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Difference between revisions of "1965 AHSME Problems/Problem 24"

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

Latest revision as of 18:25, 18 July 2024

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 $\boxed{\textbf{(E) }11}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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