What would be the least total no. which give the remainder 1,2 and 3 when divided by 7,9 and 11?
1,2 and 3 when divided by 7,9 and 11?
hence
x= 1 mod 7
x= 2 mod 9
x =3 mod 11
now 7 ,9 and 11 are pair wise coprime
x = 1 a1b1 + 2 a2b2 + 3 a3b3 mod 7*9*11(or 693)
where a1 = 9*11 = 99
a2 = 7 * 11 = 77
a3 = 7 * 9 = 63
and
a1b1 = 1 mod 7
a2b2 = 1 mod 9
a3b3 = 1 mod 11
a1 = 99 so b1 = 99 mod 7 or 1 mod 7 so b1 = 1
77b2 mod 9 = 1 so 5b2 mod 9 = 1 so b2= 2 (as 5*2 = 10 mod 9)
63b3 = 1 mod 11 or 8b3 =1 mod 11 so b3 = 7 as 7 * 8 = 56 = 1 mod 11
So x = 1 * 99 * 1 + 2 * 77 * 2 + 3 * 63 * 7 mod 693 = 1730 mod 693
= 344 mod 693
So x if of the form 693n + 344 or lowest x = 344
7 comments:
wowwwwwww! great blog :D
Big brother, you mis-spelled Chinese!
:D
the spelling is corrected (Chinese)
I tried your method with remainders = 1,2,4 and moduli (5,7,9) and it doesn't work! What you illustrated must be an exemption I think.
here is the solution (steps are not explained same as above)
X =1 mod 5
X = 2 mod 7
X = 4 mod 9
5 , 7, 9 are pairwise co-prime
5 * 7 * 9 = 315
Now 7 * 9 = 63 = 3 mod 5 so inversion wrt 5 = 2 as 3 * 2 = 6 = 1 mod 5
5 * 9 = 45 = 3 mod 7 so inversion = 5
5 * 7 = 35 = - 1 mod 9 so inversion = -1 or 8 mod 9
(applying above logic)
So x= 1 * 63 * 2 + 2 * 45 * 5 + 4 * 35 * 8 mod 315
= 126 + 450 + 1120 mod 315
= 1696 mod 315 = 121
121 satisfies the condition
find the number which divided by 7,9,13,15,17 give remainder 1,2,3,4,5..i m not getting the answer ...
good example..thank you
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