Is $N_{10}$ even or odd?

  • Thread starter Thread starter Ackbach
  • Start date Start date
  • Tags Tags
    2016
Click For Summary
SUMMARY

The discussion centers on determining whether $N_{10}$, the number of ordered 10-tuples of positive integers $(a_1,a_2,\ldots,a_{10})$ satisfying the equation $1/a_1 + 1/a_2 + \ldots + 1/a_{10} = 1$, is even or odd. This problem is derived from Problem A-5 of the 1997 William Lowell Putnam Mathematical Competition. The solution is attributed to Kiran Kedlaya and his associates, although no participants provided answers in the forum.

PREREQUISITES
  • Understanding of ordered tuples in combinatorial mathematics
  • Familiarity with the concept of positive integers
  • Knowledge of the harmonic series and its properties
  • Experience with mathematical competitions and problem-solving techniques
NEXT STEPS
  • Study the properties of ordered tuples in combinatorial contexts
  • Explore the harmonic series and its applications in number theory
  • Review solutions to previous Putnam Competition problems for advanced problem-solving strategies
  • Investigate the work of Kiran Kedlaya and his contributions to combinatorial mathematics
USEFUL FOR

Mathematics students, competitive problem solvers, and educators interested in combinatorial number theory and mathematical competitions.

Ackbach
Gold Member
MHB
Messages
4,148
Reaction score
94
Here is this week's POTW:

-----

Let $N_n$ denote the number of ordered $n$-tuples of positive integers $(a_1,a_2,\ldots,a_n)$ such that $1/a_1 + 1/a_2 +\ldots +1/a_n=1$. Determine whether $N_{10}$ is even or odd.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
Re: Problem Of The Week # 230 - Aug 25, 2016

This was Problem A-5 in the 1997 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

We may discard any solutions for which $a_1 \neq a_2$, since those come in pairs; so assume $a_1 = a_2$. Similarly, we may assume that $a_3 = a_4$, $a_5 = a_6$, $a_7 = a_8$, $a_9=a_{10}$. Thus we get the equation
\[
2/a_1 + 2/a_3 + 2/a_5 + 2/a_7 + 2/a_9 = 1.
\]
Again, we may assume $a_1 = a_3$ and $a_5 = a_7$, so we get $4/a_1 + 4/a_5 + 2/a_9 = 1$; and $a_1 = a_5$, so $8/a_1 + 2/a_9 = 1$. This implies that $(a_1-8)(a_9-2) = 16$, which by counting has 5 solutions. Thus $N_{10}$ is odd.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K