Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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==Solution 2== | ==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, | + | 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 | ~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 == | ==Video Solution 1 == | ||
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~Math-X | ~Math-X | ||
− | == | + | ==Video Solution 4== |
− | + | https://youtu.be/5KAiNlqbrsQ | |
− | |||
== See Also == | == See Also == |
Latest revision as of 13:29, 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
from which
and
- From the greatest common divisor condition, we have
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.