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# I don't understand this problem

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If $$4x\equiv 8\pmod{20}$$ and $$3x\equiv 16\pmod{20}$$, then what is the remainder when $$x^2$$ is divided by 20?

Jul 13, 2020

#1
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4, i think.

Jul 13, 2020
#2
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solve (4 x) mod 20 = 8
(3 x) mod 20 = 16 for x

x =20*n +12, where n =0, 1, 2, 3......etc.

The smallest x = 12

12^2 mod 20 =144 mod 20 =4 - remainder.

Jul 13, 2020
#3
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If

$$4x\equiv 8\pmod{20}$$ and $$3x\equiv 16\pmod{20}$$, then what is the remainder when $$x^2$$ is divided by $$20$$?

$$\begin{array}{|lrcll|} \hline (1) & 4x &\equiv& 8\pmod{20} \\ & 4x &=& 8 + 20n \qquad n\in \mathbb{Z}\\ (2) & 3x &\equiv& 16\pmod{20} \\ & 3x &=& 16+20m \qquad m\in \mathbb{Z}\\ \hline (1)-(2): & 4x-3x &=& 8-16 + 20(n-m) \\ & x &=& -8 + 20(n-m) \\ & x &\equiv& -8\pmod{20} \\ & x &\equiv& -8+20 \pmod{20} \\ &\mathbf{ x } &\equiv& \mathbf{ 12 \pmod{20} } \\ \hline & x &=& 12+20z \qquad z\in \mathbb{Z}\\ & x^2 &=& (12+20z)^2 \\ & x^2 &=& 144+ 2*12*20z + 20*20z^2 \\ & x^2 &\equiv& 144 \pmod{20} \\ & x^2 &\equiv& 144-7*20 \pmod{20} \\ & \mathbf{ x^2 } &\equiv& \mathbf{ 4 \pmod{20}} \\ \hline \end{array}$$

Jul 14, 2020