1965 AHSME Problems/Problem 24

Problem

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 $100000$ is:

$\textbf{(A)}\ 7 \qquad \textbf{(B) }\ 8 \qquad \textbf{(C) }\ 9 \qquad \textbf{(D) }\ 10 \qquad \textbf{(E) }\ 11$

Solution

Note that the given sequence is a geometric sequence with a common ratio $10^{\frac{1}{11}}$ . Let the product of the first $n$ terms of the sequence be denoted $P_n$ . It is a consequence of the laws of exponents that $P_1=(10^{\frac{1}{11}})^1$ , $P_2=(10^{\frac{1}{11}})^{1+2}$ , and, in general, $P_n=(10^{\frac{1}{11}})^{T_n}$ , where $T_n$ denotes the $n$ th triangular number. Setting $P_n$ equal to $100,000$ , we see that: \begin{align*} \\ (10^{\frac{1}{11}})^{\frac{n(n+1)}{2}}&=10^5 \\ \frac{1}{11}*\frac{n(n+1)}{2}&=5 \\ n(n+1)&=110 \\ n^2+n-110&=0 \\ (n-10)(n+11)&=0 \\ \end{align*} Because $n$ must be positive, we are left with $n=10$ . Given this information, choice (D) may seem appealing. Do not be fooled. The product of the first $n$ terms is exactly $100,000$ , but the problem asks for the smallest $n$ such that $P_n$ exceeds 100,000. Thus, the minimum value of $n$ which satsifies the problem is $\boxed{\textbf{(E) }11}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1965 AHSME Problems/Problem 24

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