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
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. 
