Double sequences with GCD

Number Theory Level 5

Given a sequence called $A$ including $50$ terms ( from $1$ to $50$ ), which can be described as ${ A }_{ n }=20+{ n }^{ 2 }$

Given another sequence called $B$ including $49$ terms, which have a property that ${ B }_{ n }=GCD({ A }_{ n },{ A }_{ n+1 })$

Given that the largest term in $B$ is ${B}_{x}$ . Find the value of $\sqrt { { B }_{ x }+x }$ .

Note : $GCD(a,b)$ is denoted as the greatest common divisor of $a$ and $b$ .

This question is not original.

This problem belongs to this set

The answer is 11.

1 solution

Andrea Palma
Apr 2, 2015

$GCD(A_n,A_{n+1}) = GCD(20+n^2, 20+n^2+2n+1) =$

$= GCD(20+n^2, 2n+1)$

Since $2n+1$ is odd, for all $z, \in \mathbb Z, m \in \mathbb N$ we have $GCD(z,2n+1) = GCD(2^mz, 2n+1)$

$GCD(A_n,A_{n+1}) = GCD(4(20+n^2), 2n+1)$

$= GCD(80+4n^2, 2n+1) =GCD(81+4n^2-1, 2n+1) =$

$= GCD(81+(2n+1)(2n-1), 2n+1) = GCD(81, 2n+1) \leq 81$ .

For $n = 40$ (and assuming that $1 \leq n \leq 50$ as I think the intention of the writer of the problem meant) we have $GCD(81,81)=81$ so it reaches the max value possible.

So $B = 81$ and $x = 40$ and the answer is $\sqrt{121} = 11$ .

Double sequences with GCD

This question is not original.

This problem belongs to this set

The answer is 11.

1 solution

0 pending reports