Math For Elementary Teachers

Part 3iv: Least Common Multiple

Further Discussion:

GCF and LCM, A More Indepth Look

If you would like to take the discussion of GCF and LCM one step further, there is a nice relationship between whole numbers a and b, GCF(a,b), and LCM(a,b).

Let’s use the same numbers we used in the previous clip, 150 and 140. If we use exponential form, we can express the prime factorizations of these numbers as follows:

150= {2} \cdot {3} \cdot {5^2}
140= {2^2} \cdot {5} \cdot {7}

Now the procedure for computing GCF(150,140) is to take each prime the smaller number of times it appears in any of the prime factorizations, as these will be common factors. To obtain the greatest common factor, we multiply all of these powers of primes together:

GCF(150,140) = {2} \cdot {5}

Since the prime factor 2 occurs once in the prime factorization of 150 and twice in the prime factorization of 140, we take one factor of 2 when computing the GCF (as only one factor is common). Similarly, we take no factors of 3, one factor of 5 and no factors of 7.

But when we compute LCM(150,140), we take each prime the greater number of times that it appears in any of the prime factorizations. When we multiply all of these powers of primes together, we will obtain the least common multiple:

LCM(150,140)= {2^2} \cdot {3} \cdot {5^2} \cdot {7}

Since the prime factor 2 occurs once in the prime factorization of 150 and twice in the prime factorization of 140, when building the LCM we need to take two factors of 2 (if we take fewer than two factors of 2 the number we eventually build will not be a multiple of both 150 and 140, and if we take more than two factors of 2 the number we eventually build will not be the least common multiple). Similarly, we need one factor of 3, two factors of 5, and one factor of 7.

Now comes the interesting part, and we will see why some of the above numbers are green while some others are red.  Multiply  150 and 140, but do so with the prime factorizations, and notice that we can use the Any-Order property of multiplication to multiply numbers in any order we like:

150 \cdot 140 = ( {2} \cdot {3} \cdot {5^2} ) \cdot (  {2^2} \cdot {5} \cdot {7} ) = ( {2} \cdot {5} ) \cdot ( {2^2} \cdot {3} \cdot {5^2} \cdot {7} ) = GCF(150,140) \cdot LCM (150,140)

There is nothing special about the numbers  150 and 140; for any two whole numbers a and b, the following relationship holds (the general proof is based on the above explanation):

a \cdot b = GCF(a,b) \cdot LCM(a,b)

So if we manipulate this equation ever so slightly, we obtain the following equations:

GCF(a,b) = \dfrac{a \cdot b}{LCM(a,b)}   and   LCM(a,b) = \dfrac{a \cdot b}{GCF(a,b)}

Therefore, we can use GCF(a,b) to find LCM(a,b) and vice versa!

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