
Using the clues provided, determine the 'password'.
501 — Two correct numbers in wrong places135 — One correct number in the right place483 — All numbers are wrong167 — One correct number in wrong place430 — One correct number in the wrong place
mastermind
edited Feb 1 at 9:46
JonMark Perry
20k64096
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
$endgroup$
Using the clues provided, determine the 'password'.
501 — Two correct numbers in wrong places135 — One correct number in the right place483 — All numbers are wrong167 — One correct number in wrong place430 — One correct number in the wrong place
mastermind
mastermind
edited Feb 1 at 9:46
JonMark Perry
20k64096
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
edited Feb 1 at 9:46
JonMark Perry
20k64096
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
edited Feb 1 at 9:46
JonMark Perry
20k64096
edited Feb 1 at 9:46
JonMark Perry
20k64096
edited Feb 1 at 9:46
JonMark Perry
20k64096
20k64096
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
asked Jan 29 at 15:42
Brenda ScottBrenda Scott
713
713
1
$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 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 at 21:10
add a comment |
1
$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 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 at 21:10
1
1
$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 at 19:06
$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 at 19:06
1
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 at 21:10
$begingroup$
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 at 21:10
add a comment |
9 Answers
9
active
oldest
votes
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[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.
edited Jan 29 at 16:55
SteveV
6,2502632
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
$endgroup$
$begingroup$
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 at 22:20
add a comment |
$begingroup$
The following are my musings to find the answer:
Starting with 430 (Rule 5):
4??Eliminated by rule 5?4?Eliminated by rule 3??4Eliminated by rule 33??Eliminated by rule 3?3?Eliminated by rule 5??3Eliminated by rule 30??Remaining - Now Rule 6?0?Remaining - Now Rule 6??0Eliminated by rule 5
Starting with 501 (Rule 1):
5?1Eliminated by rule 151?Eliminated by rule 115?Eliminated by rule 61?5Eliminated by rule 2?15Eliminated by rule 6?51Eliminated by rule 150?Eliminated by rule 15?0Eliminated by rule 1?50Eliminated by rule 2?05Eliminated by rule 10?5Remaining - Now Rule 705?Eliminated by rule 2?01Eliminated by rule 1?10Eliminated by rule 21?0Eliminated by rule 610?Eliminated by rule 101?Eliminated by rule 20?1Eliminated by rule 1
Starting with 167 (Rule 4):
1??Elimated by rule 4?1?Elimated by rule 1??1Elimated by rule 76??Elimated by rule 7?6?Elimated by rule 4??6Elimated by rule 77??Elimated by rule 7?7?Remaining - Now Rule 8??7Elimated by rule 4
Combining the two new rules, 7 and 8 gives:
Result
075
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
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Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
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– North
Jan 30 at 3:06
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Excellent observation about the 3 to get it down to three rules!
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– par
Jan 30 at 20:31
add a comment |
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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.)
edited Jan 30 at 9:51
Flying_whale
1,909425
answered Jan 29 at 17:05
CrackItCrackIt
212
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add a comment |
$begingroup$
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
answered Jan 29 at 20:10
parpar
1113
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add a comment |
$begingroup$
I will share my process:
- Notes of format
!5 | !0 | !1means 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 | !1Possible: 0,1,5Conditions: !(0+1+5)
Step Two
!3!5 | !0!1!5 | !1!3Possible: 0,1,3,5Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)
Step Three
!5 | !0!1!5 | !1Possible: 0,1,5Impossible: 3,4,8 -> as definedConditions: !(0+1+5)
Step Four
!1!5 | !0!1!5!6 | !1!7 -> 1 is eliminatedPossible: 0,5,6,7Impossible: 1,3,4,8Conditions: !(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: 0,5,6,7Impossible: 1,3,4,8Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed
Conclusion
Possible: 0,5,7First Term: !5 -> 0 (since 7 is second term)Second Term: 7Third Term: 5
.0 7 5
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
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add a comment |
$begingroup$
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
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
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add a comment |
$begingroup$
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
answered Jan 30 at 13:24
LukStormsLukStorms
1013
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add a comment |
$begingroup$
The required password is
075 .
edited Feb 1 at 8:32
Dmihawk
2,493828
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
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please explain your approach to this answer. thanks and happy puzzling ;)
$endgroup$
– Omega Krypton
Feb 1 at 9:20
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
[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.
edited Jan 29 at 16:55
SteveV
6,2502632
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
$endgroup$
$begingroup$
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 at 22:20
add a comment |
$begingroup$
[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.
edited Jan 29 at 16:55
SteveV
6,2502632
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
$endgroup$
$begingroup$
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 at 22:20
add a comment |
$begingroup$
[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.
edited Jan 29 at 16:55
SteveV
6,2502632
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
$endgroup$
[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.
edited Jan 29 at 16:55
SteveV
6,2502632
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
edited Jan 29 at 16:55
SteveV
6,2502632
edited Jan 29 at 16:55
SteveV
6,2502632
edited Jan 29 at 16:55
SteveV
6,2502632
6,2502632
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
answered Jan 29 at 15:49
Ian MacDonaldIan MacDonald
11.2k2562
11.2k2562
$begingroup$
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 at 22:20
add a comment |
$begingroup$
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 at 22:20
$begingroup$
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 at 22:20
$begingroup$
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 at 22:20
add a comment |
$begingroup$
The following are my musings to find the answer:
Starting with 430 (Rule 5):
4??Eliminated by rule 5?4?Eliminated by rule 3??4Eliminated by rule 33??Eliminated by rule 3?3?Eliminated by rule 5??3Eliminated by rule 30??Remaining - Now Rule 6?0?Remaining - Now Rule 6??0Eliminated by rule 5
Starting with 501 (Rule 1):
5?1Eliminated by rule 151?Eliminated by rule 115?Eliminated by rule 61?5Eliminated by rule 2?15Eliminated by rule 6?51Eliminated by rule 150?Eliminated by rule 15?0Eliminated by rule 1?50Eliminated by rule 2?05Eliminated by rule 10?5Remaining - Now Rule 705?Eliminated by rule 2?01Eliminated by rule 1?10Eliminated by rule 21?0Eliminated by rule 610?Eliminated by rule 101?Eliminated by rule 20?1Eliminated by rule 1
Starting with 167 (Rule 4):
1??Elimated by rule 4?1?Elimated by rule 1??1Elimated by rule 76??Elimated by rule 7?6?Elimated by rule 4??6Elimated by rule 77??Elimated by rule 7?7?Remaining - Now Rule 8??7Elimated by rule 4
Combining the two new rules, 7 and 8 gives:
Result
075
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
$endgroup$
add a comment |
$begingroup$
The following are my musings to find the answer:
Starting with 430 (Rule 5):
4??Eliminated by rule 5?4?Eliminated by rule 3??4Eliminated by rule 33??Eliminated by rule 3?3?Eliminated by rule 5??3Eliminated by rule 30??Remaining - Now Rule 6?0?Remaining - Now Rule 6??0Eliminated by rule 5
Starting with 501 (Rule 1):
5?1Eliminated by rule 151?Eliminated by rule 115?Eliminated by rule 61?5Eliminated by rule 2?15Eliminated by rule 6?51Eliminated by rule 150?Eliminated by rule 15?0Eliminated by rule 1?50Eliminated by rule 2?05Eliminated by rule 10?5Remaining - Now Rule 705?Eliminated by rule 2?01Eliminated by rule 1?10Eliminated by rule 21?0Eliminated by rule 610?Eliminated by rule 101?Eliminated by rule 20?1Eliminated by rule 1
Starting with 167 (Rule 4):
1??Elimated by rule 4?1?Elimated by rule 1??1Elimated by rule 76??Elimated by rule 7?6?Elimated by rule 4??6Elimated by rule 77??Elimated by rule 7?7?Remaining - Now Rule 8??7Elimated by rule 4
Combining the two new rules, 7 and 8 gives:
Result
075
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
$endgroup$
add a comment |
$begingroup$
The following are my musings to find the answer:
Starting with 430 (Rule 5):
4??Eliminated by rule 5?4?Eliminated by rule 3??4Eliminated by rule 33??Eliminated by rule 3?3?Eliminated by rule 5??3Eliminated by rule 30??Remaining - Now Rule 6?0?Remaining - Now Rule 6??0Eliminated by rule 5
Starting with 501 (Rule 1):
5?1Eliminated by rule 151?Eliminated by rule 115?Eliminated by rule 61?5Eliminated by rule 2?15Eliminated by rule 6?51Eliminated by rule 150?Eliminated by rule 15?0Eliminated by rule 1?50Eliminated by rule 2?05Eliminated by rule 10?5Remaining - Now Rule 705?Eliminated by rule 2?01Eliminated by rule 1?10Eliminated by rule 21?0Eliminated by rule 610?Eliminated by rule 101?Eliminated by rule 20?1Eliminated by rule 1
Starting with 167 (Rule 4):
1??Elimated by rule 4?1?Elimated by rule 1??1Elimated by rule 76??Elimated by rule 7?6?Elimated by rule 4??6Elimated by rule 77??Elimated by rule 7?7?Remaining - Now Rule 8??7Elimated by rule 4
Combining the two new rules, 7 and 8 gives:
Result
075
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
$endgroup$
The following are my musings to find the answer:
Starting with 430 (Rule 5):
4??Eliminated by rule 5?4?Eliminated by rule 3??4Eliminated by rule 33??Eliminated by rule 3?3?Eliminated by rule 5??3Eliminated by rule 30??Remaining - Now Rule 6?0?Remaining - Now Rule 6??0Eliminated by rule 5
Starting with 501 (Rule 1):
5?1Eliminated by rule 151?Eliminated by rule 115?Eliminated by rule 61?5Eliminated by rule 2?15Eliminated by rule 6?51Eliminated by rule 150?Eliminated by rule 15?0Eliminated by rule 1?50Eliminated by rule 2?05Eliminated by rule 10?5Remaining - Now Rule 705?Eliminated by rule 2?01Eliminated by rule 1?10Eliminated by rule 21?0Eliminated by rule 610?Eliminated by rule 101?Eliminated by rule 20?1Eliminated by rule 1
Starting with 167 (Rule 4):
1??Elimated by rule 4?1?Elimated by rule 1??1Elimated by rule 76??Elimated by rule 7?6?Elimated by rule 4??6Elimated by rule 77??Elimated by rule 7?7?Remaining - Now Rule 8??7Elimated by rule 4
Combining the two new rules, 7 and 8 gives:
Result
075
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
answered Jan 29 at 19:02
Tom CarpenterTom Carpenter
436210
436210
add a comment |
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$begingroup$
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.
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
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Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
$endgroup$
– North
Jan 30 at 3:06
$begingroup$
Excellent observation about the 3 to get it down to three rules!
$endgroup$
– par
Jan 30 at 20:31
add a comment |
$begingroup$
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.
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
$endgroup$
$begingroup$
Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
$endgroup$
– North
Jan 30 at 3:06
$begingroup$
Excellent observation about the 3 to get it down to three rules!
$endgroup$
– par
Jan 30 at 20:31
add a comment |
$begingroup$
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.
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
$endgroup$
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.
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
answered Jan 30 at 2:54
TooLazyToLogInMyAccTooLazyToLogInMyAcc
211
211
$begingroup$
Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
$endgroup$
– North
Jan 30 at 3:06
$begingroup$
Excellent observation about the 3 to get it down to three rules!
$endgroup$
– par
Jan 30 at 20:31
add a comment |
$begingroup$
Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
$endgroup$
– North
Jan 30 at 3:06
$begingroup$
Excellent observation about the 3 to get it down to three rules!
$endgroup$
– par
Jan 30 at 20:31
$begingroup$
Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
$endgroup$
– North
Jan 30 at 3:06
$begingroup$
Welcome to Puzzling SE! I'm a little confused on your answer for 501, can you elaborate more?
$endgroup$
– North
Jan 30 at 3:06
$begingroup$
Excellent observation about the 3 to get it down to three rules!
$endgroup$
– par
Jan 30 at 20:31
$begingroup$
Excellent observation about the 3 to get it down to three rules!
$endgroup$
– par
Jan 30 at 20:31
add a comment |
$begingroup$
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.)
edited Jan 30 at 9:51
Flying_whale
1,909425
answered Jan 29 at 17:05
CrackItCrackIt
212
$endgroup$
add a comment |
$begingroup$
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.)
edited Jan 30 at 9:51
Flying_whale
1,909425
answered Jan 29 at 17:05
CrackItCrackIt
212
$endgroup$
add a comment |
$begingroup$
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.)
edited Jan 30 at 9:51
Flying_whale
1,909425
answered Jan 29 at 17:05
CrackItCrackIt
212
$endgroup$
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.)
edited Jan 30 at 9:51
Flying_whale
1,909425
answered Jan 29 at 17:05
CrackItCrackIt
212
edited Jan 30 at 9:51
Flying_whale
1,909425
edited Jan 30 at 9:51
Flying_whale
1,909425
edited Jan 30 at 9:51
Flying_whale
1,909425
1,909425
answered Jan 29 at 17:05
CrackItCrackIt
212
answered Jan 29 at 17:05
CrackItCrackIt
212
answered Jan 29 at 17:05
CrackItCrackIt
212
212
add a comment |
add a comment |
$begingroup$
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
answered Jan 29 at 20:10
parpar
1113
$endgroup$
add a comment |
$begingroup$
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
answered Jan 29 at 20:10
parpar
1113
$endgroup$
add a comment |
$begingroup$
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
answered Jan 29 at 20:10
parpar
1113
$endgroup$
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
answered Jan 29 at 20:10
parpar
1113
answered Jan 29 at 20:10
parpar
1113
answered Jan 29 at 20:10
parpar
1113
answered Jan 29 at 20:10
parpar
1113
1113
add a comment |
add a comment |
$begingroup$
I will share my process:
- Notes of format
!5 | !0 | !1means 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 | !1Possible: 0,1,5Conditions: !(0+1+5)
Step Two
!3!5 | !0!1!5 | !1!3Possible: 0,1,3,5Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)
Step Three
!5 | !0!1!5 | !1Possible: 0,1,5Impossible: 3,4,8 -> as definedConditions: !(0+1+5)
Step Four
!1!5 | !0!1!5!6 | !1!7 -> 1 is eliminatedPossible: 0,5,6,7Impossible: 1,3,4,8Conditions: !(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: 0,5,6,7Impossible: 1,3,4,8Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed
Conclusion
Possible: 0,5,7First Term: !5 -> 0 (since 7 is second term)Second Term: 7Third Term: 5
.0 7 5
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
$endgroup$
add a comment |
$begingroup$
I will share my process:
- Notes of format
!5 | !0 | !1means 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 | !1Possible: 0,1,5Conditions: !(0+1+5)
Step Two
!3!5 | !0!1!5 | !1!3Possible: 0,1,3,5Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)
Step Three
!5 | !0!1!5 | !1Possible: 0,1,5Impossible: 3,4,8 -> as definedConditions: !(0+1+5)
Step Four
!1!5 | !0!1!5!6 | !1!7 -> 1 is eliminatedPossible: 0,5,6,7Impossible: 1,3,4,8Conditions: !(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: 0,5,6,7Impossible: 1,3,4,8Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed
Conclusion
Possible: 0,5,7First Term: !5 -> 0 (since 7 is second term)Second Term: 7Third Term: 5
.0 7 5
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
$endgroup$
add a comment |
$begingroup$
I will share my process:
- Notes of format
!5 | !0 | !1means 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 | !1Possible: 0,1,5Conditions: !(0+1+5)
Step Two
!3!5 | !0!1!5 | !1!3Possible: 0,1,3,5Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)
Step Three
!5 | !0!1!5 | !1Possible: 0,1,5Impossible: 3,4,8 -> as definedConditions: !(0+1+5)
Step Four
!1!5 | !0!1!5!6 | !1!7 -> 1 is eliminatedPossible: 0,5,6,7Impossible: 1,3,4,8Conditions: !(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: 0,5,6,7Impossible: 1,3,4,8Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed
Conclusion
Possible: 0,5,7First Term: !5 -> 0 (since 7 is second term)Second Term: 7Third Term: 5
.0 7 5
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
$endgroup$
I will share my process:
- Notes of format
!5 | !0 | !1means 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 | !1Possible: 0,1,5Conditions: !(0+1+5)
Step Two
!3!5 | !0!1!5 | !1!3Possible: 0,1,3,5Conditions: !(1+3+5)!(0+1+5)!(1+3)!(1+5)!(3+5)
Step Three
!5 | !0!1!5 | !1Possible: 0,1,5Impossible: 3,4,8 -> as definedConditions: !(0+1+5)
Step Four
!1!5 | !0!1!5!6 | !1!7 -> 1 is eliminatedPossible: 0,5,6,7Impossible: 1,3,4,8Conditions: !(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: 0,5,6,7Impossible: 1,3,4,8Conditions: !(6+7) -> since the middle term requires 7 (!0!5!6), and !(6+7), 6 is removed
Conclusion
Possible: 0,5,7First Term: !5 -> 0 (since 7 is second term)Second Term: 7Third Term: 5
.0 7 5
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
edited Jan 30 at 4:13
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
answered Jan 29 at 20:22
Brandon DixonBrandon Dixon
1212
1212
add a comment |
add a comment |
$begingroup$
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
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
$endgroup$
add a comment |
$begingroup$
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
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
$endgroup$
add a comment |
$begingroup$
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
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
$endgroup$
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
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
edited Jan 29 at 19:10
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
answered Jan 29 at 19:05
J.-S. PayetteJ.-S. Payette
213
213
add a comment |
add a comment |
$begingroup$
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
answered Jan 30 at 13:24
LukStormsLukStorms
1013
$endgroup$
add a comment |
$begingroup$
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
answered Jan 30 at 13:24
LukStormsLukStorms
1013
$endgroup$
add a comment |
$begingroup$
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
answered Jan 30 at 13:24
LukStormsLukStorms
1013
$endgroup$
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
answered Jan 30 at 13:24
LukStormsLukStorms
1013
answered Jan 30 at 13:24
LukStormsLukStorms
1013
answered Jan 30 at 13:24
LukStormsLukStorms
1013
answered Jan 30 at 13:24
LukStormsLukStorms
1013
1013
add a comment |
add a comment |
$begingroup$
The required password is
075 .
edited Feb 1 at 8:32
Dmihawk
2,493828
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
$endgroup$
$begingroup$
please explain your approach to this answer. thanks and happy puzzling ;)
$endgroup$
– Omega Krypton
Feb 1 at 9:20
add a comment |
$begingroup$
The required password is
075 .
edited Feb 1 at 8:32
Dmihawk
2,493828
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
$endgroup$
$begingroup$
please explain your approach to this answer. thanks and happy puzzling ;)
$endgroup$
– Omega Krypton
Feb 1 at 9:20
add a comment |
$begingroup$
The required password is
075 .
edited Feb 1 at 8:32
Dmihawk
2,493828
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
$endgroup$
The required password is
075 .
edited Feb 1 at 8:32
Dmihawk
2,493828
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
edited Feb 1 at 8:32
Dmihawk
2,493828
edited Feb 1 at 8:32
Dmihawk
2,493828
edited Feb 1 at 8:32
Dmihawk
2,493828
2,493828
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
answered Feb 1 at 8:24
Manish MasiwalManish Masiwal
1
1
$begingroup$
please explain your approach to this answer. thanks and happy puzzling ;)
$endgroup$
– Omega Krypton
Feb 1 at 9:20
add a comment |
$begingroup$
please explain your approach to this answer. thanks and happy puzzling ;)
$endgroup$
– Omega Krypton
Feb 1 at 9:20
$begingroup$
please explain your approach to this answer. thanks and happy puzzling ;)
$endgroup$
– Omega Krypton
Feb 1 at 9:20
$begingroup$
please explain your approach to this answer. thanks and happy puzzling ;)
$endgroup$
– Omega Krypton
Feb 1 at 9:20
add a comment |
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$begingroup$
+1 for reminding me that I haven't played Mastermind in a while...such a fun game.
$endgroup$
– Brandon_J
Jan 29 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 at 21:10