Find the last two digits of $7^{1} + 7^{2} + 7^{3} + . . . . . . . . . . . . . . . . . {.7}^{342} ?$
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Solution

$7^{1} = 07, 7^{5} = - - 07$

$7^{2} = 49, 7^{6} = - - 49$

$7^{3} = 43, 7^{7} = - - 43$

$7^{4} = 01, 7^{8} = - - 01$

There is a cyclicity of 4 as 07, 49 , 43 ,01 is the last 2 digits for the powers of 7.

Here $342 = 4 k + 2 \Rightarrow$ Required answer is $07 + 49 = 56$

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