Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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Revision as of 01:13, 13 August 2023
- The following problem is from both the 2022 AMC 10A #7 and 2022 AMC 12A #4, so both problems redirect to this page.
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
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, thus 
.
~MrThinker
Video Solution 1
~Education, the Study of Everything
Video Solution 2
~savannahsolver
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. 
