Q259 / 270 · NDA 2024 (Phase 1) · Mathematics
Let $m=77^n$. The index $n$ is given a positive integral value at random. What is the probability that the value of $m$ will have 1 in the units place?
- A$\frac{1}{2}$
- B$\frac{1}{3}$
- Correct answerC$\frac{1}{4}$
- D$\frac{1}{n}$
Solution
$\therefore \quad m=77^n$
$\begin{aligned}
Put, n & =1,2,3,4 \\
For, n & =1 \\
m & =77
\end{aligned}$
$For, n=2, m=77^2=5929$
$\begin{aligned}
For, n & =3, m=77^3=456543 \\
For, n & =4, m=77^4=35153041 \\
& =\text { last digit will be } 1
\end{aligned}$
Hence, the required probability $=\frac{1}{4}$
Hence, the correct answer is option: 'C'.