An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using digits 2,3,4,5 and 7 such that each n-digit number has the
(i) first and last digits are the same
(ii) first and third are prime numbers
(iii) second should be greater than or equal to 4.
What should be the minimum value of n such that the above is possible?

Let's consider an n-digit number.

$$_{-------------n\ digits}$$

The 900 distinct numbers have to be formed using digits 2, 3, 4, 5 and 7.

The first and last digit numbers are the same & the first and third are prime numbers

$$abc_{-----------}a\ total\ 'n'\ digits.$$

the values of a & c can be anything from 2,3,5, and 7. i.e 4 values.

The second digit should be greater than or equal to 4. Hence, b can take any value from 4, 5 & 7. i.e. 3 values.

There are no conditions from the 4th digit to the 'n-1' digit.

Hence those digits can take any value from 2, 3, 4, 5 and 7, i.e. 5 values.

For n = 4, the number of digits that can be formed $$=4\times\ 4\times\ 3=48 < 900$$

For n = 5, the number of digits that can be formed $$=4\times\ 4\times\ 3\times\ 5 = 240 < 900$$

For n = 6, the number of digits that can be formed $$=4\times\ 4\times\ 3\times\ 5\times\ 5=1200 > 900$$

Hence a minimum of 6-digit number is required to form 900 distinct numbers with given conditions.

Option (B) is correct.