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 "1987 AIME Problems/Problem 11"

(→‎Solution 2)
m (→‎Solutions)
Line 10: Line 10:
  
 
=== Solution 2===
 
=== Solution 2===
−
First note that if <math>k</math> is odd, and <math>n</math> is the middle term, the sum is equal to kn. If <math>k</math> is even, then we have the sum equal to <math>kn+k/2</math> which is going to be even. Since <math>3^{11}</math> is odd, we see that <math>k</math> is odd.  
+
First note that if <math>k</math> is odd, and <math>n</math> is the middle term, the sum is equal to kn. If <math>k</math> is even, then we have the sum equal to <math>kn+k/2</math>, which will be even. Since <math>3^{11}</math> is odd, we see that <math>k</math> is odd.  
  
 
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.
 
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.

Revision as of 14:02, 3 January 2024

Problem

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

Solutions

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 will 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_11&oldid=209731"