How do you find the Least common multiple of #18m^2, 24nm#?

Question

Emery Allen · Accepted Answer

#lcm(18m^2,24nm) = 2^3*3^2*m^2*n=72m^2n#.

The #lcm# is the product of all the factors raised to the highest power they have in the given terms. We can rewrite #18m^2# as #2*3^2*m^2#, and we can rewrite #24nm# as #2^3*3*n*m#. The highest power of #2# appears in the 2nd term: #2^3#. The highest power of #3# appears in the 1st term: #3^2#. The highest power of #m# appears in the 1st term: #m^2#. The highest power of #n# appears in the 2nd term: #n#. The #lcm(18m^2,24nm)# is just the product of all of these pieces: #lcm(18m^2,24nm) = 2^3*3^2*m^2*n=72m^2n#. If we just multiply these two numbers together, what would happen? #18m^2*24nm=(2*3^2*m^2)*(2^3*3*n*m)# We can reduce this product by dividing out those common pieces. We'll take this product and cross out the divisors the two terms have in common. We know this will be a multiple of both terms, but it won't be the lowest common one because they both have some (prime) factors in common. #(cancel(2)*3^2*m^2)*(2^3*3*n*m)# They both have at least one #2#, so we cross a 2 out. #(color(grey)(cancel(2))*3^2*m^2)*(2^3*cancel(3)*n*m)# Same goes for the #3#. #(color(grey)cancel(2)*3^2*m^2)*(2^3*color(grey)cancel(3)*n*cancel(m))# Only one has an #n#, but they both have at least one #m#. Cross this out. We're left with #2^3*3^2*m^2*n#, which is the #lcm#. What we ended up crossing out is #2*3*m#, the product of the divisors the two terms had in common. In fact, this product is called the greatest common divisor (#gcd#). This leads us to an interesting formula: #lcm(a,b)=(atimesb)/gcd(a,b)#, or #gcd(a,b)timeslcm(a,b)=atimesb#. Basically, this says "the pieces that are crossed out #times# the pieces that are not crossed out #=# all the pieces". This formula is useful because the #gcd# is often easier to see.

Emery Allen · Answer

undefined To find the least common multiple (LCM) of $18m^2$ and $24nm$, first, we need to factorize each expression:

$18m^2 = 2 \times 3^2 \times m^2$  
$24nm = 2^3 \times 3 \times n \times m$

Then, the LCM is the product of the highest powers of all the prime factors involved. So, the LCM of $18m^2$ and $24nm$ is $2^3 \times 3^2 \times m^2 \times n$, which simplifies to $72m^2n$.