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Don't really understand this question. If this is asking to find an $x$ for each mod then the answer would be just be $x+m \dots$. If this is asking to find an $x$ that satisfies all mods, then this cant be since every $x$ mod $1$ would be $0$ and no $x$ would satisfy for all. Any help?

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  • $\begingroup$ Every $x$ is also congruent to $1$ modulo $1$. There is no smallest integer, but there is a smallest positive integer, namely $1$. $\endgroup$ – André Nicolas Nov 7 '13 at 3:49
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Every integer $x$ is also congruent to $1$ modulo $1$. There is no smallest integer satisfying your conditions, but there is a smallest positive integer, namely $1$. For certainly every integer from $1$ to $10$ divides $1-1$.

To see that there is no smallest integer, let $m$ be the lcm of the numbers $1$ to $10$. Then the integers that satisfy your condition are precisely the integers of the form $1+km$. Take $k=-1$.

If you want the smallest integer $x$ greater than $1$, the answer is $1+m$. Note that $m=2^3\cdot 3^2\cdot 5\cdot 7=2520$.

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  • $\begingroup$ It is this, or the lcm(2, 3, 4, 5, 6, 7, 8, 9, 10) + 1. I only state this because I'm not exactly sure the question intended to be answered by the trivial case, 1. $\endgroup$ – David Nov 7 '13 at 3:55
  • $\begingroup$ what integers divide 1-1? no integers divide 0. $\endgroup$ – Al Jebr Nov 7 '13 at 3:57
  • $\begingroup$ At the current time, the OP says smallest integer, and the answer says there is no smallest integer. If smallest positive integer is the actual question, the answer is $1$. If the actual question says smallest integer $\gt 1$, then the answer contains enough information to solve the problem. $\endgroup$ – André Nicolas Nov 7 '13 at 3:58
  • $\begingroup$ @Juan: Every integer divides $0$. For recall that we say that $a$ divides $b$ if there is an integer $x$ such that $ax=b$. If $b=0$, we can take $x=0$. $\endgroup$ – André Nicolas Nov 7 '13 at 4:00
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HINT:

As the remainder is same $=1,$ so $x-1$ needs to be divisible by $1,2,\cdots,9,10$

Now, if $1,2,\cdots,9,10$ divides $x-1,$ the later must be divisible by lcm $(1,2,\cdots,9,10)$

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