Difference between revisions of "2000 PMWC Problems/Problem I7"
(Created page with "==Problem== <math>a</math> and <math>b</math> are two numbers that have prime factors <math>3</math> and <math>5</math> only. <math>a</math> has <math>12</math> factors (<math>1<...") |
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==Solution== | ==Solution== | ||
| + | Notice that in prime factorizations, one of <math>a</math> or <math>b</math> has to equal <math>3^15^x</math>, and the other must equal <math>3^y5^2</math>. This is due to <math>75=3^1\cdot5^2</math>. Notice that the number equal to <math>3^15^x</math> has <math>2(x+1)</math> factors, and <math>3^y5^2</math> has <math>3(y+1)</math> factors. Since only the latter is divisible by <math>3</math>, <math>3(y+1)=12</math>, so <math>y=3</math>. Thus <math>2(x+1)=10</math>, so <math>x=4</math>. Then our numbers are <math>a=3^35^2</math> and <math>b=3^15^4</math>. The LCM is thus <math>3^35^4=27\cdot625=\boxed{16875}</math>. | ||
| + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ||
==See Also== | ==See Also== | ||
Latest revision as of 22:45, 19 March 2025
Problem
and 
are two numbers that have prime factors 
and 
only. has 
factors (
and itself are included), has 
factors ( and itself are included), and their HCF (Highest Common Factor) is 
. What is the LCM (Least Common Multiple of and ?
Solution
Notice that in prime factorizations, one of or has to equal 
, and the other must equal 
. This is due to 
. Notice that the number equal to has 
factors, and has 
factors. Since only the latter is divisible by , 
, so 
. Thus 
, so 
. Then our numbers are 
and 
. The LCM is thus 
.
~ eevee9406
