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Chinese Remainder Thm/Simult. congruences: X ≡ 2 mod 3, X ≡ 2 mod 4, X ≡ 1 mod 5
"Find an integer X such that
X ≡ 2 mod 3,
X ≡ 2 mod 4, and
X ≡ 1 mod 5."
So this is what I did for the first two congruences:
x = 3a + 2
x = 4b +2
3a + 2 = 4b +2
Possible value for a is 1 and b is 3/4.
Therefore x = 3(1) + 2 = 4(3/4) + 2 = 5 (mod 12).
So x = 12c + 5
And the 4th congruence has equation:
x = 5d +1
12c + 5 = 5d + 1
5d  12c = 4.
Possible value of d is 4 and c is 4/3.
12(4/3) + 5 = 5(4) + 1 = 21 (mod 60)
But 21 is clearly wrong.
It seems like I even got the calculation for the first two congruences wrong.
I have seen this method work before so I'm not sure where I've gone wrong here.
Any help?
Last edited by sktsasus; 01292018 at 09:24 PM.
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