Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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<math>(3) c = 1.</math> | <math>(3) c = 1.</math> | ||
From the gcf information, <math>gcf(n,45)</math> = gcf(n, <math>3^2 \cdot 5</math>) = <math>gcf(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5) </math> | From the gcf information, <math>gcf(n,45)</math> = gcf(n, <math>3^2 \cdot 5</math>) = <math>gcf(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5) </math> | ||
− | = <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 c = 1, 3^min(b,2) \ | + | 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 a = 2, b = 1, c = 1 gives n = 60 and the sum of digits is hence 6. | + | Hence, multiplying using <math>a = 2</math>, <math>b = 1</math>, <math>c = 1</math> gives n = 60 and the sum of digits is hence 6. |
== See Also == | == See Also == |
Revision as of 01:41, 12 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 that
Therefore, we have The sum of its digits is
~MRENTHUSIASM
Solution 2
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 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
From the gcf information,
= gcf(n,
) =
=
This means, that since
,
, so
=
and
.
Hence, multiplying using
,
,
gives n = 60 and the sum of digits is hence 6.
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.