1976 AHSME Problems/Problem 21

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Problem 21

What is the smallest positive odd integer $n$ such that the product $2^{1/7}2^{3/7}\cdots2^{(2n+1)/7}$ is greater than $1000$? (In the product the denominators of the exponents are all sevens, and the numerators are the successive odd integers from $1$ to $2n+1$.)

$\textbf{(A) }7\qquad \textbf{(B) }9\qquad \textbf{(C) }11\qquad \textbf{(D) }17\qquad \textbf{(E) }19$

Solution

Combine the terms in the product to get $2^{\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7}}$.

The exponent can be simplified to $$\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7} \Rightarrow \frac{\frac{n(1+(2n+1))}{2}}{7} \Rightarrow \frac{n^2}{7}.$$

We want this inequality to be true with the smallest positive odd integer value of $n$: $$2^{\frac{n^2}{7}} > 1000.$$

Now, let's test the answer choices. For $n=7$, we have $2^{49/7}=2^{7}<1000$. For $n=9$, we have $2^{81/7}>2^{10}>1000$.

So our answer is $\boxed{\textbf{(B) }9}$. ~jiang147369

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