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So I have this problem. I need help with the process in finding the LCM and GCF. I need a technique which is fast but yet accurate.

My problem is find the the GCF and LCM for (24, 56, 96).

LCM = 672 GCF = 8.

I used "Factors Grid" and "Prime Factorization" but can't derive on the answers above. I've manage to get the GCF using "Factors Grid".

Need help how the answer were derive.

Thank You.

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See this also: math-only-math.com/… –  B. S. Jan 27 '13 at 16:14

3 Answers 3

Here are the prime factorizations of your three numbers: $$ \begin{array}{rlll} 24 = & 2^3&\cdot 3^1&\cdot 7^0\\ 56 = & 2^3& \cdot 3^0& \cdot 7^1\\ 96 = & 2^5&\cdot 3^1&\cdot 7^0 \end{array} $$

To find the GCF, take the minimum exponent for each prime; this gives $2^3\cdot3^0\cdot7^0 = 8$.

To find the LCM, take the maximum exponent for each prime; this gives $2^5\cdot3^1\cdot7^1 = 672$.

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First notice that each of $24, 56, 96$ is even. This means that each has a factor of 2. If a set of numbers has a common factor, this will also be a factor of the GCF. You can divide the original numbers by this common factor and work with the resulting numbers instead; they will be smaller, so easier to handle. Then you can repeat this process until the numbers are so small that the answer is apparent.

In this example, since each number is divisible by 2, you can use: $$\begin{array}{rrrrl} GCF(&24,& 56,& 96) & = \\ GCF(&2\cdot12,&2\cdot 28,&2\cdot 48) & = \\ 2\cdot GCF(&12,&28,&48)&= (\ldots\text{do it again}\ldots)\\ 4\cdot GCF(&6,&14,&24)&= (\ldots\text{once more}\ldots)\\ 8\cdot GCF(&3,&7,&12)& = \\ 8\cdot 1 &&&&=\\8 \end{array} $$

The last step being that the greatest common factor of 3, 7, and anything is evidently 1, since 3 and 7 have no common factor besides 1.

You can apply a similar process for the LCM.

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gcd: $\rm\ (24,56,24n) = (24, 56\:mod\:24, 24n) = (24,8,24n) = 8(3,1,3n) = \bf\color{#C00}8$

lcm: $\rm\ [24,56,24n] = {\bf \color{#C00}8}[3,7,3n] = 8[7,3n]\ \ (= 8\cdot 7\cdot 3n\ $ if $\rm\ 7\nmid n,\:$ e.g. your $\rm\: n = 4).$

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