When $102^{29}$ is divided by $5$ we get remainder

2

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Solution

The correct option is C $2$
$N = {(2 \times 51)}^{29}$
$= 2^{29} . {(50 + 1)}^{29}$
$= 2^{29} (50 λ + 1)$ $; λ \in I$
$= 2^{29} 50 λ + 2^{29}$
$= 50 λ^{1} + {{2. (2}^{2})}^{14}$ ${; λ}^{1} \in I$
$= 50 λ^{1} + 2. (5 - {1)}^{14}$
$= 50 λ^{1} + 2. (5 k + 1)$
$= 50 λ^{1} + 10 K + 2$
$= 5 (10 λ^{1} + 2 K) + 2$
$∴$ In dividing by $5$ ; we get the remainder $= 2.$
Hence the answer is $2.$

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When 10229 is divided by 5 we get remainder

The correct option is C 2N=(2×51)29 =229.(50+1)29 =229(50λ+1) ;λ∈I =22950λ+229 =50λ1+2.(22)14 ;λ1∈I =50λ1+2.(5−1)14 =50λ1+2.(5k+1) =50λ1+10K+2 =5(10λ1+2K)+2∴ In dividing by 5; we get the remainder =2.Hence the answer is 2.

When $102^{29}$ is divided by $5$ we get remainder