Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1971 AHSME Problems/Problem 29"

(→‎Solution)
(→‎Solution)
Line 12: Line 12:
 
==Solution==
 
==Solution==
  
−
The product of the sequence <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math> is equal to <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}</math> since we are looking for the smallest value <math>n</math> that will create <math>100,000</math>, or <math>10^5</math>, we set up the equation <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5</math>
+
The product of the sequence <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math> is equal to <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}</math> since we are looking for the smallest value <math>n</math> that will create <math>100,000</math>, or <math>10^5</math>, we set up the equation <math>10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5</math>, which simplified to <math>\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}=5</math>, or <math>1+2+3\dots n=55</math> This can be converted to <math>\frac{n(1+n)}{2}=5</math> This simplified to the quadratic <math>n^2+n-10=0</math>

Revision as of 17:16, 22 August 2019

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 $10^5$, we 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 $\frac{n(1+n)}{2}=5$ This simplified to the quadratic $n^2+n-10=0$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1971_AHSME_Problems/Problem_29&oldid=109247"