This Strange Arithmetic Sequence Problem

Let the common difference of sequences $a_n$ and $b_n$ be $d_a$ and $d_b$ respectively. We know $d_a$ and $d_b$ are integers greater than $1$ . We want to maximise $n$ .

We can write sequence $a_n$ as $a_n = a_1 + d_a(n-1)$ . However, $a_1 = 1$ , therefore $a_n = 1 + d_a(n-1)$

Similarly, $b_n = 1 + d_b(n-1)$ .

Note that $\gcd(a_n -1, b_n - 1 )= n-1$ , if $\gcd(d_a, d_b) = 1,$ . Therefore, to maximise $n-1$ we must maximise the $\gcd(a_n -1, b_n - 1)$ . $\gcd(d_a, d_b)$ must be $1$ , otherwise $n -1$ wouldn't reach its maximum value.

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It is given $a_nb_n = 2010$ . We get the following unordered pairs of $\{a_n, b_n\}$ possible.

$\{2010, 1\}, \{1005, 2\}, \{670, 3\}, \{402, 5\}, \{335, 6\}, \{201, 10\}, \{134, 15\}, \{67, 30\}$

From these, we can get the following unordered pairs of $\{a_n -1, b_n -1\}$

$\{2009, 0\}, \{1004, 1\}, \{669, 2\}, \{401, 4\}, \{334, 5\}, \{200, 9\}, \{133, 14\}, \{66, 29\}$

The highest $\gcd$ is of the pair $\{133, 14\}$ , i.e. $7$ .

Therefore $n-1 = 7$

We get $\boxed{n = 8}$

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We can verify this by back substitution. If $a_n = 134, b_n = 15$ (or vice-versa), for $n= 8$ , we get $d_a = 19, d_b = 2$ .

We get $a_n = 19 n - 18$ and $b_n = 2n - 1$ .

From these equations, we do get $a_1 =1, b_1 = 1,$ $a_8b_8 = 2010,$ and $\gcd(19, 2) = 1$ , therefore $\boxed{n=8}$ is the correct answer.

Given a1 = b1 = 1 let d1 & d2 be the respective common differences we know that d1 and d2 are positive integers

Now, an*bn = 2010

=> [1+(n-1)d1]*[1+(n-1)d2] = 2010

=> (n-1)[d1+d2+(n-1)d1d2] = 2009

the only possibility would be when n-1 = 7

Hence n = 8