2
$\begingroup$

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.

$\endgroup$
3
  • 4
    $\begingroup$ your assumptions say $a_2>a_1$ $\endgroup$ Aug 8 at 12:42
  • $\begingroup$ oh i didn't see the a_2 >a_1, my bad $\endgroup$ Aug 8 at 12:54
  • $\begingroup$ @MichalAdamaszek post it as an answer if you want reputation $\endgroup$ Aug 8 at 12:54

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $n=k+1=8$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.