Difference between revisions of "1998 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
| + | For how many values of <math>\displaystyle k</math> is <math>\displaystyle 12^{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math> and <math>8^8</math>? | ||
| + | == Solution == | ||
| + | It is evident that <math>\displaystyle k</math> has only 2s and 3s in its prime factorization, or <math>\displaystyle k = 2^a3^b</math>. | ||
| − | == | + | *<math>\displaystyle 6^6 = 2^6\cdot3^6</math> |
| + | *<math>\displaystyle 8^8 = 2^24</math> | ||
| + | *<math>\displaystyle 12^{12} = 2^{24}\cdot3^{12}</math> | ||
| + | |||
| + | The lcm of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. The <math>\displaystyle lcm(6^6,8^8) = 2^{24}3^6</math>. Therefore <math>\displaystyle 12^{12} = 2^{24}\cdot3^{12} = lcm(2^{24}3^6,2^a3^b) = 2^{max(24,a)}3^{max(6,b)}</math>, and <math>b \displaystyle = 12</math>. Since <math>0 \le \displaystyle a \le 24</math>, there are <math>\displaystyle 025</math> values of <math>\displaystyle k</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1986|before=First question|num-a=2}} | |
Revision as of 15:42, 7 September 2007
Problem
For how many values of 
is 
the least common multiple of the positive integers 
and 
?
Solution
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. The 
. Therefore 
, and 
. Since 
, there are 
values of .
See also
| 1986 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 | ||
