# 2009 AIME I Problems/Problem 7

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## Problem

The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$. Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$.

## Solution

The best way to solve this problem is to get the iterated part out of the exponent: $$5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1$$ $$5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}$$ $$5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}$$ $$a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}$$ $$a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}$$ Plug in $n = 1, 2, 3, 4$ to see the first few terms of the sequence: $$\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.$$ We notice that the terms $5, 8, 11, 14$ are in arithmetic progression. Since $a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}$, we can easily use induction to show that $a_n = \log_5{(3n + 2)}$. So now we only need to find the next value of $n$ that makes $\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$. We test $25$: $$3n + 2 = 25$$ $$3n = 23$$ This has no integral solutions, so we try $125$: $$3n + 2 = 125$$ $$3n = 123$$ $$n = \boxed{041}$$

## Solution 2 (Telescoping)

We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$, we get: $$5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}$$ This simplifies to $$5^{a_n}=3n+2.$$ We can now test powers of $5$.

$5$ - that gives us $n=1$, which is useless.

$25$ - that gives a non-integer $n$.

$125$ - that gives $n=\boxed{41}$.

-integralarefun