Chinese Remainder Theorem Modular I have this problem, I need to find the smallest possible solution
$$x \equiv 3  \pmod{10}, \\
x \equiv 11 \pmod{13}, \\
x \equiv 15 \pmod{17}.$$
I used Chinese remainder theorem and found that: $x_1=3, x_2=11$ and $x_3= -45$??
The solutions would be congruent to 
$v= (13\times17)\times3 + (10\times17)\times11 + (10\times13)\times (-45) = -3317 \pmod{10\times13\times17}$. Is that the smaller solution? Is $x_3= -45$ correct? 
 A: Let's see: $10 \times 13 \times 17 = 2210$. Then we find $$1 \times \frac{2210}{10} = 221 \equiv 1 \pmod{10},$$ $$1 \times \frac{2210}{13} = 170 \equiv 1 \pmod{13},$$ $$14 \times \frac{2210}{10} = 1820 \equiv 1 \pmod{17}.$$ Then we add up $$3 \times 1 \times 221 + 11 \times 1 \times 170 + 15 \times 14 \times 130 = 663 + 1870 + 27300 = 29833.$$ Now, $29833$ is a solution of the congruences above, but it's not the smallest positive solution. It's the remainder of this: $$29833 = \frac{28730}{2210} + 1103.$$ Now check that $$1103 \equiv 3 \pmod{10},$$ $$1103 \equiv 11 \pmod{13},$$ $$1103 \equiv 15 \pmod{17}.$$ 
So you did your calculations kind of different from what I would have done, but you can still get at the right answer just the same. Note that $1103 - (-3317) = 4420$, which is twice $2210$. This means that $-3317 + 4420 = 1103$ still gives you the smallest positive answer which I found with a different $x_3$.
A: First, we deal with the last two, which state
$x\equiv-2\pmod{13}$
$x\equiv-2\pmod{17}$
So by the Chinese Remainder Theorem,
$x\equiv-2\pmod{221}.$ We have to solve this for
$x\equiv3\pmod{10}.$
First, we set it to $x\equiv219\pmod{221}$ and we need to add enough $221$'s (note that $219$ is the least integer that satisfies the last two equations) so that we finally have the first modular equation satisfied.
This leads to $x\equiv219+221+221+221+221=119+884=\boxed{1103}\pmod{2210}.$ The smallest positive integer that satisfies this is $\boxed{1103},$ evidently.
