Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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<math>\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12</math> | <math>\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12</math> | ||
| − | == Solution == | + | == Solution 1 == |
Note that | Note that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| Line 24: | Line 24: | ||
~MRENTHUSIASM ~USAMO333 | ~MRENTHUSIASM ~USAMO333 | ||
| − | == | + | ==Solution 2== |
| + | The options for <math>\text{lcm}(x, 18)=180</math> are <math>20</math>, <math>60</math>, and <math>180</math>. The options for <math>\text{gcd}(y, 45)=15</math> are <math>15</math>, <math>30</math>, <math>60</math>, <math>75</math>, etc. We see that <math>60</math> appears in both lists; therefore, <math>6+0=\boxed{\textbf{(B) } 6}</math>. | ||
| + | |||
| + | ~MrThinker | ||
| + | |||
| + | == Remark == | ||
| + | |||
| + | If you ignore or mess up the LCM, and get <math>n=15</math>, you'll still get the correct answer. | ||
| + | |||
| + | ==Video Solution 1 == | ||
https://youtu.be/YI1E8C3ZX-U | https://youtu.be/YI1E8C3ZX-U | ||
| Line 33: | Line 42: | ||
~savannahsolver | ~savannahsolver | ||
| + | |||
| + | ==Video Solution 3 (Smart and Simple) == | ||
| + | https://youtu.be/7yAh4MtJ8a8?si=o-aImrVOwbH1HoZv&t=673 | ||
| + | |||
| + | ~Math-X | ||
| + | |||
| + | ==Video Solution 4== | ||
| + | https://youtu.be/5KAiNlqbrsQ | ||
== See Also == | == See Also == | ||
Latest revision as of 12:29, 8 February 2024
- The following problem is from both the 2022 AMC 10A #7 and 2022 AMC 12A #4, so both problems redirect to this page.
Contents
Problem
The least common multiple of a positive integer 
and 
is 
, and the greatest common divisor of and 
is 
. What is the sum of the digits of ?
Solution 1
Note that
Let 
It follows that:
- From the least common multiple condition, we have
![\[\operatorname{lcm}(n,18) = \operatorname{lcm}(2^a\cdot3^b\cdot5^c,2\cdot3^2) = 2^{\max(a,1)}\cdot3^{\max(b,2)}\cdot5^{\max(c,0)} = 2^2\cdot3^2\cdot5,\]](//latex.artofproblemsolving.com/0/a/8/0a8b85578b0aaa3d09f855daf23059df7f690af7.png)
from which 
and 
- From the greatest common divisor condition, we have
![\[\gcd(n,45) = \gcd(2^2\cdot3^b\cdot5,3^2\cdot5) = 2^{\min(2,0)}\cdot3^{\min(b,2)}\cdot5^{\min(1,1)} = 3\cdot5,\]](//latex.artofproblemsolving.com/4/0/7/4078791e2e129becb18f8794af14ff7fac594ec7.png)
from which 
![\[\operatorname{lcm}(n,18) = \operatorname{lcm}(2^a\cdot3^b\cdot5^c,2\cdot3^2) = 2^{\max(a,1)}\cdot3^{\max(b,2)}\cdot5^{\max(c,0)} = 2^2\cdot3^2\cdot5,\]](http://latex.artofproblemsolving.com/0/a/8/0a8b85578b0aaa3d09f855daf23059df7f690af7.png)
from which 
and 
![\[\gcd(n,45) = \gcd(2^2\cdot3^b\cdot5,3^2\cdot5) = 2^{\min(2,0)}\cdot3^{\min(b,2)}\cdot5^{\min(1,1)} = 3\cdot5,\]](http://latex.artofproblemsolving.com/4/0/7/4078791e2e129becb18f8794af14ff7fac594ec7.png)
from which 
Together, we conclude that 
The sum of its digits is 
~MRENTHUSIASM ~USAMO333
Solution 2
The options for 
are 
, 
, and . The options for 
are , 
, , 
, etc. We see that appears in both lists; therefore, 
.
~MrThinker
Remark
If you ignore or mess up the LCM, and get 
, you'll still get the correct answer.
Video Solution 1
~Education, the Study of Everything
Video Solution 2
~savannahsolver
Video Solution 3 (Smart and Simple)
https://youtu.be/7yAh4MtJ8a8?si=o-aImrVOwbH1HoZv&t=673
~Math-X
Video Solution 4
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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. 
