- ✓$7$
- B$14$
- C$21$
- D$28$
$=(2030-7)^{2023}$
$=(35 K -7)^{2023}$
$={ }^{2023} C _0(35 K )^{2023}(-7)^0+{ }^{2023} C _1(35 K )^{2022}(-7)+\ldots .+$
$\ldots \ldots+{ }^{2023} C _{2023}(-7)^{2023}$
$=35 N -7^{2023}$
$\text { Now, }-7^{2023}=-7 \times 7^{2022}=-7\left(7^2\right)^{1011}$
$=-7(50-1)^{1011}$
$=-7\left({ }^{1011} C _0 50^{1011}-{ }^{1011} C _1(50)^{1010}+\ldots \ldots .{ }^{1011} C _{1011}\right)$
$=-7(5 \lambda-1)$
$=-35 \lambda+7$
$\therefore$ when $(2023)^{2023}$ is divided by $35$ remainder is $7$
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$\left(\log _e 2\right)-\frac{1}{7}$ then the value of $7 a-3$ is equal to: