problems with LCM.?

Question

I have a hard time doing lcm math problems; is there an easy way without having to factor every number.

Answer 1

Yes,there is a process to find GCF/LCM without actually factoring the numbers.but that depends on the particular skill of the student himself or herself.
let us start with 2# say 6 and 15.Remember a fact that the LCM will always be a multiple of the larger no.hence we know that the whatevr may be the LCM,it must be a higher multiple of 15.As 15 is not exactly divisible by 6,we multiply 15 by 2 and get 30.now check up mentally if 30 is exactly divisible by 6 or not.It is.hence 30 is the LCM
Take two more # 12 and 15.Go on multiplying 15 by 2,3,4 etc till such time the product is exactly divisible by the smaller no. and that product is the LCM
There is one more still easier method but I shall discuss it some other time

Answer 2

Yes, Dave, there is an easier way and
it deserves to be better known.
Let's start with 2 numbers.
The LCM of 2 numbers, a and b equals
ab/(a,b), i.e., their product divided by their GCD.
(Here (a,b) is the greatest common divisor of a and b.)
Example: The LCM of 12 and 15 is 12*15/3 = 60.
I like to arrange the work as follows.
Write the numbers in a row:
..... 12 15
Divide their GCD into both and make a new row
with the GCD on the side:
3 .....4 5
Since 4 and 5 have no common factor
the LCM is the product of the numbers on
the side times all those in the bottom row.
3*4*5 = 60.
You may ask: Why go through all this?
Well, it easily generalises to more than 2 numbers:
If we want the LCM of a, b and c
we can get it as follows:
LCM(a,b,c) = abc/[(a,b,c)*(a,b)*(a,c)*(b,c)]
Rather than remember such a complicated
formula, do the following, which we show by example:
Let's find LCM(12,15,18)
Write the numbers in a row
............. 12 15 18
Next take out the GCD of all 3 numbers and make a new row
with the GCD on the side.
If it is 1, proceed to the next step.
Here it is 3, so we get
3..............4 5 6
Now look for the GCD of pairs. Here there is
only one pair with a common factor, namely 4 and 6.
So take it out and make a new row.
2 ............ 2 5 3
Now no 2 numbers in the last row have a common factor
so the answer is the product of the numbers
on the side times all the numbers in the bottom row.
3*2*2*5*3 = 180.
This easily generalises to more than 3 numbers:
If we have 4 numbers, say, first take out
the gcd of all the numbers, then those of
all triples, then those of all pairs.
The LCM is still the product of all
the numbers on the side times those
in the final row!
For a final example let's find the LCM
of 2,3,4,5,6,7,8,9 and 10.
.................2 3 4 5 6 7 8 9 10
Here the largest group of numbers with
a common factor is the quintuple 2,4,6, 8 and 10.
So take out their GCD of 2:
2.................1 3 2 5 3 7 4 9 5
No quadruple has a common factor, but we do
have the triple 3, 3, 9. So we get
3..................1 1 2 5 1 7 4 3 5
Next, we have 2 pairs to consider: 2 and 4 and 5 and 5
Do these 1 at a time:
2..................1 1 1 5 1 7 2 3 5
5..................1 1 1 1 1 7 2 3 1
We have reached the stage where no 2 of the
numbers have a common factor.
So the answer is
2*3*2*5*7*2*3= 2520.

Answer 3

When comparing the numbers, there are three possibilities. For simplicity, consider two numbers.
If they have no common factors, then the LCM is the product,
Ex LCM(2, 3) = 6
If one is a multiple of the other, then the LCM is the larger number.
Ex. LCM(4, 16) = 16
Otherwise, the LCM is a product of distinct primes, raised to the highest power.
Ex. LCM(8, 18) = 8*9 = 72
8 = 2^3
18 = 2*3^2

Ex. LCM(5, 15, 35) = 3*5*7
[35 is not a multiple of 15]

Recognizing the first two conditions can save time, but prime factoring is a necessary skill & more often required than not.

Answer 4

In a word, no. The lcm is the smallest number that has all the given numbers as factors. To figure out what this number is, you can't get around knowing what the factors of the given numbers are.

Answer 5

Not that I'm aware of. There are tricks however. Obviously if the number is even it's divisible by 2. If it ends with 5 or 0 it's divisible by 5 and if all it's digits add up to a number divisible by 3 then it's divisible by 3. So those 3 take care of numbers divisible by 2,3,4,5,6,8,and 9. That only leaves 7 to check for.

Answer 6

it all depends on the #'s.