Difference between revisions of "2009 AMC 10B Problems/Problem 6"
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== Solution == | == Solution == | ||
The age of each person is a factor of <math>128 = 2^7</math>. So the twins could be <math>2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8</math> years of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their ages is <math>2 + 8 + 8 = \boxed{18}</math>. The answer is <math>\mathrm{(D)}</math>. | The age of each person is a factor of <math>128 = 2^7</math>. So the twins could be <math>2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8</math> years of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their ages is <math>2 + 8 + 8 = \boxed{18}</math>. The answer is <math>\mathrm{(D)}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/F8k7r3LDXoA | ||
+ | |||
+ | ~savannahsolver | ||
== See also == | == See also == |
Revision as of 20:59, 21 January 2021
- The following problem is from both the 2009 AMC 10B #6 and 2009 AMC 12B #5, so both problems redirect to this page.
Contents
[hide]Problem
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
Solution
The age of each person is a factor of . So the twins could be years of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their ages is . The answer is .
Video Solution
~savannahsolver
See also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.