Sequence and divisibility

For discussing Olympiad Level Algebra (and Inequality) problems
Katy729
Posts:47
Joined:Sat May 06, 2017 2:30 am
Sequence and divisibility

Unread post by Katy729 » Thu Aug 24, 2017 10:27 pm

Let $ n$ be a positive integer and let $ a_1,a_2,a_3,\ldots,a_k$ $ ( k\ge 2)$ be distinct integers in the set $ { 1,2,\ldots,n}$ such that $ n$ divides $ a_i(a_{i + 1} - 1)$ for $ i = 1,2,\ldots,k - 1$. Prove that $ n$ does not divide $ a_k(a_1 - 1).$

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: Sequence and divisibility

Unread post by tanmoy » Tue Sep 05, 2017 12:25 am

For the sake of the contradiction, let's assume that it does. Then,
$a_1 \equiv a_1. a_2 \equiv a_1. a_2. a_3$ $\equiv........\equiv a_1. a_2......a_{k-2}. a_k$
$\equiv.....\equiv a_1. a_k \equiv a_k \pmod n$.

So, $n$ divides $ |a_1 - a_k|$. But $0 < |a_1 - a_k| < n$, which is a contradiction.
So, $n$ doesn't divide $a_k(a_1 - 1)$.
"Questions we can't answer are far better than answers we can't question"

Katy729
Posts:47
Joined:Sat May 06, 2017 2:30 am

Re: Sequence and divisibility

Unread post by Katy729 » Wed Sep 06, 2017 5:52 pm

Thanks tanmoy. :)

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