Difference between revisions of "2002 AIME II Problems/Problem 3"
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==Solution 2(similar to Solution 1)== | ==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>. <math>27*r^2= c = 48</math>. Therefore, the answer is <math>27+36+48 = \boxed{111}</math> | + | 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 | ~idk12345678 |
Latest 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
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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.