Question on Greatest Common Factor

[Wwc]

I am trying to find out different ways to find the greatest common factor and was looking at math.com

At this page:

Factoring - Greatest Common Factor (GCF) - First Glance

www.math.com

Ok so that way works for those numbers there but how does that method work for my question of gcf of: 128 and 96

I understand the gcf is 32, but how is that found using their method?

Can someone write out how it would work using that method shown for 128 and 96?

Also is attached the top of google results showing the same thing.

Using this method I cant seek to get the correct 32 gcf of 128 and 96.

Thanks

 

[MarkFL]

I would write:

[MATH]96=2^5\cdot3[/MATH]
[MATH]128=2^7[/MATH]
We see that only a power of 2 is present in both factorizations, and the smaller power is 5, and so:

[MATH]\text{gcf}(96,128)=2^5=32[/MATH]

 

[ksdhart2]

Well, without seeing what answer(s) you do get, or any of your workings, it's impossible to troubleshoot and figure out what went wrong. However, I can say that I am able to get the correct answer by following their example. Begin by noting:

• \(\displaystyle 96 = 2 \cdot 48\)
• \(\displaystyle 128 = 2 \cdot 64\)

And:

• \(\displaystyle 96 = 2 \cdot 2 \cdot 24\)
• \(\displaystyle 128 = 2 \cdot 2 \cdot 32\)

Continuing this process to its logical conclusion, we get:

• \(\displaystyle 96 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3\)
• \(\displaystyle 128 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2\)

Now let's do exactly what they do in the example, circling any numbers that appear in both factorizations:

• \(\displaystyle 96 = \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot 3\)
• \(\displaystyle 128 = \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot \enclose{circle}[mathcolor="red"]{\color{black}{2}} \cdot 2 \cdot 2\)

We circled five 2's in each factorization, so the GCF will be these five 2's multiplied together. When we do that, we get \(\text{GCF}(96, 128) = 2^5 = 32\).

 

[Wwc]

Yep that makes sense... thanks