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Prove that the smallest integer producing remainders 2,4,6,1 when divided by 3,5,7,11 respectively is 419.

Here's what I did. x/3 --->Remainder 2. x/5 --->Remainder 4. x/7 --->Remainder 6. x/11 ---->Remainder 1. Notice that adding 1 to x makes things perfectly divisible by 3,5,7,11. So, x+1=LCM(3,5,7,11)=>x=1154, which is not the smallest integer.

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    $\begingroup$ Is it $2,4,6,1$ оr $2,4,6,10$ if it's the first then $x+1$ is divisible by $3,5,7$ but not by $11$. $\endgroup$ – kingW3 Jun 28 '17 at 19:21
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You correctly noted that $x+1$ is divisible by $3,5,7$ but it isn't with $11$ because $x$ gives $1$ as the remainder and $x+1$ gives $2$ as a remainder not $0$.

So it is divisible by $105$ and it must give a remainder of $2$ when divided by $11$ by trial and error it's easy to notice that the first such number is $420$ hence $x=419$.

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To prove it, you just need to compute the remainders to show that $419$ works. Then note that solutions recur every $3\cdot 5 \cdot 7 \cdot 11=1155$ so there is no positive integer less than $419$ that works. As stated, the question is wrong because $419-1155=-736$ is smaller than $419$ and it works, too.

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