1965 AMC 12 Problem 24

Attempt Problem 24 of the 1965 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1965 AMC 12 solutions, or check the answer key.

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24.

Given the sequence

10^{\frac{1}{11}},

10^{\frac{2}{11}},

10^{\frac{3}{11}},

\ldots,

10^{\frac{n}{11}},

the smallest value of

n

such that the product of the first

n

members of this sequence exceeds

100{,}000

is:

7

8

9

10

11

Add the exponents in the product

Require

\frac{n(n+1)}{22}\gt5

The product is

10^{\frac{1+2+\cdots+n}{11}} =10^{\frac{n(n+1)}{22}}.

To exceed

100{,}000=10^5,

we need

n(n+1)\gt110.

n=10

equality holds, while

n=11

exceeds it.

Therefore, the correct answer is E.

Problem 24 in other years