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 1)
m (→‎Solution 3)
 
(7 intermediate revisions by 2 users not shown)
Line 9: Line 9:
 
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>.
  
−
== 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 equals <math>kn</math>. 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.
 +
 +
=== Solution 3===
 +
Proceed as in Solution 1 until it is noted that <math>m</math> is a divisor of <math>2\cdot 3^{11}</math>. The divisors of <math>2\cdot 3^{11}</math> are <math>3^{1} , 2\cdot 3^{1} , 3^{2} , 2\cdot 3^{2} , \ldots , 2\cdot 3^{10} , 3^{11}</math>. Note that the factors of <math>m(2n + m + 1)</math> are of opposite parity (if <math>m</math> is odd, then <math>(2n + m + 1)</math> is even and vice versa). Thus, one of the two factors will be a power of three, and the other will be twice a power of three. <math>(2n + m + 1)</math> will represent the greater factor while <math>m</math> will represent the lesser factor. Given this information, we need to find the factor pair that maximizes the lesser of the two factors, as this will maximize the value of <math>m</math>. The factor pair which maximizes the lesser factor is <math>2\cdot 3^{5}</math> and <math>3^{6}</math>. It follows that <math>m</math> = <math>2\cdot 3^{5}</math> = <math>\boxed{486}</math>.
 +
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
 
== See also ==
 
== See also ==

Latest revision as of 18:30, 20 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 equals $kn$. 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.

Solution 3

Proceed as in Solution 1 until it is noted that $m$ is a divisor of $2\cdot 3^{11}$. The divisors of $2\cdot 3^{11}$ are $3^{1} , 2\cdot 3^{1} , 3^{2} , 2\cdot 3^{2} , \ldots , 2\cdot 3^{10} , 3^{11}$. Note that the factors of $m(2n + m + 1)$ are of opposite parity (if $m$ is odd, then $(2n + m + 1)$ is even and vice versa). Thus, one of the two factors will be a power of three, and the other will be twice a power of three. $(2n + m + 1)$ will represent the greater factor while $m$ will represent the lesser factor. Given this information, we need to find the factor pair that maximizes the lesser of the two factors, as this will maximize the value of $m$. The factor pair which maximizes the lesser factor is $2\cdot 3^{5}$ and $3^{6}$. It follows that $m$ = $2\cdot 3^{5}$ = $\boxed{486}$.

~ cxsmi

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=211589"