n-digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is:
A. 6
B. 8
C. 9
D. 7

Hint: We will arrange the given three-digits to n-places of the n-digit number and by equating it to the given such distinct number that has to be formed. Hence we will get the value of n.

Complete step-by-step answer:
We have been given the three digits as 2, 5 and 7.
 Now, the n-digit numbers have total n places that have to be filled by the given three digit numbers.
Hence, the total number of ways to make an n-digit number with 2, 5 and 7 is going to be \[{{3}^{n}}\].
According to the question,
\[\begin{align}
  & {{3}^{n}}\ge 900 \\
 & {{3}^{n}}\ge 9\times 100 \\
 & {{3}^{n}}\ge {{3}^{2}}\times 100 \\
 & \dfrac{{{3}^{n}}}{{{3}^{2}}}\ge 100 \\
 & {{3}^{n-2}}\ge 100 \\
\end{align}\]
Now we have to find the value of n in which \[{{3}^{n-2}}\] is greater than 100.
 Let us consider the power of 3 as 4 and 5 respectively. Then we will get,
\[{{3}^{4}}=81\] and \[{{3}^{5}}=243\]
Now we can clearly see that \[{{3}^{5}}\] has a greater value than 100 which means that,
\[\begin{align}
  & n-2\ge 5 \\
 & n\ge 7 \\
\end{align}\]
The smallest value of n for which this would be possible, is 7.
Therefore, the correct option of the above question is option D.

Note: Be careful while choosing the option because we have \[n\ge 7\] which means n can be 7. Don’t choose n = 8 instead of n = 7.
We have been asked to find out the smallest value of n, so we must choose n = 7 as the right answer.