5 Digit Code Puzzle
$begingroup$
We have a 5 digit code, any place can take 1 2 or 3.
How do I find where the 2 is placed (if placed) knowing this criteria?
1) If the 1st digit is not a 3 then the 2nd is.
2) If the 1st digit is 3 then the 3rd digit is 2.
3) If 2nd digit is 3 and 4th is 2 then 5th is 1.
4) If the 3rd digit is not 2 then the 4th is 2.
5) If 3rd digit is not 2 then the 5th digit is not 1.
mathematics logical-deduction calculation-puzzle
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
$endgroup$
add a comment |
$begingroup$
We have a 5 digit code, any place can take 1 2 or 3.
How do I find where the 2 is placed (if placed) knowing this criteria?
1) If the 1st digit is not a 3 then the 2nd is.
2) If the 1st digit is 3 then the 3rd digit is 2.
3) If 2nd digit is 3 and 4th is 2 then 5th is 1.
4) If the 3rd digit is not 2 then the 4th is 2.
5) If 3rd digit is not 2 then the 5th digit is not 1.
mathematics logical-deduction calculation-puzzle
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
$endgroup$
$begingroup$
Welcome to Puzzling! Where is the source of this puzzle? Please note that questions from other places must be credited correctly (and make sure you have permission to use them!)
$endgroup$
– Deusovi♦
Dec 15 '18 at 14:04
$begingroup$
Hello there,the puzzle is made by me so i am the source(?),my inspiration started after starting computer science logic.
$endgroup$
– Agaeus
Dec 15 '18 at 14:13
1
$begingroup$
May I suggest using the letters A,B and C instead of numbers for the code, that would make them a lot easier to distinguish from the ubiquitous "Nth digit" numbers.
$endgroup$
– Bass
Dec 15 '18 at 16:57
$begingroup$
I think real answer is: Any place!! ;).
$endgroup$
– shA.t
Dec 16 '18 at 12:40
add a comment |
$begingroup$
We have a 5 digit code, any place can take 1 2 or 3.
How do I find where the 2 is placed (if placed) knowing this criteria?
1) If the 1st digit is not a 3 then the 2nd is.
2) If the 1st digit is 3 then the 3rd digit is 2.
3) If 2nd digit is 3 and 4th is 2 then 5th is 1.
4) If the 3rd digit is not 2 then the 4th is 2.
5) If 3rd digit is not 2 then the 5th digit is not 1.
mathematics logical-deduction calculation-puzzle
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
$endgroup$
We have a 5 digit code, any place can take 1 2 or 3.
How do I find where the 2 is placed (if placed) knowing this criteria?
1) If the 1st digit is not a 3 then the 2nd is.
2) If the 1st digit is 3 then the 3rd digit is 2.
3) If 2nd digit is 3 and 4th is 2 then 5th is 1.
4) If the 3rd digit is not 2 then the 4th is 2.
5) If 3rd digit is not 2 then the 5th digit is not 1.
mathematics logical-deduction calculation-puzzle
mathematics logical-deduction calculation-puzzle
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
edited Dec 15 '18 at 16:22
JonMark Perry
18k63786
18k63786
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
asked Dec 15 '18 at 13:58
AgaeusAgaeus
233
233
$begingroup$
Welcome to Puzzling! Where is the source of this puzzle? Please note that questions from other places must be credited correctly (and make sure you have permission to use them!)
$endgroup$
– Deusovi♦
Dec 15 '18 at 14:04
$begingroup$
Hello there,the puzzle is made by me so i am the source(?),my inspiration started after starting computer science logic.
$endgroup$
– Agaeus
Dec 15 '18 at 14:13
1
$begingroup$
May I suggest using the letters A,B and C instead of numbers for the code, that would make them a lot easier to distinguish from the ubiquitous "Nth digit" numbers.
$endgroup$
– Bass
Dec 15 '18 at 16:57
$begingroup$
I think real answer is: Any place!! ;).
$endgroup$
– shA.t
Dec 16 '18 at 12:40
add a comment |
$begingroup$
Welcome to Puzzling! Where is the source of this puzzle? Please note that questions from other places must be credited correctly (and make sure you have permission to use them!)
$endgroup$
– Deusovi♦
Dec 15 '18 at 14:04
$begingroup$
Hello there,the puzzle is made by me so i am the source(?),my inspiration started after starting computer science logic.
$endgroup$
– Agaeus
Dec 15 '18 at 14:13
1
$begingroup$
May I suggest using the letters A,B and C instead of numbers for the code, that would make them a lot easier to distinguish from the ubiquitous "Nth digit" numbers.
$endgroup$
– Bass
Dec 15 '18 at 16:57
$begingroup$
I think real answer is: Any place!! ;).
$endgroup$
– shA.t
Dec 16 '18 at 12:40
$begingroup$
Welcome to Puzzling! Where is the source of this puzzle? Please note that questions from other places must be credited correctly (and make sure you have permission to use them!)
$endgroup$
– Deusovi♦
Dec 15 '18 at 14:04
$begingroup$
Welcome to Puzzling! Where is the source of this puzzle? Please note that questions from other places must be credited correctly (and make sure you have permission to use them!)
$endgroup$
– Deusovi♦
Dec 15 '18 at 14:04
$begingroup$
Hello there,the puzzle is made by me so i am the source(?),my inspiration started after starting computer science logic.
$endgroup$
– Agaeus
Dec 15 '18 at 14:13
$begingroup$
Hello there,the puzzle is made by me so i am the source(?),my inspiration started after starting computer science logic.
$endgroup$
– Agaeus
Dec 15 '18 at 14:13
1
1
$begingroup$
May I suggest using the letters A,B and C instead of numbers for the code, that would make them a lot easier to distinguish from the ubiquitous "Nth digit" numbers.
$endgroup$
– Bass
Dec 15 '18 at 16:57
$begingroup$
May I suggest using the letters A,B and C instead of numbers for the code, that would make them a lot easier to distinguish from the ubiquitous "Nth digit" numbers.
$endgroup$
– Bass
Dec 15 '18 at 16:57
$begingroup$
I think real answer is: Any place!! ;).
$endgroup$
– shA.t
Dec 16 '18 at 12:40
$begingroup$
I think real answer is: Any place!! ;).
$endgroup$
– shA.t
Dec 16 '18 at 12:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The only thing we can say for sure is that
the $3^{rd}$ digit is a $2$. The other positions can be anything.
Proof:
If the $1^{st}$ digit is a $3$ then we are done.
If not, then digit:$2$ is a $3$. Now assume digit:$3$ is not a $2$, therefore digit:$4$ is a $2$ (by (4)), and so digit:$5$ is a $1$ (by (3)).
However (5) tells us that if digit:$2$=3 and digit:$4$=2, then digit:$5$ is not $1$, which is a contradiction, so digit:$3$=2.
For the other digit positions, these codes, $13231,23213,31211,32222$, contain a $1$, $2$ or $3$ in every other position.
Using some JavaScript:
s=;
for (a=1;a<4;a++)
for (b=1;b<4;b++)
for (c=1;c<4;c++)
for (d=1;d<4;d++)
for (e=1;e<4;e++) {
if (a!=3 && b!=3) continue;
if (a==3 && c!=2) continue;
if (b==3 && d==2 && e!=1) continue;
if (c!=2 && d!=2) continue;
if (c!=2 && e==1) continue;
s.push(''+a+b+c+d+e);
}
console.log(s);
we get:
$13211,13212,13213,13221,13231,\13232,13233,23211,23212,23213,\ 23221,23231,23232,23233,31211,\31212,31213,31221,31222,31223,\ 31231,31232,31233,32211,32212,\32213,32221,32222,32223,32231,\ 32232,32233,33211,33212,33213,\33221,33231,33232,33233$
which is $39$ codes in total.
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
You mean rot13("vs gur svefg qvtvg vf guerr") at the start of the proof? May I also suggest putting the proof first, then picking two examples to show rot13(gjb pna nyfb tb va nal bgure cbfvgvba), then the program as a postscript.
$endgroup$
– deep thought
Dec 15 '18 at 18:26
add a comment |
$begingroup$
Using the corrected program by @JonMarkPerry,
i.e. by changing if (c!=3 && d!=2) continue; to if (c!=2 && d!=2) continue;
We get
13211, 13212, 13213, 13221, 13231, 13232, 13233, 23211, 23212, 23213, 23221, 23231, 23232, 23233, 31211, 31212, 31213, 31221, 31222, 31223, 31231, 31232, 31233, 32211, 32212, 32213, 32221, 32222, 32223, 32231, 32232, 32233, 33211, 33212, 33213, 33221, 33231, 33232, 33233
Which means
the third position is always 2, and others may be also.
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
The only thing we can say for sure is that
the $3^{rd}$ digit is a $2$. The other positions can be anything.
Proof:
If the $1^{st}$ digit is a $3$ then we are done.
If not, then digit:$2$ is a $3$. Now assume digit:$3$ is not a $2$, therefore digit:$4$ is a $2$ (by (4)), and so digit:$5$ is a $1$ (by (3)).
However (5) tells us that if digit:$2$=3 and digit:$4$=2, then digit:$5$ is not $1$, which is a contradiction, so digit:$3$=2.
For the other digit positions, these codes, $13231,23213,31211,32222$, contain a $1$, $2$ or $3$ in every other position.
Using some JavaScript:
s=;
for (a=1;a<4;a++)
for (b=1;b<4;b++)
for (c=1;c<4;c++)
for (d=1;d<4;d++)
for (e=1;e<4;e++) {
if (a!=3 && b!=3) continue;
if (a==3 && c!=2) continue;
if (b==3 && d==2 && e!=1) continue;
if (c!=2 && d!=2) continue;
if (c!=2 && e==1) continue;
s.push(''+a+b+c+d+e);
}
console.log(s);
we get:
$13211,13212,13213,13221,13231,\13232,13233,23211,23212,23213,\ 23221,23231,23232,23233,31211,\31212,31213,31221,31222,31223,\ 31231,31232,31233,32211,32212,\32213,32221,32222,32223,32231,\ 32232,32233,33211,33212,33213,\33221,33231,33232,33233$
which is $39$ codes in total.
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
You mean rot13("vs gur svefg qvtvg vf guerr") at the start of the proof? May I also suggest putting the proof first, then picking two examples to show rot13(gjb pna nyfb tb va nal bgure cbfvgvba), then the program as a postscript.
$endgroup$
– deep thought
Dec 15 '18 at 18:26
add a comment |
$begingroup$
The only thing we can say for sure is that
the $3^{rd}$ digit is a $2$. The other positions can be anything.
Proof:
If the $1^{st}$ digit is a $3$ then we are done.
If not, then digit:$2$ is a $3$. Now assume digit:$3$ is not a $2$, therefore digit:$4$ is a $2$ (by (4)), and so digit:$5$ is a $1$ (by (3)).
However (5) tells us that if digit:$2$=3 and digit:$4$=2, then digit:$5$ is not $1$, which is a contradiction, so digit:$3$=2.
For the other digit positions, these codes, $13231,23213,31211,32222$, contain a $1$, $2$ or $3$ in every other position.
Using some JavaScript:
s=;
for (a=1;a<4;a++)
for (b=1;b<4;b++)
for (c=1;c<4;c++)
for (d=1;d<4;d++)
for (e=1;e<4;e++) {
if (a!=3 && b!=3) continue;
if (a==3 && c!=2) continue;
if (b==3 && d==2 && e!=1) continue;
if (c!=2 && d!=2) continue;
if (c!=2 && e==1) continue;
s.push(''+a+b+c+d+e);
}
console.log(s);
we get:
$13211,13212,13213,13221,13231,\13232,13233,23211,23212,23213,\ 23221,23231,23232,23233,31211,\31212,31213,31221,31222,31223,\ 31231,31232,31233,32211,32212,\32213,32221,32222,32223,32231,\ 32232,32233,33211,33212,33213,\33221,33231,33232,33233$
which is $39$ codes in total.
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
You mean rot13("vs gur svefg qvtvg vf guerr") at the start of the proof? May I also suggest putting the proof first, then picking two examples to show rot13(gjb pna nyfb tb va nal bgure cbfvgvba), then the program as a postscript.
$endgroup$
– deep thought
Dec 15 '18 at 18:26
add a comment |
$begingroup$
The only thing we can say for sure is that
the $3^{rd}$ digit is a $2$. The other positions can be anything.
Proof:
If the $1^{st}$ digit is a $3$ then we are done.
If not, then digit:$2$ is a $3$. Now assume digit:$3$ is not a $2$, therefore digit:$4$ is a $2$ (by (4)), and so digit:$5$ is a $1$ (by (3)).
However (5) tells us that if digit:$2$=3 and digit:$4$=2, then digit:$5$ is not $1$, which is a contradiction, so digit:$3$=2.
For the other digit positions, these codes, $13231,23213,31211,32222$, contain a $1$, $2$ or $3$ in every other position.
Using some JavaScript:
s=;
for (a=1;a<4;a++)
for (b=1;b<4;b++)
for (c=1;c<4;c++)
for (d=1;d<4;d++)
for (e=1;e<4;e++) {
if (a!=3 && b!=3) continue;
if (a==3 && c!=2) continue;
if (b==3 && d==2 && e!=1) continue;
if (c!=2 && d!=2) continue;
if (c!=2 && e==1) continue;
s.push(''+a+b+c+d+e);
}
console.log(s);
we get:
$13211,13212,13213,13221,13231,\13232,13233,23211,23212,23213,\ 23221,23231,23232,23233,31211,\31212,31213,31221,31222,31223,\ 31231,31232,31233,32211,32212,\32213,32221,32222,32223,32231,\ 32232,32233,33211,33212,33213,\33221,33231,33232,33233$
which is $39$ codes in total.
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
$endgroup$
The only thing we can say for sure is that
the $3^{rd}$ digit is a $2$. The other positions can be anything.
Proof:
If the $1^{st}$ digit is a $3$ then we are done.
If not, then digit:$2$ is a $3$. Now assume digit:$3$ is not a $2$, therefore digit:$4$ is a $2$ (by (4)), and so digit:$5$ is a $1$ (by (3)).
However (5) tells us that if digit:$2$=3 and digit:$4$=2, then digit:$5$ is not $1$, which is a contradiction, so digit:$3$=2.
For the other digit positions, these codes, $13231,23213,31211,32222$, contain a $1$, $2$ or $3$ in every other position.
Using some JavaScript:
s=;
for (a=1;a<4;a++)
for (b=1;b<4;b++)
for (c=1;c<4;c++)
for (d=1;d<4;d++)
for (e=1;e<4;e++) {
if (a!=3 && b!=3) continue;
if (a==3 && c!=2) continue;
if (b==3 && d==2 && e!=1) continue;
if (c!=2 && d!=2) continue;
if (c!=2 && e==1) continue;
s.push(''+a+b+c+d+e);
}
console.log(s);
we get:
$13211,13212,13213,13221,13231,\13232,13233,23211,23212,23213,\ 23221,23231,23232,23233,31211,\31212,31213,31221,31222,31223,\ 31231,31232,31233,32211,32212,\32213,32221,32222,32223,32231,\ 32232,32233,33211,33212,33213,\33221,33231,33232,33233$
which is $39$ codes in total.
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
edited Dec 15 '18 at 18:44
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
answered Dec 15 '18 at 16:11
JonMark PerryJonMark Perry
18k63786
18k63786
$begingroup$
You mean rot13("vs gur svefg qvtvg vf guerr") at the start of the proof? May I also suggest putting the proof first, then picking two examples to show rot13(gjb pna nyfb tb va nal bgure cbfvgvba), then the program as a postscript.
$endgroup$
– deep thought
Dec 15 '18 at 18:26
add a comment |
$begingroup$
You mean rot13("vs gur svefg qvtvg vf guerr") at the start of the proof? May I also suggest putting the proof first, then picking two examples to show rot13(gjb pna nyfb tb va nal bgure cbfvgvba), then the program as a postscript.
$endgroup$
– deep thought
Dec 15 '18 at 18:26
$begingroup$
You mean rot13("vs gur svefg qvtvg vf guerr") at the start of the proof? May I also suggest putting the proof first, then picking two examples to show rot13(gjb pna nyfb tb va nal bgure cbfvgvba), then the program as a postscript.
$endgroup$
– deep thought
Dec 15 '18 at 18:26
$begingroup$
You mean rot13("vs gur svefg qvtvg vf guerr") at the start of the proof? May I also suggest putting the proof first, then picking two examples to show rot13(gjb pna nyfb tb va nal bgure cbfvgvba), then the program as a postscript.
$endgroup$
– deep thought
Dec 15 '18 at 18:26
add a comment |
$begingroup$
Using the corrected program by @JonMarkPerry,
i.e. by changing if (c!=3 && d!=2) continue; to if (c!=2 && d!=2) continue;
We get
13211, 13212, 13213, 13221, 13231, 13232, 13233, 23211, 23212, 23213, 23221, 23231, 23232, 23233, 31211, 31212, 31213, 31221, 31222, 31223, 31231, 31232, 31233, 32211, 32212, 32213, 32221, 32222, 32223, 32231, 32232, 32233, 33211, 33212, 33213, 33221, 33231, 33232, 33233
Which means
the third position is always 2, and others may be also.
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
$endgroup$
add a comment |
$begingroup$
Using the corrected program by @JonMarkPerry,
i.e. by changing if (c!=3 && d!=2) continue; to if (c!=2 && d!=2) continue;
We get
13211, 13212, 13213, 13221, 13231, 13232, 13233, 23211, 23212, 23213, 23221, 23231, 23232, 23233, 31211, 31212, 31213, 31221, 31222, 31223, 31231, 31232, 31233, 32211, 32212, 32213, 32221, 32222, 32223, 32231, 32232, 32233, 33211, 33212, 33213, 33221, 33231, 33232, 33233
Which means
the third position is always 2, and others may be also.
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
$endgroup$
add a comment |
$begingroup$
Using the corrected program by @JonMarkPerry,
i.e. by changing if (c!=3 && d!=2) continue; to if (c!=2 && d!=2) continue;
We get
13211, 13212, 13213, 13221, 13231, 13232, 13233, 23211, 23212, 23213, 23221, 23231, 23232, 23233, 31211, 31212, 31213, 31221, 31222, 31223, 31231, 31232, 31233, 32211, 32212, 32213, 32221, 32222, 32223, 32231, 32232, 32233, 33211, 33212, 33213, 33221, 33231, 33232, 33233
Which means
the third position is always 2, and others may be also.
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
$endgroup$
Using the corrected program by @JonMarkPerry,
i.e. by changing if (c!=3 && d!=2) continue; to if (c!=2 && d!=2) continue;
We get
13211, 13212, 13213, 13221, 13231, 13232, 13233, 23211, 23212, 23213, 23221, 23231, 23232, 23233, 31211, 31212, 31213, 31221, 31222, 31223, 31231, 31232, 31233, 32211, 32212, 32213, 32221, 32222, 32223, 32231, 32232, 32233, 33211, 33212, 33213, 33221, 33231, 33232, 33233
Which means
the third position is always 2, and others may be also.
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
answered Dec 15 '18 at 16:59
SteveVSteveV
5,6052629
5,6052629
add a comment |
add a comment |
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Welcome to Puzzling! Where is the source of this puzzle? Please note that questions from other places must be credited correctly (and make sure you have permission to use them!)
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– Deusovi♦
Dec 15 '18 at 14:04
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Hello there,the puzzle is made by me so i am the source(?),my inspiration started after starting computer science logic.
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– Agaeus
Dec 15 '18 at 14:13
1
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May I suggest using the letters A,B and C instead of numbers for the code, that would make them a lot easier to distinguish from the ubiquitous "Nth digit" numbers.
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– Bass
Dec 15 '18 at 16:57
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I think real answer is: Any place!! ;).
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– shA.t
Dec 16 '18 at 12:40