Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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= <math>3^{min(b,2)} \cdot 5^{min(c,1)} = 15</math> | = <math>3^{min(b,2)} \cdot 5^{min(c,1)} = 15</math> | ||
This means, that since <math>c = 1</math>, <math> 3^{min(b,2)} \cdot 5 = 15</math>, so <math>min(b,2)</math> = <math>1</math> and <math>b = 1</math>. | This means, that since <math>c = 1</math>, <math> 3^{min(b,2)} \cdot 5 = 15</math>, so <math>min(b,2)</math> = <math>1</math> and <math>b = 1</math>. | ||
− | Hence, multiplying using <math>a = 2</math>, <math>b = 1</math>, <math>c = 1</math> gives <math>n = 60</math> and the sum of digits is hence <math>\boxed{\textbf{(B)} ~6}</math> | + | Hence, multiplying using <math>a = 2</math>, <math>b = 1</math>, <math>c = 1</math> gives <math>n = 60</math> and the sum of digits is hence <math>\boxed{\textbf{(B)} ~6}</math> |
~USAMO333 | ~USAMO333 |
Revision as of 02:16, 17 November 2022
Problem
The least common multiple of a positive divisor and
is
, and the greatest common divisor of
and
is
. What is the sum of the digits of
?
Solution 1
Note that
From the least common multiple condition, we conclude that
where
From the greatest common divisor condition, we conclude that
Therefore, we have The sum of its digits is
~MRENTHUSIASM
Solution 2
Since the contains only factors of
,
, and
,
cannot be divisible by any other prime.
Let n =
, where
,
, and
are nonnegative integers.
We know that
=
=
=
=
=
Thus
(1)=
so
![]()
(2)=
so
![]()
(3)![]()
From the gcf information, = gcf(n,
) =
=
This means, that since
,
, so
=
and
.
Hence, multiplying using
,
,
gives
and the sum of digits is hence
~USAMO333
Video Solution 1(Quick and Easy)
~Education, the Study of Everything
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.