Difference between revisions of "2009 AMC 10B Problems/Problem 6"
Whitelisted (talk | contribs) m (→Video Solution) |
m (Reverted edits by Whitelisted (talk) to last revision by Savannahsolver) (Tag: Rollback) |
||
Line 14: | Line 14: | ||
==Video Solution== | ==Video Solution== | ||
− | https:// | + | https://youtu.be/F8k7r3LDXoA |
− | ~ | + | ~savannahsolver |
== See also == | == See also == |
Revision as of 16:38, 28 June 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.