If the number of integral terms in the expansion of $$\left(3^{\frac{1}{2}} + 5^{\frac{1}{8}}\right)^n$$ is exactly 33, then the least value of $$n$$ is

We start with the binomial expansion

$$\left(3^{\tfrac12}+5^{\tfrac18}\right)^n=\sum_{k=0}^{n}\binom{n}{k}\left(3^{\tfrac12}\right)^{\,n-k}\left(5^{\tfrac18}\right)^{\,k}.$$

The general (k-th) term of this expansion is therefore

$$T_k=\binom{n}{k}\,3^{\tfrac{n-k}{2}}\;5^{\tfrac{k}{8}}.$$

For this term to be an integral (i.e., non-fractional) power of both 3 and 5, the exponents of 3 and 5 must be non-negative integers. So we impose two separate conditions:

1. The exponent of 3 is $$(n-k)/2,$$ which must be an integer.

2. The exponent of 5 is $$k/8,$$ which must also be an integer.

Now translate these statements into congruences:

• From $$\dfrac{k}{8}\in\mathbb Z$$ we get $$k\equiv 0\pmod{8}.$$ Hence we may write $$k=8m,$$ where $$m$$ is a non-negative integer.

• From $$\dfrac{n-k}{2}\in\mathbb Z$$ we get $$n-k\equiv 0\pmod{2}.$$ Substituting $$k=8m$$ gives $$n-8m\equiv 0\pmod{2}.$$ Because $$8m$$ is always even, this reduces to $$n\equiv 0\pmod{2},$$ so $$n$$ itself must be even.

Therefore every integer term corresponds to a non-negative integer $$m$$ satisfying $$k=8m\le n.$$ Equivalently, $$m$$ can take all integer values from $$0$$ up to

$$\left\lfloor\frac{n}{8}\right\rfloor.$$

Counting these values, we get exactly

$$\left\lfloor\frac{n}{8}\right\rfloor+1$$

integral terms in the whole expansion.

The statement of the problem tells us that the number of integral terms is $$33.$$ Hence

$$\left\lfloor\frac{n}{8}\right\rfloor+1=33.$$

Subtracting 1 from both sides gives

$$\left\lfloor\frac{n}{8}\right\rfloor=32.$$

This means that $$n/8$$ lies between $$32$$ and $$33$$, but is not equal to $$33$$. In inequality form,

$$32\le\frac{n}{8}\lt 33.$$

Multiplying through by $$8$$, we obtain

$$256\le n\lt264.$$

Because $$n$$ must be an integer, the possible values are $$256,257,258,259,260,261,262,263.$$ But we also discovered earlier that $$n$$ must be even, so the admissible set narrows down to $$256,258,260,262.$$

We are asked for the least value of $$n,$$ which is clearly $$n=256.$$

Hence, the correct answer is Option C.