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REDUNDANT PRIME

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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REDUNDANT PRIME

by Gurpinder » Tue Aug 10, 2010 9:11 am
hello world,

i came across this concept in the mgmat number properties book about redundant primes.
the question was, if J is divisible by 12 and 10, is J divisible by 24?
so the prime factors of 12 =3x2x2 and prime factors of 10 =5x2 and for 24 = 3x2x2x2

the book said that since 12 has 2 2s and 10 has a single 2, the single 2 might be one of the 2 2s we have for 12. therefore, since we need 3 2s for the 24, we cannot determine whether J is divisible by 24.

First of all, does this make sense?

secondly, I applied this principle to a GMAT problem and got it wrong!

the question from the GMAT was:

If the positive integer x is a multiple of 4 and the
positive integer y is a multiple of 6, then xy must be a
multiple of which of the following?

I. 8
II. 12
III. 18

(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III

OA: B
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by 4GMAT_Mumbai » Tue Aug 10, 2010 11:09 am
Gurpinder wrote:hello world,

i came across this concept in the mgmat number properties book about redundant primes.
the question was, if J is divisible by 12 and 10, is J divisible by 24?
so the prime factors of 12 =3x2x2 and prime factors of 10 =5x2 and for 24 = 3x2x2x2

the book said that since 12 has 2 2s and 10 has a single 2, the single 2 might be one of the 2 2s we have for 12. therefore, since we need 3 2s for the 24, we cannot determine whether J is divisible by 24.

First of all, does this make sense?
Yes, it does make sense - Let J be 60 = 2 * 2 * 3 * 5. It is divisible by 12 as there is a 2,2 and 3. It is also divisible by 10 as it has a 2 and a 5. However, it is not divisible by 24
secondly, I applied this principle to a GMAT problem and got it wrong!

the question from the GMAT was:

If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?

I. 8
II. 12
III. 18

(B) I and II only
OA: B
The approach here is a little different because in the previous case, it was the same integer J. Here, we are talking about two different integers - x and y. x brings in two 2s as prime factors and y brings in a 2 and a 3 as prime factors. In xy, the 2 brought in by y is entirely different from the two 2s brought in by x.

Net - Net, what was mentioned in the book was right indeed - but we have to remember that they were talking about the same integer J whereas the question talks about two integers x and y.

Hope this helps. Thanks.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by Gurpinder » Tue Aug 10, 2010 2:43 pm
heyyyyy,

ok i see the difference. BUT i still don't get the principle of redundant primes. How can primes overlap?!?!?!

so are you saying that when you have only 1 variable and is divided by 2 diff #'s the primes can overlap. but when you have DIFF vaiables the primes don't overlap??

please explain.
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by Brian@VeritasPrep » Tue Aug 10, 2010 3:05 pm
Hey Gurpinder,

Good question - let me see if I can help:

Primes factors are "redundant" for purposes of calculating common multiples, common denominators, etc. when they're repeated in the factorization of both numbers.

Consider the question:

Is x divisible by 36?

1) x is divisible by 6

2) x is divisible by 12

Many examinees would simply multiply 6 * 12 to get 72, which is divisible by 36, and answer that the statements together are sufficient. However, let's look at the prime factorization of both:

6 = 2 * 3
12 = 2 * 2 * 3

One 2 and one 3 are "redundant" in that 6 only provides us with one 2 and one 3, each of which is replicated in the factor set for 12.

12 is the least common multiple of 6 and 12 - it contains all of the unique factors of each in the minimum required occurrences: 2, 2, and 3. Note that multiples of 12 include 12, 24, 36, 48, etc. - in order to have a multiple of 12, we need to have 2*2*3*something, and the required 2, 2, and 3 are all provided independently by the 12 - the 6 adds nothing new.

Now consider a different problem:

Is x divisible by 42?

1) x is divisible by 6

2) x is divisible by 14

Here, if we do the prime factorizations, we get:

6 = 2 * 3
14 = 2 * 7

to reach 42, we need 42 = 2 * 3 * 7

The statements, together, contribute one unique 2, one unique 3, and one unique 7, so we know that we're guaranteed to have a multiple of 42.
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by Gurpinder » Tue Aug 10, 2010 3:33 pm
Brian@VeritasPrep wrote:Hey Gurpinder,
6 = 2 * 3
12 = 2 * 2 * 3

One 2 and one 3 are "redundant" in that 6 only provides us with one 2 and one 3, each of which is replicated in the factor set for 12.
Hey Brian,

I almost get it. So when you say this, is it simply that if a number is a multiple of another number, that the second number will have the prime numbers of the first number? This is what I am making of that.

With an example,

4 = 2*2

8 = 2*2*2

total 2s = 2^5 but we are saying that the 2 2s from 4 are redundant and in fact the total 2s we have is 2^3?
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by Brian@VeritasPrep » Tue Aug 10, 2010 3:59 pm
Exactly right - it's not that the primes in 4 (2*2) "don't count", it's just that they're redundant in determining a common multiple when we're considering another number that contains at least two 2s (like 8: 2*2*2).

And, yes, if x is a multiple of y, then that means that x contains all of y's prime factors. So:

if x is a multiple of 6, then x has 2, 3, and whatever other factors it has, but it definitely is divisible by 2*3.

You'll see this concept on questions that ask for common multiples or divisibility. When you're dealing with two potential factors of a larger number, know that you can only count the "unique" prime factors of those two factors when you're checking for that divisibility.

So, back to your hypothetical, 32 is divisible by both 8 and 4, but just knowing that a number is divisible by 8 and 4 doesn't make it necessarily divisible by 32, since those two 2s in 4 are redundant with the factors of 8.
Brian Galvin
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