# To determine: The smallest nonnegative integer x that satisfies the given system of congruences. x\equiv 1\pmod {4} x\equiv 8\pmod {9} x\equiv 10\pmod{25}

To determine: The smallest nonnegative integer x that satisfies the given system of congruences.
$$\displaystyle{x}\equiv{1}\pm{o}{d}{\left\lbrace{4}\right\rbrace}$$
$$\displaystyle{x}\equiv{8}\pm{o}{d}{\left\lbrace{9}\right\rbrace}$$
$$\displaystyle{x}\equiv{10}\pm{o}{d}{\left\lbrace{25}\right\rbrace}$$

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Bentley Leach

$$\displaystyle{x}\equiv{1}\pm{o}{d}{\left\lbrace{4}\right\rbrace}$$
$$\displaystyle{x}\equiv{8}\pm{o}{d}{\left\lbrace{9}\right\rbrace}$$
$$\displaystyle{x}\equiv{10}\pm{o}{d}{\left\lbrace{25}\right\rbrace}$$
First let us solve the first two congruences.
$$\displaystyle{x}\equiv{1}\pm{o}{d}{\left\lbrace{4}\right\rbrace}$$
$$\displaystyle{x}\equiv{8}\pm{o}{d}{\left\lbrace{9}\right\rbrace}$$
We see that the solution x is unique modulo $$4.9=36$$.
Now, $$9-4(2)=1$$.
Thus,
$$x=1.9-8.4(2)$$
$$x=9-64$$
$$x=-55$$
$$\displaystyle{x}=-{19}\pm{o}{d}{\left\lbrace{36}\right\rbrace}$$
$$\displaystyle{x}={17}\pm{o}{d}{\left\lbrace{36}\right\rbrace}$$
Therefore, $$\displaystyle{x}={17}\pm{o}{d}{\left\lbrace{36}\right\rbrace}$$.
Now, let us solve the below equations.
$$\displaystyle{x}\equiv{17}\pm{o}{d}{\left\lbrace{36}\right\rbrace}$$
$$\displaystyle{x}\equiv{10}\pm{o}{d}{\left\lbrace{25}\right\rbrace}$$
We see that the solution x is unique modulo $$36.25=900$$.
Now, $$25(13)-36(9)=1$$.
Thus,
$$x=17.25(13)-10.36(9)$$
$$x=5525-3240$$
$$x=2285$$
$$\displaystyle{x}={485}\pm{o}{d}{\left\lbrace{900}\right\rbrace}$$
Therefore, $$x = 485.$$