Let the first term of a geometric sequence be $\frac{3}{4}$, and let the second term be $15$. What is the smallest $n$ for which the $n$th term of the sequence is divisible by one million?
The common ratio is $$\frac{15}{\frac{3}{4}} = 20$$Therefore, the $n$th term is $(20^{n-1}) \left(\frac{3}{4}\right)$.


If one million (aka $10^6$) divides the $n$th term, then it must be divisible by $5^6$. This can only happen if $n-1$ is at least $6$, or $n \ge 7$.


The $7$th term is $$\left(20^6\right) \left(\frac{3}{4}\right) = \left(4\right)^6\left(5\right)^6\left(\frac{3}{4}\right) = (2)^{10}(5)^6(3),$$which is divisible by $(2)^6(5)^6=10^6$, so the answer is indeed $\boxed{7}$.
7
Level 5
Algebra
Problem: Let the first term of a geometric sequence be $\frac{3}{4}$, and let the second term be $15$. What is the smallest $n$ for which the $n$th term of the sequence is divisible by one million?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we first identify the common ratio of the geometric sequence. The first term \( a_1 = \frac{3}{4} \) and the second term \( a_2 = 15 \). The common ratio \( r \) can be calculated as follows:

1. Set up the equation for the second term: 
   \( a_2 = a_1 \cdot r \).
2. Substitute the known values: 
   \( 15 = \frac{3}{4} \cdot r \).
3. Solve for \( r \):
   \( r = 15 \div \frac{3}{4} = 15 \cdot \frac{4}{3} = 20 \).

Now we have the first term \( a_1 = \frac{3}{4} \) and the common ratio \( r = 20 \). The general formula for the \( n \)th term of a geometric sequence is given by:

\( a_n = a_1 \cdot r^{n-1} \).

Substituting the known values into this formula gives:

\( a_n = \frac{3}{4} \cdot 20^{n-1} \).

Next, we need to determine the smallest \( n \) such that \( a_n \) is divisible by one million. We express one million in terms of its prime factorization:

\( 1,000,000 = 10^6 = (2 \cdot 5)^6 = 2^6 \cdot 5^6 \).

Now, we analyze the term \( a_n \):

1. The term can be rewritten as:
   \( a_n = \frac{3}{4} \cdot 20^{n-1} = \frac{3}{4} \cdot (2^2 \cdot 5)^{n-1} = \frac{3}{4} \cdot 2^{2(n-1)} \cdot 5^{n-1} \).
2. Simplifying \( \frac{3}{4} \):
   \( \frac{3}{4} = \frac{3}{2^2} \).
3. Thus, we can express \( a_n \) as:
   \( a_n = 3 \cdot 2^{2(n-1) - 2} \cdot 5^{n-1} = 3 \cdot 2^{2n - 4} \cdot 5^{n-1} \).

For \( a_n \) to be divisible by \( 2^6 \) and \( 5^6 \), we set up the following inequalities:

1. For the power of 2:
   \( 2n - 4 \geq 6 \) leads to \( 2n \geq 10 \) or \( n \geq 5 \).
2. For the power of 5:
   \( n - 1 \geq 6 \) leads to \( n \geq 7 \).

The more restrictive condition is \( n \geq 7 \). Therefore, the smallest \( n \) for which the \( n \)th term of the sequence is divisible by one million is \( 7 \).<|END_OF_LATENT|> The final answer is $7$.