1976 AHSME Problems/Problem 21

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

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1976 AHSME Problems/Problem 21

Problem 21

Solution

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