Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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+ | {{duplicate|[[2022 AMC 10A Problems/Problem 7|2022 AMC 10A #7]] and [[2022 AMC 12A Problems/Problem 4|2022 AMC 12A #4]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | The least common multiple of a positive | + | The least common multiple of a positive integer <math>n</math> and <math>18</math> is <math>180</math>, and the greatest common divisor of <math>n</math> and <math>45</math> is <math>15</math>. What is the sum of the digits of <math>n</math>? |
<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> | ||
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15 &= 3\cdot5. | 15 &= 3\cdot5. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | From the least common multiple condition, we | + | Let <math>n = 2^a\cdot3^b\cdot5^c.</math> It follows that: |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>From the least common multiple condition, we have <cmath>\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,</cmath> from which <math>a=2, b\in\{0,1,2\},</math> and <math>c=1.</math></li><p> | ||
+ | <li>From the greatest common divisor condition, we have <cmath>\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,</cmath> from which <math>b=1.</math></li><p> | ||
+ | </ol> | ||
+ | Together, we conclude that <math>n=2^2\cdot3\cdot5=60.</math> The sum of its digits is <math>6+0=\boxed{\textbf{(B) } 6}.</math> | ||
− | + | ~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 | ||
− | + | ~Education, the Study of Everything | |
− | + | ==Video Solution 2== | |
− | = | + | https://youtu.be/q2y-Wfdi4q8 |
− | |||
− | |||
− | ~ | + | ~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 == | ||
{{AMC10 box|year=2022|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2022|ab=A|num-b=6|num-a=8}} | ||
{{AMC12 box|year=2022|ab=A|num-b=3|num-a=5}} | {{AMC12 box|year=2022|ab=A|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
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.