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

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 $$a_1=b_1=1 and $$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.

• your assumptions say $a_2>a_1$ Aug 8 at 12:42
• oh i didn't see the a_2 >a_1, my bad Aug 8 at 12:54
• @MichalAdamaszek post it as an answer if you want reputation Aug 8 at 12:54

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.

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.

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 $$n=k+1=8$$.