Where is the flaw in my approach? : AMC12A 2010 Problem 20

Question

On the problem below, I am unsure why my answer was wrong.

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

$$ \mathbf{A)} \ \ \ 2 \quad \quad \mathbf{B)} \ \ \ 3 \quad \quad\mathbf{C)} \ \ \ 8 \quad \quad \mathbf{D)} \ \ \ 288 \quad \quad \mathbf{E)} \ \ \ 2009 $$

My answer was 2009. The 2 sequences I thought of were $a_1 = 1$ and the difference is 0, as well as $b_1 = 1$ and the difference being 1 all the way up to 2010, which is 2009. Where is the flaw in my logic? I googled if arithmetic sequences could have a common difference of 0 and it said yes.

Answer 1

The assumption is that the increments are positive integers. So we have $$[a_1+(n-1)k][b_1+(n-1)\ell] = 2010, $$ where $k,\ell$ are the increments of $(a_n)$ and $(b_n)$ respectively. It follows that $n-1\mid 2009 = 7^2 \cdot 41$ Obviously, $n=288$ is out of the question. For $n=50$ and $n=42$, one of the increments must be zero. For $n=8$, $k=2$ and $\ell=19$ is suitable, so (C) $n=8$ is correct.

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For $n=50$, from $$(1+49k)(1+49\ell) = 2010 \Leftrightarrow (49k+1)\ell = 41-k $$ it immediately follows that $k=0$.

For $n=42$, $$(1+41k)(1+41\ell) = 2010 \Leftrightarrow (41k+1)\ell = 49-k $$ it is again clear that $k=0$ is forced.

For $n=8$, given that $$ (1+7k)(1+7\ell) = 2010 \Leftrightarrow (7k+1)\ell = 287-k $$ one can find appropriate $k,\ell$ with small effort.

Answer 2

Let $c=a_2-a_1>0$, $d=b_2-b_1>0$, then $a_n=a_1+(n-1)c=1+(n-1)c$, $b_n=1+(n-1)d$. Let $n-1=k$, then $a_nb_n=(1+kc)(1+kd)=1+k(c+d)+k^2 cd=2010$. $c+d\geq 2$, $cd\geq 1$, then $a_nb_n\geq 1+2k+k^2=(k+1)^2$, then $(k+1)^2\leq 2010 \Rightarrow k+1\leq \sqrt{2010}\Rightarrow$ $k\leq \sqrt{2010}-1 < \sqrt{2025}-1=44$, then $k\leq 43$. $1+k(c+d)+k^2 cd=2010\Rightarrow$ $k(c+d+kcd)=2009=7^2\cdot 41$. $k$ divides 2009, then $k$ can be 1, 7 or 41.

At $k=41$: $c+d+kcd=2009/k \Rightarrow c+d+41cd=49 \Rightarrow$ $41cd<49$, $cd=1$, $c=d=1$ but then $c+d+41cd\neq 49$, then $k=41$ is impossible.

At $k=7$: $c+d+kcd=2009/k \Rightarrow c+d+7cd=287 \Rightarrow$ $c+d=7(41-cd)$ is divisible by 7. Let $c+d=7m$, then $cd=41-m$. $d=7m-c$, then $c(7m-c)=41-m$, then $m=\frac{c^2+41}{7c+1}$, then $7m=\frac{7c^2+287}{7c+1}=\frac{7c^2+c+287-c}{7c+1}=c+\frac{287-c}{7c+1}$, then $49m=7c+\frac{2009-7c}{7c+1}=7c+\frac{2010-7c-1}{7c+1}=7c-1+\frac{2010}{7c+1}$. Integer divisors of 2010 are {1, 2, 3, 5, 6, 10, 15, 30, 67, 134, 201, 335, 402, 670, 1005, 2010}, then $7c+1$ is one of 15, 134 and 2010. $cd$ is less than 41, then $c$ is less than 41, then $7c+1$ is less than 288, then $7c+1$ is one of 15 and 134, then $c$ is one of 2 and 19. At $c=2$, $m=\frac{c^2+41}{7c+1}=3$, then $d=7m-c=19$. At $c=19$, $m=3$, $d=2$. Then case $k=7$ is possible. Then answer is $n=k+1=8$.