IMO 2012 #6
by Wolstenholme, Oct 29, 2014, 2:15 PM
Find all positive integers 
for which there exist non-negative integers 
such that
![\[
\frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} =
\frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1.
\]](//latex.artofproblemsolving.com/0/9/b/09b176b3a1bc7d1d73fab5c7b0b5ed8d36c5fe1c.png)
Solution:
So it looks like the key to this problem will be some god-awful induction that I'm not willing to do. However, I'm hoping what I wrote below would scrap at least
points at the IMO.
First, assume for some
we have found a working -tuple 
. Let 
be the largest integer in this -tuple. Then we have that 
. Taking this equation modulo yields 
so we must have 
Moreover, it is easy to see that if we have a working-tuple where is odd, the 
-tuple 
works as well.
I conjecture that all that are congruent to 
or modulo 
work as well.
for which there exist non-negative integers 
such that![\[
\frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} =
\frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1.
\]](http://latex.artofproblemsolving.com/0/9/b/09b176b3a1bc7d1d73fab5c7b0b5ed8d36c5fe1c.png)
Solution:
So it looks like the key to this problem will be some god-awful induction that I'm not willing to do. However, I'm hoping what I wrote below would scrap at least
points at the IMO.First, assume for some
we have found a working -tuple 
. Let 
be the largest integer in this -tuple. Then we have that 
. Taking this equation modulo yields 
so we must have 
Moreover, it is easy to see that if we have a working
-tuple where is odd, the 
-tuple 
works as well.I conjecture that all
that are congruent to 
or modulo 
work as well.
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