Question
Find the last two digits of 71 + 72 + 73 +……………..7342 ?
Answer: Option E
:
E
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:
E
option (e)
71=0775=−−07
72=4976=−−49
73=4377=−−43
74=0178=−−01
There is a cyclicity of 4 as 07, 49 , 43 ,01 is the last 2 digits for the powers of 7
Here 342=4k+2⇒Required answer is 07+49=56
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