2009 AIME I Problems/Problem 7

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: $5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1$ $5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}$ $5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}$ $a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}$ $a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}$ 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 $125$ : $3n + 2 = 125$ $3n = 123$ $n = \boxed{041}$

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2009 AIME I Problems/Problem 7

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