1971 AMC 12 Problem 29

Attempt Problem 29 of the 1971 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 1971 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

29.

Given the progression 10111,10211,10311,10411,…,10n11, \begin{gathered} 10^{\frac{1}{11}},10^{\frac{2}{11}},10^{\frac{3}{11}},\\ 10^{\frac{4}{11}},\ldots,10^{\frac{n}{11}}, \end{gathered} the least positive integer nn such that the product of the first nn terms exceeds 100,000100{,}000 is:
77
88
99
1010
1111
Answer: E
Concepts:geometric sequenceexponenttriangular numberinequality
Small hint:
Add the exponents when multiplying the terms
Big hint:
Require n(n+1)22>5\frac{n(n+1)}{22}\gt5, noting that equality is not enough
Solution:
The product is 101+2+⋯+n11=10n(n+1)22. 10^{\frac{1+2+\cdots+n}{11}}=10^{\frac{n(n+1)}{22}}. It exceeds 100,000=105100{,}000=10^5 exactly when n(n+1)>110.n(n+1)\gt110. For n=10n=10 there is equality, while n=11n=11 works.
Therefore, the correct answer is E.
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