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 
divides 
.
consecutive positive integers is divisible by 
such that 
divides 
.![\[A=\frac{(m+3)^{n}+1}{3m}\]](http://latex.artofproblemsolving.com/f/5/8/f58ed8713c887510943d543c89f83640a546401f.png)
is an integer. Prove that 
is odd.
be divisible by 
.
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 
.
.
Y by Adventure10, Mango247
In an increasing infinite sequence of positive integers, every term starting from the 
-th term divides the sum of all preceding terms. Prove that every term starting from some term is equal to the sum of all preceding terms.
-th term divides the sum of all preceding terms. Prove that every term starting from some term is equal to the sum of all preceding terms.
is suth sequence. Then ![\[ x_n|\sum_{i=1}^{n-1}x_i\]](http://latex.artofproblemsolving.com/a/4/5/a45f2c452f60d22a3a02414d3ad6d21d2adba839.png)
for big n, or ![\[ \sum_{i=1}^nx_i=y_ny_n\]](http://latex.artofproblemsolving.com/a/1/1/a1154b7afdf1e747ff6f84feb755ff19bf4960a6.png)
for big n 
![\[ \sum_{i=1}^nx_n=y_nx_n+x_n=y_{n+1}x_{n+1}\]](http://latex.artofproblemsolving.com/4/e/4/4e4dae260682f2188ae3b6ea179ed5d8e418af6c.png)
for all big n. Because 
we get 
or 
.
and
.
,
give contradition. Therefore 
.
.
