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:

We are asked to build n-digit numbers using only the three digits 2, 5 and 7. Because none of these digits is zero, any string of length n made from them will automatically be an n-digit positive integer, even when the first position is filled. Thus every position in the number (units, tens, hundreds, …, up to the n-th place) can be occupied independently by any of the three choices 2, 5 or 7.

We now state the basic counting principle: if a first task can be done in $$p$$ ways and, independently, a second task can be done in $$q$$ ways, then the two tasks together can be done in $$p\times q$$ ways. Extending this idea, if each of n independent positions can be filled in $$k$$ ways, the total number of ways is $$k^{\,n}$$.

Here $$k=3$$ (the digits 2, 5, 7) and we have n independent positions, so the total number of distinct n-digit numbers that can be formed is

$$3^{\,n}.$$

Our goal is to find the smallest n for which at least 900 such numbers exist. Symbolically we require

$$3^{\,n}\;\ge\;900.$$

Instead of logarithms, we can test successive powers of 3 until we exceed 900, showing every step:

For $$n=5$$ we have $$3^{5}=243,$$ which is less than 900.

For $$n=6$$ we have $$3^{6}=729,$$ still less than 900.

For $$n=7$$ we have $$3^{7}=3^{6}\times3=729\times3=2187,$$ which is greater than 900.

Thus $$3^{\,6}=729<900$$ while $$3^{\,7}=2187\ge900.$$ The inequality $$3^{\,n}\ge900$$ is satisfied for the first time at $$n=7$$.

Therefore, the smallest value of n that allows the formation of at least 900 distinct numbers is $$n=7$$.

Hence, the correct answer is Option D.