When 599 is divided by 13, the remainder is

We have, 599=53⋅596=(125)(625)24 =[13×9+8](1+48×13)24 =(13×9+8)[1+24C1×(48×13)+24C2(48×13)2+…+(48×13)24]= 8 + terms containing powers of 13. Hence remainder = 8.

When 599 is divided by 13, the remainder is

We have, 599=53⋅596=(125)(625)24 =[13×9+8](1+48×13)24 =(13×9+8)[1+24C1×(48×13)+24C2(48×13)2+…+(48×13)24]= 8 + terms containing powers of 13. Hence remainder = 8.

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When 5⁹⁹ is divided by 13, the remainder is

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Detailed Solution

We have,

$\begin{aligned} 5^{99} = 5^{3} \cdot 5^{96} = (125) (625)^{24} \\ = [13 \times 9 + 8] (1 + 48 \times 13)^{24} \\ = (13 \times 9 + 8) [1 +^{24} C_{1} \times (48 \times 13) +^{24} C_{2} (48 \times 13)^{2} + \dots + (48 \times 13)^{24}] \end{aligned}$