# The smallest positive integer x that satisfies x\equiv 3\pmod 5 x\equiv 5\pmod 7 x\equiv 7\pmod 11

The smallest positive integer x that satisfies
$$\displaystyle{x}\equiv{3}\pm{o}{d}{5}$$
$$\displaystyle{x}\equiv{5}\pm{o}{d}{7}$$
$$\displaystyle{x}\equiv{7}\pm{o}{d}{11}$$

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$$\displaystyle{x}\equiv{3}\pm{o}{d}{5}$$
$$\displaystyle{x}\equiv{5}\pm{o}{d}{7}$$
We know gcd(5,7)=1.
Also, 1=(3)5-2(7).
Thus,
x=5(3)5-3.2(7)
=75-42
$$\displaystyle={33}\pm{o}{d}{35}$$
Now, let us find the solution for the congruences given below:
$$\displaystyle{x}={33}\pm{o}{d}{\left\lbrace{35}\right\rbrace}$$
$$\displaystyle{x}\equiv{7}\pm{o}{d}{\left\lbrace{11}\right\rbrace}$$
We know gcd(35,11)=1.
Also, 1=-5(35)+16(11).
Thus,
x=7(-5)(35)+33(16)(11)
=-1225+5808
=4583