1971 AHSME Problems/Problem 29

Problem 29

Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ . The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is

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

Solution

The product of the sequence $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ is equal to $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}$ since we are looking for the smallest value $n$ that will create $100,000$ , or $10^5$ . From there, we can set up the equation $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5$ , which simplified to $\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}=5$ , or $1+2+3\dots n=55$ This can be converted to $\frac{n(1+n)}{2}=55$ This simplified to the quadratic $n^2+n-110=0$ Or $(n+11)(n-10)=0$ So $n=-11$ or $10$ Since only positive values of $n$ work, $n=10$ makes the expression equal $100000$ . However, we have to exceed $100000$ , so our answer is $\boxed{\textbf{(E) }11}.$

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1971 AHSME Problems/Problem 29

Problem 29

Solution