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Number theory Difficulty 5.6 AIME, harder Find the answer

2. 求下列模为素数幂的同余方程的解:
(i) x3+x2+10x+1≡0( mod 33)x^{3}+x^{2}+10 x+1 \equiv 0\left(\bmod 3^{3}\right);
(ii) x3+25x+3≡0( mod 33)x^{3}+25 x+3 \equiv 0\left(\bmod 3^{3}\right);
(iii) x3−5x2+3≡0( mod 34)x^{3}-5 x^{2}+3 \equiv 0\left(\bmod 3^{4}\right);
(iv) x5+x4+1≡0( mod 34)x^{5}+x^{4}+1 \equiv 0\left(\bmod 3^{4}\right);
(v) x3−2x+4≡0( mod 53)x^{3}-2 x+4 \equiv 0\left(\bmod 5^{3}\right);
(vi) x3+x+57≡0( mod 53)x^{3}+x+57 \equiv 0\left(\bmod 5^{3}\right);
(vii) x3+x2−4≡0( mod 73)x^{3}+x^{2}-4 \equiv 0\left(\bmod 7^{3}\right);
(viii) x3+x2−5≡0( mod 73)x^{3}+x^{2}-5 \equiv 0\left(\bmod 7^{3}\right);
(ix) x2+5x+13≡0( mod 33)x^{2}+5 x+13 \equiv 0\left(\bmod 3^{3}\right);
(x) x2+5x+13≡0( mod 34)x^{2}+5 x+13 \equiv 0\left(\bmod 3^{4}\right);
(xi) x2≡3( mod 113)x^{2} \equiv 3\left(\bmod 11^{3}\right);
(xii) x2≡−2( mod 194)x^{2} \equiv-2\left(\bmod 19^{4}\right).

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Solution

2. (i) x≡−10( mod 33)x \equiv-10\left(\bmod 3^{3}\right);
(ii) x≡−12( mod 33)x \equiv-12\left(\bmod 3^{3}\right);
(iii) x≡11( mod 34)x \equiv 11\left(\bmod 3^{4}\right);
(iv) 无解;
(v) x≡−56,−2+25⋅j,8+25j( mod x \equiv-56,-2+25 \cdot j, 8+25 j(\bmod
53),j=0,±1,±2\left.5^{3}\right), j=0, \pm 1, \pm 2;
(vi) x≡4( mod 53)x \equiv 4\left(\bmod 5^{3}\right);
(vii) 无解;
(viii) x≡23( mod 73)x \equiv 23\left(\bmod 7^{3}\right);
(ix) x≡2+9j( mod 33),j=0,±1x \equiv 2+9 j\left(\bmod 3^{3}\right), j=0, \pm 1 ；
(x) 无解;
(xi) x≡±578( mod 113)x \equiv \pm 578\left(\bmod 11^{3}\right);
(xii) x≡±(2590+4⋅193)≡±30026( mod 194)x \equiv \pm\left(2590+4 \cdot 19^{3}\right) \equiv \pm 30026\left(\bmod 19^{4}\right).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.