MCQ
The remainder, when $7^{103}$ is divided by $17$ is $..........$.
- A$11$
- ✓$12$
- C$13$
- D$14$
$=7 \times(49)^{51}$
$=7 \times(51-2)^{51}$
Remainder :- $7 \times(-2)^{51}$
$\Rightarrow-7\left(2^3 \cdot(16)^{12}\right)$
$\Rightarrow-56(17-1)^{12}$
$\text { Remainder }=-56 \times(-1)^{12}=-56+68=12$
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$( S 1): \lim _{ n \rightarrow \infty} \frac{1}{ n ^2}(2+4+6+\ldots \ldots \ldots+2 n)=1$
(S2) : $\lim _{ n \rightarrow \infty} \frac{1}{ n ^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots \ldots .+ n ^{15}\right)=\frac{1}{16}$