HMMT February

Algebra Difficulty 5.4 AIME, harder Prove it United States

Problem:

Suppose $a_{1}, a_{2}, \ldots, a_{100}$ are positive real numbers such that
$a_{k}=\frac{k a_{k-1}}{a_{k-1}-(k-1)}$
for $k=2,3, \ldots, 100$ . Given that $a_{20}=a_{23}$ , compute $a_{100}$ .

Solution

Solution:

If we cross multiply, we obtain $a_{n} a_{n-1}=n a_{n-1}+(n-1) a_{n}$ , which we can rearrange and factor as $\left(a_{n}-n\right)\left(a_{n-1}-(n-1)\right)=n(n-1)$ .

Let $b_{n}=a_{n}-n$ . Then, $b_{n} b_{n-1}=n(n-1)$ . If we let $b_{1}=t$ , then we have by induction that $b_{n}=n t$ if $n$ is odd and $b_{n}=n / t$ if $n$ is even. So we have
$a_{n}= \begin{cases}n t+n & \text{ if } n \text{ odd } \\ n / t+n & \text{ if } n \text{ even }\end{cases}$
for some real number $t$ . We have $20 / t+20=23 t+23$ , so $t \in\{-1,20 / 23\}$ . But if $t=-1$ , then $a_{1}=0$ which is not positive, so $t=20 / 23$ and $a_{100}=100 / t+100=215$ .

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