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>. | ||
| + | == Solution 2 == | ||
| + | Since the price of one ticket is a whole number, it must be a divisor of both <math>48</math> and <math>64</math>. An integer is a common divisor of two numbers if and only if it is divisible by their difference. Hence, all the possible prices are positive divisors of <math>16</math>, or <math>1</math>, <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math>, making the answer <math>5</math>, or E. - By Monkey_King | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 01:46, 29 August 2024
- The following problem is from both the 2010 AMC 12B #3 and 2010 AMC 10B #8, so both problems redirect to this page.
Problem
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 
.
Solution 2
Since the price of one ticket is a whole number, it must be a divisor of both and . An integer is a common divisor of two numbers if and only if it is divisible by their difference. Hence, all the possible prices are positive divisors of , or 
, 
, 
, 
, and , making the answer , or E. - By Monkey_King
Video Solution
~Education, the Study of Everything
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 | |
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
