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 
.
is called a Niven number, named for Ivan Niven, if it is divisible by the sum of its digits. For example, is a Niven number. Show that it is not possible to have more than 
consecutive Niven numbers.
. Let 
denote the sum of digits of 
. Now if we have two odd Niven numbers 
and 
whose tens digits differ by one (as opposed to them being 9 and 0), then 
. Therefore one of 
is even, which is a contradiction since and are odd.
; we could take the pair 
or 
. Therefore some number in our list is divisible by ; call it 
. The list cannot have both 
and 
, so is not in the list. Similarly, we cannot have both 
and 
, so is not in the list. This restricts our list to be 
through 
.
, we have 
. Then one of 
is divisible by 10. This number must be 
since none of 
are divisible by 
. Since cannot be 
(as dividing a number is not well-defined), we must have that 
.
. Then 
and 
, so
which means that 
.
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 
.
.