AoPS Academy
consecutive positive integers is divisible by 
for which there exists an integer 
such that 
divides 
.
and be natural numbers such that ![\[A=\frac{(m+3)^{n}+1}{3m}\]](http://latex.artofproblemsolving.com/f/5/8/f58ed8713c887510943d543c89f83640a546401f.png)
is an integer. Prove that 
is odd.
and be natural numbers and let 
be divisible by 
. Show that 
is divisible by .
and 
be natural numbers having the property ![\[n_{2}\; \vert \; 2^{n_{1}}-1, n_{3}\; \vert \; 2^{n_{2}}-1, \cdots, n_{k}\; \vert \; 2^{n_{k-1}}-1, n_{1}\; \vert \; 2^{n_{k}}-1.\]](http://latex.artofproblemsolving.com/2/0/7/20707ef12d0e08bec8067225a9e9128bbc1d2d45.png)
Show that 
.
.
. Assume that there is a partition of 
into two subsets 
and 
such that no subset contains three number 
for which 
. Without loss of generality, assume 
, thus 
, 
, 
. Note that 
so 
. But then 
, contradiction.
which is partitioned into 
and 
. We have 
, 
, 
, so there cannot be 
for which 
. Also for all 
, 
, so 
. Our proof is now complete.
be a rational number with 
and 
.
.
![\[\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}\]](http://latex.artofproblemsolving.com/d/4/5/d4511807f1e04ac2fa8aae36d65f9e0f5ba11c6f.png)
is not divisible by 
for any integer 
.
is composite, then 
.
.
be an integer. Find the smallest integer 
such that for any partition of the set 
into two subsets, at least one subset contains three numbers 
such that 
.