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Determine the number of solutions in \(x\) of the congruence \(64x\equiv 2\pmod {66}\) such that \(0< x\le 100\).

RektTheNoob  Aug 2, 2018
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Determine the number of solutions in \(x\) of the congruence \(64x\equiv 2\pmod {66}\) such that \(0< x\le 100\).

\(\text{Determine the number of solutions in $x$ of the congruence $64x\equiv 2\pmod {66} \\$such that $0< x\le 100$.}\)

 

\(\begin{array}{|rcll|} \hline 64x &\equiv& 2\pmod {66} \quad \text{or} \quad 64x= 2 + 66n \\\\ 64x&=& 2 + 66n \quad | \quad : 2 \\ 32x&=& 1 + 33n \\\\ \mathbf{x}& \mathbf{=}& \mathbf{\dfrac{1 + 33n}{32}} \\\\ x &=& \dfrac{1 + 32n + n }{32} \\\\ x &=& \dfrac{32n+ (1 + n) }{32} \\\\ x &=& n+ \underbrace{\dfrac{1 + n}{32}}_{=a} \\\\ a &=& \dfrac{1 + n}{32} \\\\ 32a &=& 1+n \\ \mathbf{n}& \mathbf{=}& \mathbf{32a-1} \\\\ \hline \mathbf{x}& \mathbf{=}& \mathbf{\dfrac{1 + 33n}{32}} \quad | \quad \mathbf{n=32a-1} \\\\ x &=& \dfrac{1 + 33(32a-1)}{32} \\\\ x &=& \dfrac{1 + 33\cdot 32a- 33}{32} \\\\ x &=& \dfrac{-32 + 33\cdot 32a}{32} \\\\ x &=& -1 +33a \\ \mathbf{x}& \mathbf{=}& \mathbf{-1 +33a} \quad a\in N \\ \hline \end{array}\)

 

\(\begin{array}{|r|r|c|} \hline a & x = -1 +33a \\ \hline 1 & 32 & 0< x\le 100 & \checkmark \\ \hline 2 & 65 & 0< x\le 100 & \checkmark \\ \hline 3 & 98 & 0< x\le 100 & \checkmark \\ \hline 4 & 131 && \text{no solution } \\ \hline \gt 3 & && \text{no solution } \\ \hline \end{array} \)

 

\(x=32,65,98\)

 

laugh

heureka  Aug 2, 2018

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