Difference between revisions of "2002 AIME II Problems/Problem 3"

Revision as of 02:31, 6 December 2019

Problem

It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$

Solution

$abc=6^6$ . Since they form an increasing geometric sequence, $b$ is the geometric mean of the product $abc$ . $b=\sqrt[3]{abc}=6^2=36$ .

Since $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$ . Out of these, the only value of $a$ that works is $a=27$ , from which we can deduce that $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48$ .

Thus, $a+b+c=27+36+48=\boxed{111}$

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 == See also ==
 {{AIME box|year=2002|n=II|num-b=2|num-a=4}}
+[[Category: Intermediate Algebra Problems]]
 {{MAA Notice}}

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Difference between revisions of "2002 AIME II Problems/Problem 3"

Revision as of 02:31, 6 December 2019

Problem

Solution

See also