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hello every one,
just wanted to ask one question :-
which number , when divides by 2,3,4,5,6, gives remainder 1, when divided by 7 gives 0 .
very confusing hehe
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61 fits the first bit, but you've confused me with the part where it says 'when divided by 7 gives 0'.
61/7 = 8[sup]5[/sup]/[sub]7[/sub]
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when divided by 7 gives remainder 0.
The smallest such number is 301. The next is 721. Then 1141, 1561. 1981, etc.
In fact, any number of the form 420k+301 (k = 0, 1, 2, 3 ) will work.
Last edited by JaneFairfax (2007-05-20 05:12:07)
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Wow , it's strange , that I got 420k-161 by extended Euclidean algorithm for Chinese remainder theorem.
Numbers are the essence of the Universe
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I dont think the algorithm for the Chinese remainder theorem works because 2, 3, 4, 5, 6 and 7 are not pairwise coprime. You probably need to use the method of successive substitution.
In my case, I did it using Excel.
Last edited by JaneFairfax (2007-05-20 21:07:58)
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I turned it into the number divided by has 3,4,5 remainder 1 .
Last edited by Stanley_Marsh (2007-05-20 15:46:42)
Numbers are the essence of the Universe
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You need to consider 4,5,6, not 3,4,5. Since any number that is congruent to 1 (mod 6) is also congruent to both 1 (mod 2) and 1 (mod 3), you can leave out 2 and 3.
So you want to find a number x ≡ 1 (mod 4) ≡ 1 (mod 5) ≡ 1 (mod 6). Then x−1 is divisible by 4, 5 and 6, so x−1 is divisible by lcm(4,5,6) = 60. Thus x ≡ 1 (mod 60) ≡ 0 (mod 7). Now you can apply the algorithm for the Chinese remainder theorem as 60 and 7 are coprime.
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3,4,5 should work as well though. You can ignore 2 by your reasoning, and then because any number congruent to both 1 (mod 3) and 1 (mod 4) is also congruent to 1 (mod 6), you can leave that one out as well.
Why did the vector cross the road?
It wanted to be normal.
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Yeah, all right, 3,4,5 will work as well (my mistake). The important thing is that the lcm must be 60.
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Hmm where do you learn about modulus? in university?
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Its modulo (not modulus) arithmetic. I first learned it at school. Its not that complicated or that abstract either; the real world is full of modulo-arithmetic applications.
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Modulus is a very confusing word. From my experience, a modulus is a function, written f(x) = |x|, which gets rid of the negative sign of x is there is one.
So if x ≥ 0 then |x| = x and if x < 0 then |x| = -x.
However, other people use abs(x) to define that (abs being short for absolute value) and modulus to describe the thing that Jane was doing. But we use modulo for that.
But whatever you call it, it's basically a way of describing the remainder of something.
a (mod b) is worked out by the remainder of a ÷ b.
As a random example, 14 (mod 3) = 2, because 14 ÷ 3 = 4 r2.
I've just been introduced to the definition this semester, and I'm in my first year of Uni in Britain. Other places might teach it earlier or later though.
Why did the vector cross the road?
It wanted to be normal.
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