What is the largest possible value of $n$?

Question

Arithmetic sequences $a_n$ and $b_n$ have integer terms with $a_1 = b_1 = 1 < a_2 < b_2$ and $a_n . b_n = 2010$ for some $n$. What is the largest possible value of $n$?

Answer 1

I suppose that $a_n,b_n$ are arithmetic progressions.

Write $a_n = 1 + (n-1)d_1$, and $b_n = 1 + (n-1)d_2$, where $d_1,d_2$ are the common differences of the arithmetic sequences $a_n,b_n$ respectively.

Now, note that $(1 + (n-1)d_1)(1 + (n-1)d_2) = 2010$. We want the largest value of $n$ so that there exist $d_1,d_2 \geq 1$ with the above happening. First of all, note that since the terms are integers, it follows that we must take all possible factorizations of $2010$, at the least.

The factors, in this order, are $1,2,3,5,6,10,15,30,67,134,201,335,402,670,1005,2010$.

Suppose $1 + (n-1)d_1 = a$ and $1 + (n-1)d_2 = b$. Then, $(n-1)d_1 = a-1$ and $(n-1)d_2 = b-1$. That is, $n-1$ divides the $\gcd$ of $a-1$ and $b-1$. We must therefore choose a pair $a,b$ such that the $\gcd$ of $a-1,b-1$ is as large as possible (of course, $a\neq 1, b \neq 1$). Analysing:

$ a=2,b=1005 \implies \gcd = 1 $

$ a=3,b=670 \implies \gcd = 1 $

$ a=5,b=402 \implies \gcd = 1 $

$ a=6,b=335 \implies \gcd = 1 $

$ a=10,b=201 \implies \gcd = 1 $

$ a=15,b=134 \implies \gcd = 7 $

$ a=30,b=67 \implies \gcd = 1 $

Hence, we conclude that the maximum value of $n-1$ is $7$, so that $n=8$. Indeed, if $d_1 = 2$ and $d_2 = 19$, we see that: $$ a_n = 1,3,5,7,9,11,13,\color{blue}{15} \\ b_n = 1,20,39,58,77,96,115,\color{blue}{134} $$ are the eighth terms of the progressions $a_n,b_n$, whose product is $2010$. Hence, the answer is $8$.