Crack the 5 digits code
$begingroup$
I created this "crack the code" puzzle.
I hope the members here like it.
mathematics logical-deduction no-computers mastermind
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
$endgroup$
add a comment |
$begingroup$
I created this "crack the code" puzzle.
I hope the members here like it.
mathematics logical-deduction no-computers mastermind
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
$endgroup$
$begingroup$
Are A, B, C, D, E different?
$endgroup$
– Wen1now
Jun 29 '17 at 4:57
$begingroup$
Bulls and Cows?
$endgroup$
– Weijun Zhou
Jan 27 '18 at 7:20
add a comment |
$begingroup$
I created this "crack the code" puzzle.
I hope the members here like it.
mathematics logical-deduction no-computers mastermind
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
$endgroup$
I created this "crack the code" puzzle.
I hope the members here like it.
mathematics logical-deduction no-computers mastermind
mathematics logical-deduction no-computers mastermind
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
edited Jun 29 '17 at 4:53
Wen1now
6,79822579
6,79822579
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
asked Jun 29 '17 at 4:30
Nurnani TatiNurnani Tati
51039
51039
$begingroup$
Are A, B, C, D, E different?
$endgroup$
– Wen1now
Jun 29 '17 at 4:57
$begingroup$
Bulls and Cows?
$endgroup$
– Weijun Zhou
Jan 27 '18 at 7:20
add a comment |
$begingroup$
Are A, B, C, D, E different?
$endgroup$
– Wen1now
Jun 29 '17 at 4:57
$begingroup$
Bulls and Cows?
$endgroup$
– Weijun Zhou
Jan 27 '18 at 7:20
$begingroup$
Are A, B, C, D, E different?
$endgroup$
– Wen1now
Jun 29 '17 at 4:57
$begingroup$
Are A, B, C, D, E different?
$endgroup$
– Wen1now
Jun 29 '17 at 4:57
$begingroup$
Bulls and Cows?
$endgroup$
– Weijun Zhou
Jan 27 '18 at 7:20
$begingroup$
Bulls and Cows?
$endgroup$
– Weijun Zhou
Jan 27 '18 at 7:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the number
Thus the second last digit must be a
2
since A+B+C+D+E=10D+E.
Now the third digit must therefore be a
4 or 6
and the sum of the first four digits must be
20
Therefore we must have
5762?
where
0 or 8 as the question mark are perfectly valid.
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
$endgroup$
$begingroup$
Final step; the equation is false if D is the first of your options, so you should be able to exclude that option if I'm not completely mistaken.
$endgroup$
– Jakob Pamp Bengtsson
Jun 29 '17 at 10:22
$begingroup$
@JakobPampBengtsson Wait, what?
$endgroup$
– Wen1now
Jun 29 '17 at 10:34
add a comment |
$begingroup$
Solving the last portion to Wen1now's response-
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the code
Looking at 2, we can further conclude that the third digit is
6
Now the last two digits, looking at the equation (A+B+C+D+E=D*10+E)
We can solve for D to be 2
Therefore the sum of the first four digits is
20
Solving that the last digit is
0
Code
57620
edited 2 hours ago
F1Krazy
5,80712152
answered 2 hours ago
Sasha WongSasha Wong
112
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
This is basically cut and paste of Wen1now’s answer with a final assertion that is not explained at all. How did you determine the correctness of the choice you made? How did you rule out the alternative? Without something substantive to show why your answer is more, rather than less, complete than the existing answer, this is no better than duplicative.
$endgroup$
– Rubio♦
1 hour ago
add a comment |
protected by Community♦ 1 hour ago
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the number
Thus the second last digit must be a
2
since A+B+C+D+E=10D+E.
Now the third digit must therefore be a
4 or 6
and the sum of the first four digits must be
20
Therefore we must have
5762?
where
0 or 8 as the question mark are perfectly valid.
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
$endgroup$
$begingroup$
Final step; the equation is false if D is the first of your options, so you should be able to exclude that option if I'm not completely mistaken.
$endgroup$
– Jakob Pamp Bengtsson
Jun 29 '17 at 10:22
$begingroup$
@JakobPampBengtsson Wait, what?
$endgroup$
– Wen1now
Jun 29 '17 at 10:34
add a comment |
$begingroup$
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the number
Thus the second last digit must be a
2
since A+B+C+D+E=10D+E.
Now the third digit must therefore be a
4 or 6
and the sum of the first four digits must be
20
Therefore we must have
5762?
where
0 or 8 as the question mark are perfectly valid.
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
$endgroup$
$begingroup$
Final step; the equation is false if D is the first of your options, so you should be able to exclude that option if I'm not completely mistaken.
$endgroup$
– Jakob Pamp Bengtsson
Jun 29 '17 at 10:22
$begingroup$
@JakobPampBengtsson Wait, what?
$endgroup$
– Wen1now
Jun 29 '17 at 10:34
add a comment |
$begingroup$
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the number
Thus the second last digit must be a
2
since A+B+C+D+E=10D+E.
Now the third digit must therefore be a
4 or 6
and the sum of the first four digits must be
20
Therefore we must have
5762?
where
0 or 8 as the question mark are perfectly valid.
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
$endgroup$
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the number
Thus the second last digit must be a
2
since A+B+C+D+E=10D+E.
Now the third digit must therefore be a
4 or 6
and the sum of the first four digits must be
20
Therefore we must have
5762?
where
0 or 8 as the question mark are perfectly valid.
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
answered Jun 29 '17 at 5:04
Wen1nowWen1now
6,79822579
6,79822579
$begingroup$
Final step; the equation is false if D is the first of your options, so you should be able to exclude that option if I'm not completely mistaken.
$endgroup$
– Jakob Pamp Bengtsson
Jun 29 '17 at 10:22
$begingroup$
@JakobPampBengtsson Wait, what?
$endgroup$
– Wen1now
Jun 29 '17 at 10:34
add a comment |
$begingroup$
Final step; the equation is false if D is the first of your options, so you should be able to exclude that option if I'm not completely mistaken.
$endgroup$
– Jakob Pamp Bengtsson
Jun 29 '17 at 10:22
$begingroup$
@JakobPampBengtsson Wait, what?
$endgroup$
– Wen1now
Jun 29 '17 at 10:34
$begingroup$
Final step; the equation is false if D is the first of your options, so you should be able to exclude that option if I'm not completely mistaken.
$endgroup$
– Jakob Pamp Bengtsson
Jun 29 '17 at 10:22
$begingroup$
Final step; the equation is false if D is the first of your options, so you should be able to exclude that option if I'm not completely mistaken.
$endgroup$
– Jakob Pamp Bengtsson
Jun 29 '17 at 10:22
$begingroup$
@JakobPampBengtsson Wait, what?
$endgroup$
– Wen1now
Jun 29 '17 at 10:34
$begingroup$
@JakobPampBengtsson Wait, what?
$endgroup$
– Wen1now
Jun 29 '17 at 10:34
add a comment |
$begingroup$
Solving the last portion to Wen1now's response-
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the code
Looking at 2, we can further conclude that the third digit is
6
Now the last two digits, looking at the equation (A+B+C+D+E=D*10+E)
We can solve for D to be 2
Therefore the sum of the first four digits is
20
Solving that the last digit is
0
Code
57620
edited 2 hours ago
F1Krazy
5,80712152
answered 2 hours ago
Sasha WongSasha Wong
112
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
This is basically cut and paste of Wen1now’s answer with a final assertion that is not explained at all. How did you determine the correctness of the choice you made? How did you rule out the alternative? Without something substantive to show why your answer is more, rather than less, complete than the existing answer, this is no better than duplicative.
$endgroup$
– Rubio♦
1 hour ago
add a comment |
$begingroup$
Solving the last portion to Wen1now's response-
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the code
Looking at 2, we can further conclude that the third digit is
6
Now the last two digits, looking at the equation (A+B+C+D+E=D*10+E)
We can solve for D to be 2
Therefore the sum of the first four digits is
20
Solving that the last digit is
0
Code
57620
edited 2 hours ago
F1Krazy
5,80712152
answered 2 hours ago
Sasha WongSasha Wong
112
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
This is basically cut and paste of Wen1now’s answer with a final assertion that is not explained at all. How did you determine the correctness of the choice you made? How did you rule out the alternative? Without something substantive to show why your answer is more, rather than less, complete than the existing answer, this is no better than duplicative.
$endgroup$
– Rubio♦
1 hour ago
add a comment |
$begingroup$
Solving the last portion to Wen1now's response-
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the code
Looking at 2, we can further conclude that the third digit is
6
Now the last two digits, looking at the equation (A+B+C+D+E=D*10+E)
We can solve for D to be 2
Therefore the sum of the first four digits is
20
Solving that the last digit is
0
Code
57620
edited 2 hours ago
F1Krazy
5,80712152
answered 2 hours ago
Sasha WongSasha Wong
112
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Solving the last portion to Wen1now's response-
Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that
1 and 3 are not in the code
Therefore two of
5, 7, 9
are in the number. But it can't be
7 and 9 so 5 must be the first digit.
Looking at 2 we deduce that
9 cannot be correct otherwise it contradicts 3
Hence
7 is correct so the first two digits are 5 7
Therefore looking at 1,
1, 3, 4, 9 are not in the code
Looking at 2, we can further conclude that the third digit is
6
Now the last two digits, looking at the equation (A+B+C+D+E=D*10+E)
We can solve for D to be 2
Therefore the sum of the first four digits is
20
Solving that the last digit is
0
Code
57620
edited 2 hours ago
F1Krazy
5,80712152
answered 2 hours ago
Sasha WongSasha Wong
112
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
F1Krazy
5,80712152
edited 2 hours ago
F1Krazy
5,80712152
edited 2 hours ago
F1Krazy
5,80712152
5,80712152
answered 2 hours ago
Sasha WongSasha Wong
112
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Sasha WongSasha Wong
112
answered 2 hours ago
Sasha WongSasha Wong
112
112
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sasha Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
This is basically cut and paste of Wen1now’s answer with a final assertion that is not explained at all. How did you determine the correctness of the choice you made? How did you rule out the alternative? Without something substantive to show why your answer is more, rather than less, complete than the existing answer, this is no better than duplicative.
$endgroup$
– Rubio♦
1 hour ago
add a comment |
$begingroup$
This is basically cut and paste of Wen1now’s answer with a final assertion that is not explained at all. How did you determine the correctness of the choice you made? How did you rule out the alternative? Without something substantive to show why your answer is more, rather than less, complete than the existing answer, this is no better than duplicative.
$endgroup$
– Rubio♦
1 hour ago
$begingroup$
This is basically cut and paste of Wen1now’s answer with a final assertion that is not explained at all. How did you determine the correctness of the choice you made? How did you rule out the alternative? Without something substantive to show why your answer is more, rather than less, complete than the existing answer, this is no better than duplicative.
$endgroup$
– Rubio♦
1 hour ago
$begingroup$
This is basically cut and paste of Wen1now’s answer with a final assertion that is not explained at all. How did you determine the correctness of the choice you made? How did you rule out the alternative? Without something substantive to show why your answer is more, rather than less, complete than the existing answer, this is no better than duplicative.
$endgroup$
– Rubio♦
1 hour ago
add a comment |
protected by Community♦ 1 hour ago
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
$begingroup$
Are A, B, C, D, E different?
$endgroup$
– Wen1now
Jun 29 '17 at 4:57
$begingroup$
Bulls and Cows?
$endgroup$
– Weijun Zhou
Jan 27 '18 at 7:20