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 
.
consecutive positive integers is divisible by 
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, UpvoteFarm, Mango247
Clara computed the product of the first 
positive integers and Valerid computed the product of the first 
even positive integers, where 
. They got the same answer. Prove that one of them had made a mistake.
positive integers and Valerid computed the product of the first 
even positive integers, where 
. They got the same answer. Prove that one of them had made a mistake.
for 
.
in both sides, we see that 
(otherwise 
has a factor 3 more than the 
(then 
implies 
)
(then 
implies 
)
,
or 
where 
is prime.
