Difference between revisions of "1998 AIME Problems/Problem 1"
m (→Solution) |
|||
Line 8: | Line 8: | ||
*<math>12^{12} = 2^{24}\cdot3^{12}</math> | *<math>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. <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>025</math> values of <math>k</math>. | + | 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{025}</math> values of <math>k</math>. |
== See also == | == See also == |
Revision as of 05:17, 29 June 2011
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. . Therefore , and . Since , there are values of .
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 |