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by divya23 » Wed Jun 15, 2011 6:01 am
we have to find the remainder when x is divided by 6.
wen x is divided by 2 remainder is 1 and when x is divided by 3 remainder is 0

here wen we solve this we get the value of x as 3,9,15
now both 9 and 15 when divided by 6 give the remainder as 3 but it is not posible wen x = 3

how is then this stmnt sufficient

please reply
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by Anurag@Gurome » Wed Jun 15, 2011 6:09 am
divya23 wrote:...now both 9 and 15 when divided by 6 give the remainder as 3 but it is not posible wen x = 3
When 3 is divided by 6, the remainder is 3.
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by Ashley@VeritasPrep » Wed Jun 15, 2011 11:30 am
Just adding one thing here, something that may be useful in terms of generalizing this concept to other problems. If you are ever told that "when x is divided by 2, the remainder is 1," that's the exact same thing as being told that "x is odd." (And of course, being told that "when x is divided by n, the remainder is 0" is the same as being told that "x is a multiple of n".) So in this particular problem, we are effectively being told that "x is an odd multiple of 3." Multiples of *any* odd number n will flip-flop odd to even to odd to even to odd and so on. The even ones will always be multiples of 2n; the odd ones will always yield a remainder of n when divided by 2n.

For instance, considering the remainders when multiples of 7
7 14 21 28 35 42 49 56 63 70 77 84 ...
are divided by 14, I'll get
7 0 7 0 7 0 7 0 7 0 7 0 ...

Works every time :)
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by jkati » Thu Jun 16, 2011 10:04 pm
here we have that x is divisible by 3 and when divided by 2 we will get the remainder 1 so for example choose the numbers for this condition = 3, 9,15,21,27,33,39.....


when we divide these numbers by 6 in every case we will get the remainder as 3.

so the answer should be 3
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