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 Post subject: GMAT Number Theory
PostPosted: Wed Apr 04, 2012 1:31 am 
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Joined: Sun May 30, 2010 3:15 am
Posts: 424
If n is a positive even integer, then what are the GCF and LCM of the expressions n² + 5n + 6 and n² + 3n + 2?

Solution:
Just as with numbers, you need to compare the factors.

n² + 5n + 6 = (n + 2)(n + 3)

n² + 3n + 2 = (n + 2)(n + 1)

(n + 2) is a common factor, while (n + 3) and (n + 1) are two consecutive odd integers. The GCF of two consecutive odd integers is 1, because if we divide (n + 3) by any factor of (n + 1), the remainder will be 2.

Therefore the GCF of n² + 5n + 6 and n² + 3n + 2 is (n + 2).

The least common multiple will be the product (n + 2)(n + 3)(n + 1). The answer choices may leave the expressions as factors, or may multiply the factors:
n³ + 6n² + 11n + 6
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Could you explain the solution a little better. I'm confused
because I thought the GCF was n² + 2, would you
mind explaining. Thanks.


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 Post subject: Re: a practice question: finding GCF and LCM (with a variabl
PostPosted: Wed Apr 04, 2012 2:22 am 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
I thought the GCF was n² + 2
You can check your result to see if it is at least a common factor by dividing the given expressions by (n² + 2).

(n + 2)(n + 3) / (n² + 2)
(n + 2)(n + 1) / (n² + 2)

If at least one of them is NOT divisible, then (n² + 2) is NOT a common factor. Clearly they do NOT have to be divisible by n² + 2. You may try a value to be sure, for example, n = 4 (easier n = 2 will also do it for the first expression).
(4 + 2)(4 + 3) / (4² + 2) = 42/18 = 7/3
(4 + 2)(4 + 1) / (4² + 2) = 30/18 = 5/3


On the other hand, the proper GCF, (n + 2), is a common factor indeed:
(n + 2)(n + 3) / (n + 2) = (n + 3)
(n + 2)(n + 1) / (n + 2) = (n + 1)

But how do we know that it is the GREATEST common factor? Because if we divide by the greatest common factor, the remaining numbers must be relatively prime (have no common factors greater than 1). Otherwise (if they had a common factor) we could multiply the GCF by that common factor and make it greater.

So (n + 3) and (n + 1) must be relatively prime. And indeed they are, because
n + 3 = (n + 1) + 2
n + 1 is odd, so 2 can't be a common factor. If there exists any other possible common factor of (n + 1) and (n + 3), it must be greater than 2.

If we divide n + 3 by ANY factor of n + 1 the remainder is 2, because
n + 3 = (n + 1) + 2
(n + 1) is divisible by that factor. 2 is NOT divisible, because that factor is greater than 2.

So we proved that (n + 2) is GCF of (n + 2)(n + 1) and (n + 2)(n + 3).


Here is a short explanation of the same problem with fixed numbers:
What is the GCF of (10 + 2)(10 + 3) and (10 + 2)(10 + 1)?

(10 + 2)(10 + 3) = 12 × 13 |same number × (odd + 2)
(10 + 2)(10 + 1) = 12 × 11 |same number × odd

GCF = 12, because 13 and 11 are relatively prime.


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