# 1971 AHSME Problems/Problem 29

## Problem

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

By the rules of exponents, the product of the first $n$ terms of the sequence equals $10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}$. From here, we can set up the equation $10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}=100,000=10^5$, which simplifies to $\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}=5$, or $1+2+3\dots n=55$. By the formula for triangular numbers, $\frac{n(n+1)}{2}=55$. Thus, $n^2+n-110=0$, or $(n+11)(n-10)=0$. Because $n>0$, only $n=10$ makes the product equal $100,000$. However, we have to strictly exceed $100,000$, so our answer is $\boxed{\textbf{(E) }11}.$