A problem on chinese remainder theorem (CSIR NET DEC 2015)
$begingroup$
Which of the following intervals contains an integer satisfying following three congruences
$$x=2pmod5\
x=3pmod7\
x=4pmod{11}$$
$a) [401,600] \ b)[601, 800] \ c)[801,1000] \ d)[1001,1200]$
(CSIR NET 2015 Dec)
I tried this question and I got answer but it is not in the option.
I applied Chinese remainder theorem.
$$x=2pmod5\
x=3pmod7\
x=4pmod {11}$$
$$N_1=7times11=77\
N_2=5times11=55\
N_3=7times5=35$$
$77x=1pmod5implies b_1=3\
55x=1pmod7implies b_2=6\
35x=1pmod {11} implies b_3=6$
then,
$x=2times77times3+3times55times6+6times35times4=2292$
This answer is not in the option.
If my work is wrong please correct it.
number-theory congruences chinese-remainder-theorem
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
asked Mar 3 '16 at 13:59
piepie
435
$endgroup$
add a comment |
$begingroup$
Which of the following intervals contains an integer satisfying following three congruences
$$x=2pmod5\
x=3pmod7\
x=4pmod{11}$$
$a) [401,600] \ b)[601, 800] \ c)[801,1000] \ d)[1001,1200]$
(CSIR NET 2015 Dec)
I tried this question and I got answer but it is not in the option.
I applied Chinese remainder theorem.
$$x=2pmod5\
x=3pmod7\
x=4pmod {11}$$
$$N_1=7times11=77\
N_2=5times11=55\
N_3=7times5=35$$
$77x=1pmod5implies b_1=3\
55x=1pmod7implies b_2=6\
35x=1pmod {11} implies b_3=6$
then,
$x=2times77times3+3times55times6+6times35times4=2292$
This answer is not in the option.
If my work is wrong please correct it.
number-theory congruences chinese-remainder-theorem
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
asked Mar 3 '16 at 13:59
piepie
435
$endgroup$
add a comment |
$begingroup$
Which of the following intervals contains an integer satisfying following three congruences
$$x=2pmod5\
x=3pmod7\
x=4pmod{11}$$
$a) [401,600] \ b)[601, 800] \ c)[801,1000] \ d)[1001,1200]$
(CSIR NET 2015 Dec)
I tried this question and I got answer but it is not in the option.
I applied Chinese remainder theorem.
$$x=2pmod5\
x=3pmod7\
x=4pmod {11}$$
$$N_1=7times11=77\
N_2=5times11=55\
N_3=7times5=35$$
$77x=1pmod5implies b_1=3\
55x=1pmod7implies b_2=6\
35x=1pmod {11} implies b_3=6$
then,
$x=2times77times3+3times55times6+6times35times4=2292$
This answer is not in the option.
If my work is wrong please correct it.
number-theory congruences chinese-remainder-theorem
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
asked Mar 3 '16 at 13:59
piepie
435
$endgroup$
Which of the following intervals contains an integer satisfying following three congruences
$$x=2pmod5\
x=3pmod7\
x=4pmod{11}$$
$a) [401,600] \ b)[601, 800] \ c)[801,1000] \ d)[1001,1200]$
(CSIR NET 2015 Dec)
I tried this question and I got answer but it is not in the option.
I applied Chinese remainder theorem.
$$x=2pmod5\
x=3pmod7\
x=4pmod {11}$$
$$N_1=7times11=77\
N_2=5times11=55\
N_3=7times5=35$$
$77x=1pmod5implies b_1=3\
55x=1pmod7implies b_2=6\
35x=1pmod {11} implies b_3=6$
then,
$x=2times77times3+3times55times6+6times35times4=2292$
This answer is not in the option.
If my work is wrong please correct it.
number-theory congruences chinese-remainder-theorem
number-theory congruences chinese-remainder-theorem
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
asked Mar 3 '16 at 13:59
piepie
435
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
asked Mar 3 '16 at 13:59
piepie
435
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
edited Nov 25 '17 at 16:58
देवेन्द्र prasad
1383
1383
asked Mar 3 '16 at 13:59
piepie
435
asked Mar 3 '16 at 13:59
piepie
435
asked Mar 3 '16 at 13:59
piepie
435
435
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You know that the answer you get applying the CRT is not a unique integer, right?
It is only unique modulo $5cdot 7cdot 11$.
In particular, $752$ and $1137$ are solutions.
I'll leave you with the tasks of deciding if this is complete, and how you will use it to answer your question.
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
I didn't get what you are saying. how will get answer 752 and 1137 .please give explained answe.
$endgroup$
– pie
Mar 3 '16 at 16:36
$begingroup$
@FairoosaNavas it sounds like you need to look at the Chinese remainder theorem again,for example sample at Wikipedia. It clearly says that the solutions are equivalent mod $385$ in your case, which is how I've given you these two other solutions
$endgroup$
– rschwieb
Mar 4 '16 at 4:07
$begingroup$
thank you@ rschwieb, Igot ans.......
$endgroup$
– pie
Mar 4 '16 at 12:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You know that the answer you get applying the CRT is not a unique integer, right?
It is only unique modulo $5cdot 7cdot 11$.
In particular, $752$ and $1137$ are solutions.
I'll leave you with the tasks of deciding if this is complete, and how you will use it to answer your question.
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
I didn't get what you are saying. how will get answer 752 and 1137 .please give explained answe.
$endgroup$
– pie
Mar 3 '16 at 16:36
$begingroup$
@FairoosaNavas it sounds like you need to look at the Chinese remainder theorem again,for example sample at Wikipedia. It clearly says that the solutions are equivalent mod $385$ in your case, which is how I've given you these two other solutions
$endgroup$
– rschwieb
Mar 4 '16 at 4:07
$begingroup$
thank you@ rschwieb, Igot ans.......
$endgroup$
– pie
Mar 4 '16 at 12:01
add a comment |
$begingroup$
You know that the answer you get applying the CRT is not a unique integer, right?
It is only unique modulo $5cdot 7cdot 11$.
In particular, $752$ and $1137$ are solutions.
I'll leave you with the tasks of deciding if this is complete, and how you will use it to answer your question.
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
I didn't get what you are saying. how will get answer 752 and 1137 .please give explained answe.
$endgroup$
– pie
Mar 3 '16 at 16:36
$begingroup$
@FairoosaNavas it sounds like you need to look at the Chinese remainder theorem again,for example sample at Wikipedia. It clearly says that the solutions are equivalent mod $385$ in your case, which is how I've given you these two other solutions
$endgroup$
– rschwieb
Mar 4 '16 at 4:07
$begingroup$
thank you@ rschwieb, Igot ans.......
$endgroup$
– pie
Mar 4 '16 at 12:01
add a comment |
$begingroup$
You know that the answer you get applying the CRT is not a unique integer, right?
It is only unique modulo $5cdot 7cdot 11$.
In particular, $752$ and $1137$ are solutions.
I'll leave you with the tasks of deciding if this is complete, and how you will use it to answer your question.
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
$endgroup$
You know that the answer you get applying the CRT is not a unique integer, right?
It is only unique modulo $5cdot 7cdot 11$.
In particular, $752$ and $1137$ are solutions.
I'll leave you with the tasks of deciding if this is complete, and how you will use it to answer your question.
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
answered Mar 3 '16 at 14:11
rschwiebrschwieb
108k12104253
108k12104253
$begingroup$
I didn't get what you are saying. how will get answer 752 and 1137 .please give explained answe.
$endgroup$
– pie
Mar 3 '16 at 16:36
$begingroup$
@FairoosaNavas it sounds like you need to look at the Chinese remainder theorem again,for example sample at Wikipedia. It clearly says that the solutions are equivalent mod $385$ in your case, which is how I've given you these two other solutions
$endgroup$
– rschwieb
Mar 4 '16 at 4:07
$begingroup$
thank you@ rschwieb, Igot ans.......
$endgroup$
– pie
Mar 4 '16 at 12:01
add a comment |
$begingroup$
I didn't get what you are saying. how will get answer 752 and 1137 .please give explained answe.
$endgroup$
– pie
Mar 3 '16 at 16:36
$begingroup$
@FairoosaNavas it sounds like you need to look at the Chinese remainder theorem again,for example sample at Wikipedia. It clearly says that the solutions are equivalent mod $385$ in your case, which is how I've given you these two other solutions
$endgroup$
– rschwieb
Mar 4 '16 at 4:07
$begingroup$
thank you@ rschwieb, Igot ans.......
$endgroup$
– pie
Mar 4 '16 at 12:01
$begingroup$
I didn't get what you are saying. how will get answer 752 and 1137 .please give explained answe.
$endgroup$
– pie
Mar 3 '16 at 16:36
$begingroup$
I didn't get what you are saying. how will get answer 752 and 1137 .please give explained answe.
$endgroup$
– pie
Mar 3 '16 at 16:36
$begingroup$
@FairoosaNavas it sounds like you need to look at the Chinese remainder theorem again,for example sample at Wikipedia. It clearly says that the solutions are equivalent mod $385$ in your case, which is how I've given you these two other solutions
$endgroup$
– rschwieb
Mar 4 '16 at 4:07
$begingroup$
@FairoosaNavas it sounds like you need to look at the Chinese remainder theorem again,for example sample at Wikipedia. It clearly says that the solutions are equivalent mod $385$ in your case, which is how I've given you these two other solutions
$endgroup$
– rschwieb
Mar 4 '16 at 4:07
$begingroup$
thank you@ rschwieb, Igot ans.......
$endgroup$
– pie
Mar 4 '16 at 12:01
$begingroup$
thank you@ rschwieb, Igot ans.......
$endgroup$
– pie
Mar 4 '16 at 12:01
add a comment |
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