Question

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

Answer 1

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,

& {{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{aligned}$$ Now we have to find the value of n in which $${{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{aligned} & n-2\ge 5 \\\ & n\ge 7 \\\ \end{aligned}$$ The smallest value of n for which this would be possible, is 7. Therefore, the correct option of the above question is option D. Note: Be careful while choosing the option because we have $$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.