A) \[0\]
B) \[\pm \text{ }5\]
C) \[\pm \text{ }1\]
D) \[\pm \text{ }7\]
Correct Answer: D
Solution :
Given equation is\[\Rightarrow \]. \[\frac{L.C.M.(6,2)}{H.C.F.(14,7)}=\frac{6}{7}\]and\[\text{7}\times \text{13}+\text{13}=\text{1}0\text{4}=\text{23}\times \text{13}\] \[\therefore \] (Given) \[\text{7}\times \text{13}+\text{13}\] \[\therefore \] \[\text{224}=\text{12}0\times \text{1}+\text{1}0\text{4}\]\[\text{12}0=\text{1}0\text{4}\times \text{1}+\text{16}\] \[\text{1}0\text{4}=\text{16}\times \text{6}+\text{8}\]\[16=8\times 2+0\] If\[\text{256}=\text{8}\times \text{32}+0\], and\[3465={{3}^{2}}\times 5\times 7\times 11\] \[\text{546}0=\text{22}\times \text{3}\times \text{5}\times \text{7}\times \text{l3}\]or \[\therefore \]
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