Difference between revisions of "1971 AHSME Problems/Problem 29"
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− | == Problem | + | == Problem == |
Given the progression <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math>. | Given the progression <math>10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}</math>. | ||
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==Solution== | ==Solution== | ||
− | + | By the rules of [[exponentiation|exponents]], the product of the first <math>n</math> terms of the sequence equals <math>10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}</math>. From here, we can set up the equation <math>10^{\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}}=100,000=10^5</math>, which simplifies to <math>\frac{1}{11}+\frac{2}{11}+\dots+\frac{n}{11}=5</math>, or <math>1+2+3\dots n=55</math>. By the formula for [[triangular numbers]], <math>\frac{n(n+1)}{2}=55</math>. Thus, <math>n^2+n-110=0</math>, or <math>(n+11)(n-10)=0</math>. Because <math>n>0</math>, only <math>n=10</math> makes the product equal <math>100,000</math>. However, we have to strictly exceed <math>100,000</math>, so our answer is <math>\boxed{\textbf{(E) }11}.</math> | |
+ | |||
+ | == See Also == | ||
+ | {{AHSME 35p box|year=1971|num-b=28|num-a=30}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:14, 7 August 2024
Problem
Given the progression .
The least positive integer
such that the product of the first
terms of the progression exceeds
is
Solution
By the rules of exponents, the product of the first terms of the sequence equals
. From here, we can set up the equation
, which simplifies to
, or
. By the formula for triangular numbers,
. Thus,
, or
. Because
, only
makes the product equal
. However, we have to strictly exceed
, so our answer is
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.