If \[n\]is even, then \[({{6}^{n}}-1)\]is divisible by |
A) 37
B) 30
C) 35
D) 6
Correct Answer: C
Solution :
If \[x\] is even, then let \[n=2x,\]tor some x. |
Now, \[{{6}^{n}}-1=({{6}^{2x}}-1)={{36}^{x}}-1\] |
\[=(36-1)({{36}^{x-1}}+{{36}^{x-2}}+...+1)\] |
\[=(35)({{36}^{x-1}}+{{36}^{x-2}}+...+1)\] |
Hence, the number is divisible by 35. |
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