Difference between revisions of "2010 AMC 12B Problems/Problem 3"
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== Solution == | == Solution == | ||
We find the greatest common factor of <math>48</math> and <math>64</math> to be <math>16</math>. The number of factors of <math>16</math> is <math>5</math> which is the answer <math>(E)</math>. | We find the greatest common factor of <math>48</math> and <math>64</math> to be <math>16</math>. The number of factors of <math>16</math> is <math>5</math> which is the answer <math>(E)</math>. | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/I3yihAO87CE?t=179 | ||
| + | |||
| + | ~IceMatrix | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=2|num-a=4|ab=B}} | {{AMC12 box|year=2010|num-b=2|num-a=4|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:54, 26 September 2020
- The following problem is from both the 2010 AMC 12B #3 and 2010 AMC 10B #8, so both problems redirect to this page.
Contents
Problem 3
A ticket to a school play cost 
dollars, where is a whole number. A group of 9th graders buys tickets costing a total of $
, and a group of 10th graders buys tickets costing a total of $
. How many values for are possible?
Solution
We find the greatest common factor of and to be 
. The number of factors of is 
which is the answer 
.
Video Solution
https://youtu.be/I3yihAO87CE?t=179
~IceMatrix
See also
| 2010 AMC 12B (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 • 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. 
