Difference between revisions of "2009 AIME I Problems/Problem 7"

Latest revision as of 01:22, 17 July 2025

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

Solution 3 (I did too many FE's)

The given recursion is equivalent to $\frac{5^{a_{n+1}}}{3(n+1)+2}=\frac{5^{a_n}}{3n+2}$ . By a simple inductive argument, the value $\frac{5^{a_k}}{3k+2}$ is constant for all naturals $k$ . In particular, when $k=1$ , the expression is equal to $1$ , so $\frac{5^{a_k}}{3k+2}=1$ for all naturals $k$ . Therefore, $a_k=\log_5(3k+2)$ , and testing powers of $5$ yields $k=\boxed{41}$ as the answer, which is when $a_{41}=\log_5(3\cdot41+2)=\log_5(125)=3$ .

~ eevee9406

@@ Line 24: / Line 24: @@
 -integralarefun
+==Solution 3 (I did too many FE's)==
+The given recursion is equivalent to <math>\frac{5^{a_{n+1}}}{3(n+1)+2}=\frac{5^{a_n}}{3n+2}</math>. By a simple inductive argument, the value <math>\frac{5^{a_k}}{3k+2}</math> is constant for all naturals <math>k</math>. In particular, when <math>k=1</math>, the expression is equal to <math>1</math>, so <math>\frac{5^{a_k}}{3k+2}=1</math> for all naturals <math>k</math>. Therefore, <math>a_k=\log_5(3k+2)</math>, and testing powers of <math>5</math> yields <math>k=\boxed{41}</math> as the answer, which is when <math>a_{41}=\log_5(3\cdot41+2)=\log_5(125)=3</math>.
+~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
 == See also ==

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
All AIME Problems and Solutions

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