2016 AMC 8 Problems/Problem 20: Difference between revisions
Pi is 3.14 (talk | contribs) |
|||
| (6 intermediate revisions by 6 users not shown) | |||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
We wish to find possible values of <math>a</math>, <math>b</math>, and <math>c</math>. By finding the greatest common factor of <math>12</math> and <math>15</math>, we can find that <math>b</math> is 3. Moving on to <math>a</math> and <math>c</math>, in order to minimize them, we wish to find the least such that the least common multiple of <math>a</math> and <math>3</math> is <math>12</math>, <math>\rightarrow 4</math>. Similarly, with <math>3</math> and <math>c</math>, we obtain <math>5</math>. The least common multiple of <math>4</math> and <math>5</math> is <math>20 \rightarrow \boxed{\textbf{(A)} 20}</math> | We wish to find possible values of <math>a</math>, <math>b</math>, and <math>c</math>. By finding the greatest common factor of <math>12</math> and <math>15</math>, we can find that <math>b</math> is 3. Moving on to <math>a</math> and <math>c</math>, in order to minimize them, we wish to find the least such that the least common multiple of <math>a</math> and <math>3</math> is <math>12</math>, <math>\rightarrow 4</math>. Similarly, with <math>3</math> and <math>c</math>, we obtain <math>5</math>. The least common multiple of <math>4</math> and <math>5</math> is <math>20 \rightarrow \boxed{\textbf{(A)} 20}</math> | ||
===Solution 2=== | |||
The factors of 12 are 12,6,4,3,2,1. The factors of 15 are 1,3,5,15. The 2 numbers that repeat are 1 and 3 so b either has to be 1 or 3. If b is 3 then a is 4 and c is 5 and the least common multiple of 4 and 5 are 20. We don't have to test 1 because 20 is the lowest answer, so if b equaling 1 resulted in the least common multiple being less that 20 then the correct answer won't be there. So the answer is <math>\boxed{\textbf{(A)} 20}</math>. | |||
===Solution 3=== | |||
Let's say that <math>a = 4, b = 3, c = 5</math>, this works so clearly <math>GCF(4, 5) = \boxed{\textbf{(A)} 20}</math>. - JasonDaGoat | |||
==Video Solution by Pi Academy== | |||
https://youtu.be/wvRmxjwOUHY?si=mNtAIGDHVdPaKWUX | |||
==Video Solution(CREATIVE THINKING + ANALYSIS!!!)== | |||
https://youtu.be/_-xC-qQMCbk | |||
~Education, the Study of Everything | |||
== Video Solution by OmegaLearn== | == Video Solution by OmegaLearn== | ||
| Line 21: | Line 36: | ||
{{AMC8 box|year=2016|num-b=19|num-a=21}} | {{AMC8 box|year=2016|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Number Theory Problems]] | |||
Latest revision as of 15:22, 7 September 2025
Problem
The least common multiple of 
and 
is 
, and the least common multiple of and 
is 
. What is the least possible value of the least common multiple of and ?
Solution
We wish to find possible values of , , and . By finding the greatest common factor of and , we can find that is 3. Moving on to and , in order to minimize them, we wish to find the least such that the least common multiple of and 
is , 
. Similarly, with and , we obtain 
. The least common multiple of 
and is 
Solution 2
The factors of 12 are 12,6,4,3,2,1. The factors of 15 are 1,3,5,15. The 2 numbers that repeat are 1 and 3 so b either has to be 1 or 3. If b is 3 then a is 4 and c is 5 and the least common multiple of 4 and 5 are 20. We don't have to test 1 because 20 is the lowest answer, so if b equaling 1 resulted in the least common multiple being less that 20 then the correct answer won't be there. So the answer is 
.
Solution 3
Let's say that 
, this works so clearly 
. - JasonDaGoat
Video Solution by Pi Academy
https://youtu.be/wvRmxjwOUHY?si=mNtAIGDHVdPaKWUX
Video Solution(CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=2340
~ pi_is_3.14
Video Solution
~savannahsolver
See Also
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
