Q43 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
For $n \in N$, the largest positive integer that divides $81 n+20 n-1$ is $k$. If $S$ is the sum of all positive divisors of $k$ then $S-k=$
- Correct answerA117
- B130
- C115
- D127
Solution
$81^n+20 n-1 \Rightarrow(1+80)^n+20 n-1$
$\begin{aligned}
& =1+80 n+{ }^n C_2 \times 80^2 \times 1+\ldots .+20 n-1 \\
& =100 n+100 \times 8^2 \times{ }^n C_2+{ }^n C_3 80^3 \ldots \\
& =100\left(n+{ }^n C_2 \times 64 \ldots .\right) \\
& k=100=2^2 \times 5^2
\end{aligned}$
Sum of all divisors is $\begin{aligned} & \left(1+2+2^2\right) \times(1+5+25)=7 \times 31=217=S \\ & S-K=217-100=117 . \end{aligned}$
Sum of all divisors is $\begin{aligned} & \left(1+2+2^2\right) \times(1+5+25)=7 \times 31=217=S \\ & S-K=217-100=117 . \end{aligned}$
Hence, the correct answer is option: 'A'.