Re: [sage-support] Chinese Remainder Theorem
Thank you.
On 23 September 2011 10:38, D. S. McNeil dsm...@gmail.com wrote:
On Fri, Sep 23, 2011 at 12:39 AM, Santanu Sarkar
sarkar.santanu@gmail.com wrote:
I want to find integer such that
x= 1 mod 3
x=2 mod 5
x=3 mod 7
like this system of congruences using Chinese Remainder Theorem.
In Sage, crt() function takes only 4 argument.
sage: help(CRT)
crt(a, b, m=None, n=None)
Returns a solution to a Chinese Remainder Theorem problem.
INPUT:
- ``a``, ``b`` - two residues (elements of some ring for which
extended gcd is available), or two lists, one of residues and
one of moduli.
[...]
If ``a`` and ``b`` are lists, returns a simultaneous solution to
the congruences `x\equiv a_i\pmod{b_i}`, if one exists.
.. SEEALSO::
- :func:`CRT_list`
sage: CRT([1,2,3],[3,5,7])
52
sage: x = CRT([1,2,3],[3,5,7])
sage: x % 3, x % 5, x % 7
(1, 2, 3)
Doug
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[sage-support] Chinese Remainder Theorem
I want to find integer such that x= 1 mod 3 x=2 mod 5 x=3 mod 7 like this system of congruences using Chinese Remainder Theorem. In Sage, crt() function takes only 4 argument. -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
Re: [sage-support] Chinese Remainder Theorem
On Fri, Sep 23, 2011 at 12:39 AM, Santanu Sarkar
sarkar.santanu@gmail.com wrote:
I want to find integer such that
x= 1 mod 3
x=2 mod 5
x=3 mod 7
like this system of congruences using Chinese Remainder Theorem.
In Sage, crt() function takes only 4 argument.
sage: help(CRT)
crt(a, b, m=None, n=None)
Returns a solution to a Chinese Remainder Theorem problem.
INPUT:
- ``a``, ``b`` - two residues (elements of some ring for which
extended gcd is available), or two lists, one of residues and
one of moduli.
[...]
If ``a`` and ``b`` are lists, returns a simultaneous solution to
the congruences `x\equiv a_i\pmod{b_i}`, if one exists.
.. SEEALSO::
- :func:`CRT_list`
sage: CRT([1,2,3],[3,5,7])
52
sage: x = CRT([1,2,3],[3,5,7])
sage: x % 3, x % 5, x % 7
(1, 2, 3)
Doug
--
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URL: http://www.sagemath.org
[sage-support] Chinese Remainder Theorem
I want to find x using Chinese Remainder Theorem such that x=a_1 mod b_1 x=a_2 mod b_2 x=a_3 mod b_3 x=a_4 mod b_4 -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
Re: [sage-support] Chinese Remainder Theorem
On May 15, 2011, at 22:30 , Santanu Sarkar wrote: I want to find x using Chinese Remainder Theorem such that x=a_1 mod b_1 x=a_2 mod b_2 x=a_3 mod b_3 x=a_4 mod b_4 Try sage: CRT? HTH Justin -- Justin C. Walker, Curmudgeon-at-Large () The ASCII Ribbon Campaign /\ Help Cure HTML Email -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
