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Please help me with parts b and c. ​

NO LINKS!!! Please help me with parts b and c. ​-example-1
User Petran
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1 Answer

18 votes
18 votes

9514 1404 393

Answer:

b. A × B = GCF(A, B) × LCM(A, B)

c. 15 = GCF × LCM / 20

Explanation:

b. The products of the numbers in the table are 80, 108, 216, 45.

The products of the corresponding GCF and LCM are 80, 108, 216, 45.

Conjecture: the product of the numbers is equal to the product of the GCF and LCM.

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c. GCF×LCM = 5×60 = 300 = A×B = 20×B

B = 300/20 = 15

The other number is 15, found by using the conjecture of part b.

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Additional comment

If you think for a minute about what the GCF and LCM are, you realize the relationship discussed in this problem must be the case. Consider the factors of two numbers A and B:

A = (factors in common with B) × (factors unique to A)

B = (factors in common with A) × (factors unique to B)

The GCF, by definition, are (factors in common between A and B).

The LCM, by definition, is the product of unique factors:

(common factors) × (factors unique to A) × (factors unique to B)

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Then the product of the LCM and the GCF is ...

GCF × LCM =

(factors in common) × ((factors in common) × (unique to A) × (unique to B))

Using the associative and commutative properties of multiplication, we can rearrange this product to be ...

((factors in common)×(unique to A)) × ((factors in common)×(unique to B))

= A × B

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Personally, I use a kind of diagram to represent the factorization of A and B and their LCM and GCF:

[unique to A (common] unique to B)

Then A = factors in [ ], and B = factors in ( ). The stuff in (common] is the GCF, and the overall product is the LCM.

Using the example of part c, this would look like [4 (5] 3), so A = 4·5 = 20 and B = 5·3 = 15. The GCF is 5, and the LCM is 4·5·3 = 60.

The end parts [unique to A( and ]unique to B) can have no common factors. Any common factors must reside in the (common] part.

User Saharsh
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