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Comment on Least Common Multiple (LCM)
I am having difficulty with
Question link: https:/
Question link: https://gre.myprepclub.com/forum/in-a-certain-medical-group-dr-schwartz-...
Good question.
Our goal is to find the LCM of 25, 30 and 50
25 = (5)(5)
30 = (2)(3)(5)
50 = (2)(5)(5)
Let's say N = the LCM of 25, 30 and 50
First off, all 3 number share 5 as a common prime.
So, N = (5)(other numbers)
Since N is divisible by 25 (which equals 5 x 5), we can see that N needs a second 5 in its prime factorization.
So, N = (5)(5)(?)(?)
So, now we can be sure that N is a multiple of 25
Next, since N is divisible by 30 (which equals 2 x 3 x 5), we can see that N needs a 2, a 3 and a 5 in its prime factorization. It already has a 5, we need only add a 2 and a 3 to the prime factorization.
So, N = (2)(3)(5)(5)(?)(?)
So, now we can be sure that N is a multiple of 30
Finally, since N is divisible by 50 (which equals 2 x 5 x 5), we can see that N needs a 2 and two 5's in its prime factorization. Well, the prime factorization already has a 2 and two 5's so, as it stands, N is ALREADY is a multiple of 50
So, the LCM of 25, 30 and 50 = (2)(3)(5)(5) = 150
Does that help?
Cheers,
Brent
VERY tedious... better option
Agreed.
Agreed.
That said, listing multiples could take a while for numbers with very little in common.
For example, finding the LCM of 45, 81 and 120 would take a while with this technique.
Cheers,
Brent
If LCM of x and y is 24.
Besides saying x & y equals 1 and 24, is there an easy approach to finding other possible values of x & y? Particularly if I'm not great at calculating in my head.
If you're trying to think of
If you're trying to think of two numbers that have an LCM of N, then some number pairs that always work are:
1 and N
N and N
N and any factor of N
That's a start anyway.
Hi Brent. What if there is no
Here's one approach:
Here's one approach:
Let N = the least common multiple of 2, 3, 15 and 28
This means N is divisible by 2, 3, 15 and 28
So, there must be a 2, a 3, a 15 and a 28 "hiding" within the prime factorization of N.
Let's deal with each number individually.
If N is divisible by 2, then a 2 is "hiding" within the prime factorization of N.
So, N = (2)(?)(?)(?)... [there's our required 2 "hiding"]
If N is divisible by 3, then a 3 is "hiding" within the prime factorization of N.
So, N = (2)(3)(?)(?)(?)...
15 = (3)(5)
If N is divisible by 15, then a 3 AND a 5 must be "hiding" within the prime factorization of N.
We ALREADY have a 3 hiding in the prime factorization of N, but we still need a 5.
So, N = (2)(3)(5)(?)(?)(?)...
28 = (2)(2)(7)
If N is divisible by 28, then two 2's AND one 7 must be "hiding" within the prime factorization of N.
We ALREADY have ONE 2 hiding in the prime factorization of N, so let's add one more to get...
So, N = (2)(3)(5)(2)(?)(?)(?)...
We still need one 7 in the prime factorization of N, so let's add it to get...
So, N = (2)(3)(5)(2)(7)(?)(?)(?)...
Since we're looking for the LEAST common multiple, we can say N = (2)(3)(5)(2)(7) = 420
Does that help?
Cheers,
Brent
Hi Brent,
I understood both of your approach (above mentioned in comment and finding via prime factor). But they both should be giving us the same results , when applied they tend to give us different result . Eg LCM using prime factor of number 30, 25 ,50 is 1500. and LCM using n method mentioned in above example is 150. which method should i use in exam. Please help and thank you in advance
The video only explains how
The video only explains how to find the LCM of 2 numbers.
For more than two numbers, it's best to use the strategy I describe in my post above (posted on November 22, 2018)
Cheers,
Brent