Difference between revisions of "1998 AIME Problems/Problem 1"
(AIME 1998 #1) |
Namelyorange (talk | contribs) |
||
(10 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | For how many values of <math> | + | For how many values of <math>k</math> is <math>12^{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math>, <math>8^8</math>, and <math>k</math>? |
− | == Solution == | + | == Solution 1== |
− | It is evident that <math> | + | It is evident that <math>k</math> has only 2s and 3s in its prime factorization, or <math>k = 2^a3^b</math>. |
− | *<math> | + | *<math>6^6 = 2^6\cdot3^6</math> |
− | *<math> | + | *<math>8^8 = 2^{24}</math> |
− | *<math> | + | *<math>12^{12} = 2^{24}\cdot3^{12}</math> |
− | The | + | The [[LCM]] of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. <math>[6^6,8^8] = 2^{24}3^6</math>. Therefore <math>12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}</math>, and <math>b = 12</math>. Since <math>0 \le a \le 24</math>, there are <math>\boxed{25}</math> values of <math>k</math>. |
+ | |||
+ | == Solution 2== | ||
+ | We want the number of <math>k</math> such that <math>\operatorname{lcm}(6^6,8^8,k)=12^{12}</math>. | ||
+ | |||
+ | Using <math>\operatorname{lcm}</math> properties, this is <math>\operatorname{lcm}(\operatorname{lcm}(2^6\cdot 3^6,2^{24}),k)=2^{24} \cdot 3^{12}</math>, or <math>\operatorname{lcm}(2^{24}\cdot 3^6,k)=2^{24}\cdot 3^{12}</math>. | ||
+ | |||
+ | At this point, we realize that <math>k=2^a\cdot3^b</math>, as any other prime factors would be included in the <math>\operatorname{lcm}</math>. | ||
+ | |||
+ | Also, <math>b=12</math> (or the power of <math>3</math> in the <math>\operatorname{lcm}</math> wouldn't be <math>12</math>) and <math>0\le a\le 24</math> (or the power of <math>2</math> in the <math>\operatorname{lcm}</math> would be <math>a</math> and not <math>24</math>). | ||
+ | |||
+ | Therefore, <math>a</math> can be any integer from <math>0</math> to <math>24</math>, for a total of <math>25</math> values of <math>a</math> and <math>\boxed{025}</math> values of <math>k</math>. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/HISL2-N5NVg?t=2899 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
− | {{AIME box|year= | + | {{AIME box|year=1998|before=First question|num-a=2}} |
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:50, 23 April 2024
Problem
For how many values of is
the least common multiple of the positive integers
,
, and
?
Solution 1
It is evident that has only 2s and 3s in its prime factorization, or
.
The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. . Therefore
, and
. Since
, there are
values of
.
Solution 2
We want the number of such that
.
Using properties, this is
, or
.
At this point, we realize that , as any other prime factors would be included in the
.
Also, (or the power of
in the
wouldn't be
) and
(or the power of
in the
would be
and not
).
Therefore, can be any integer from
to
, for a total of
values of
and
values of
.
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=2899
~ pi_is_3.14
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.