Difference between revisions of "2002 AIME II Problems/Problem 3"
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It is given that <math>\log_{6}a + \log_{6}b + \log_{6}c = 6,</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[positive]] [[integer]]s that form an increasing [[geometric sequence]] and <math>b - a</math> is the [[Perfect square|square]] of an integer. Find <math>a + b + c.</math> | It is given that <math>\log_{6}a + \log_{6}b + \log_{6}c = 6,</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[positive]] [[integer]]s that form an increasing [[geometric sequence]] and <math>b - a</math> is the [[Perfect square|square]] of an integer. Find <math>a + b + c.</math> | ||
− | == Solution == | + | == Solution 1== |
<math>abc=6^6</math>. Since they form an increasing geometric sequence, <math>b</math> is the [[geometric mean]] of the [[product]] <math>abc</math>. <math>b=\sqrt[3]{abc}=6^2=36</math>. | <math>abc=6^6</math>. Since they form an increasing geometric sequence, <math>b</math> is the [[geometric mean]] of the [[product]] <math>abc</math>. <math>b=\sqrt[3]{abc}=6^2=36</math>. | ||
− | Since <math>b-a</math> is the square of an integer, we can find a few values of <math>a</math> that work: 11, 20, 27, 32, and 35. | + | Since <math>b-a</math> is the square of an integer, we can find a few values of <math>a</math> that work: <math>11, 20, 27, 32,</math> and <math>35</math>. Out of these, the only value of <math>a</math> that works is <math>a=27</math>, from which we can deduce that <math>c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48</math>. |
− | < | + | Thus, <math>a+b+c=27+36+48=\boxed{111}</math> |
+ | |||
+ | ==Solution 2(similar to Solution 1)== | ||
+ | Let <math>r</math> be the common ratio of the geometric sequence. Since it is increasing, that means that <math>b = ar</math>, and <math>c = ar^2</math>. Simplifying the logarithm, we get <math>\log_6(a^3*r^3) = 6</math>. Therefore, <math>a^3*r^3 = 6^6</math>. Taking the cube root of both sides, we see that <math>ar = 6^2 = 36</math>. Now since <math>ar = b</math>, that means <math>b = 36</math>. Using the trial and error shown in solution 1, we get <math>a = 27</math>, and <math>r = \frac{4}{3}</math>. Now, <math>27*r^2= c = 48</math>. Therefore, the answer is <math>27+36+48 = \boxed{111}</math> | ||
+ | |||
+ | ~idk12345678 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2002|n=II|num-b=2|num-a=4}} | |
− | + | ||
− | + | [[Category: Intermediate Algebra Problems]] | |
+ | {{MAA Notice}} |
Revision as of 21:50, 5 April 2024
Problem
It is given that where and are positive integers that form an increasing geometric sequence and is the square of an integer. Find
Solution 1
. Since they form an increasing geometric sequence, is the geometric mean of the product . .
Since is the square of an integer, we can find a few values of that work: and . Out of these, the only value of that works is , from which we can deduce that .
Thus,
Solution 2(similar to Solution 1)
Let be the common ratio of the geometric sequence. Since it is increasing, that means that , and . Simplifying the logarithm, we get . Therefore, . Taking the cube root of both sides, we see that . Now since , that means . Using the trial and error shown in solution 1, we get , and . Now, . Therefore, the answer is
~idk12345678
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.