1976 AMC 12 Problem 21

Question

What is the smallest positive odd integer

n

such that the product

2^{\frac{1}{7}}2^{\frac{3}{7}}\cdots2^{\frac{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.

)

7

9

11

17

19

Combine the powers by summing the odd numerators

The odd-number sum

1+3+\cdots+(2n+1)

equals

(n+1)^2;

compare with

2^{10}=1024

The product is

2^{\frac{1+3+\cdots+(2n+1)}{7}} =2^{\frac{(n+1)^2}{7}}.

For

n=7,

the exponent

\frac{64}{7}

is less than

\frac{19}{2},

so the product is less than

2^{\frac{19}{2}}=512\sqrt2\lt768\lt1000.

All smaller positive odd

n

also fail. For

n=9,

the exponent is

\frac{100}{7}\gt10,

so the product exceeds

2^{10}=1024.

Thus the first allowable odd

n

is

9.

Therefore, the correct answer is B.

Problem 21 in other years