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Difference between revisions of "1987 AIME Problems/Problem 11"

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== Problem ==
 
== Problem ==
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Find the largest possible value of <math>\displaystyle k</math> for which <math>\displaystyle 3^{11}</math> is expressible as the sum of <math>\displaystyle k</math> consecutive [[positive integer]]s.
+
Find the largest possible value of <math>k</math> for which <math>3^{11}</math> is expressible as the sum of <math>k</math> consecutive [[positive integer]]s.
 
== Solution ==
 
== Solution ==
 
Let us write down one such sum, with <math>m</math> terms and first term <math>n + 1</math>:
 
Let us write down one such sum, with <math>m</math> terms and first term <math>n + 1</math>:
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{{AIME box|year=1987|num-b=10|num-a=12}}
 
{{AIME box|year=1987|num-b=10|num-a=12}}
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Revision as of 19:09, 4 July 2013

Problem

Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers.

Solution

Let us write down one such sum, with $m$ terms and first term $n + 1$:

$3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)$.

Thus $m(2n + m + 1) = 2 \cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$. However, because $n \geq 0$ we have $m^2 < m(m + 1) \leq 2\cdot 3^{11}$ so $m < \sqrt{2\cdot 3^{11}} < 3^6$. Thus, we are looking for large factors of $2\cdot 3^{11}$ which are less than $3^6$. The largest such factor is clearly $2\cdot 3^5 = 486$; for this value of $m$ we do indeed have the valid expression $3^{11} = 122 + 123 + \ldots + 607$, for which $k=\boxed{486}$.

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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