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Difference between revisions of "1987 AIME Problems/Problem 11"
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== Problem ==
 
== Problem ==
 
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.
 
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.
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== Solution ==
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== Solution 1==
 
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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Thus <math>m(2n + m + 1) = 2 \cdot 3^{11}</math> so <math>m</math> is a [[divisor]] of <math>2\cdot 3^{11}</math>.  However, because <math>n \geq 0</math> we have <math>m^2 < m(m + 1) \leq 2\cdot 3^{11}</math> so <math>m < \sqrt{2\cdot 3^{11}} < 3^6</math>.  Thus, we are looking for large factors of <math>2\cdot 3^{11}</math> which are less than <math>3^6</math>.  The largest such factor is clearly <math>2\cdot 3^5 = 486</math>; for this value of <math>m</math> we do indeed have the valid [[expression]] <math>3^{11} = 122 + 123 + \ldots + 607</math>, for which <math>k=\boxed{486}</math>.
 
Thus <math>m(2n + m + 1) = 2 \cdot 3^{11}</math> so <math>m</math> is a [[divisor]] of <math>2\cdot 3^{11}</math>.  However, because <math>n \geq 0</math> we have <math>m^2 < m(m + 1) \leq 2\cdot 3^{11}</math> so <math>m < \sqrt{2\cdot 3^{11}} < 3^6</math>.  Thus, we are looking for large factors of <math>2\cdot 3^{11}</math> which are less than <math>3^6</math>.  The largest such factor is clearly <math>2\cdot 3^5 = 486</math>; for this value of <math>m</math> we do indeed have the valid [[expression]] <math>3^{11} = 122 + 123 + \ldots + 607</math>, for which <math>k=\boxed{486}</math>.
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== Solution 2==
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First note that if k is odd, and n is the middle term, the sum is equal to kn. If k is even, then we have the sum equal to kn+k/2 which is going to be even. Since 3^11 is odd, we see that k is odd.
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Thus, we have <math>nk=3^{11} \implies n=3^{11}/k</math>. Also, note <math>n-(k+1)/2=0 \implies n=(k+1)/2.</math> Subsituting <math>n=3^{11}/k</math>, we have <math>k^2+k=2*3^{11}</math>. Proceed as in solution 1.
  
 
== See also ==
 
== See also ==

Revision as of 16:01, 3 February 2018

Problem

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

Solution 1

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 $k=\boxed{486}$.

Solution 2

First note that if k is odd, and n is the middle term, the sum is equal to kn. If k is even, then we have the sum equal to kn+k/2 which is going to be even. Since 3^11 is odd, we see that k is odd.

Thus, we have $nk=3^{11} \implies n=3^{11}/k$. Also, note $n-(k+1)/2=0 \implies n=(k+1)/2.$ Subsituting $n=3^{11}/k$, we have $k^2+k=2*3^{11}$. Proceed as in solution 1.

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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