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# Prime Boxes

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Prime Boxes [#permalink]  29 Apr 2011, 23:20
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Hi Friends

I just began my preparation by starting off with Manahattan GMAT "The Number Properties Book", in it he did explain the concept of a prime box.As it states, a box that contains all the prime factors of any number is a prime box. It helps us to find out whether a giver no is a factor of another no or not.Okie, I have a doubt here, I understood what a prime box was and how to make a prime box for a specific number. But I have this new, using primebox is the only way to solve problems related to this stuff? or can we do it by any general procedure?..I thereby request any one explain me about the alternative procedure, if available...

The question was
If 80 is a factor of r,is 15 a factor of r?

regards
SVA
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Re: Prime Boxes [#permalink]  30 Apr 2011, 02:47
Quote:
If 80 is a factor of r,is 15 a factor of r?

Enumerate the prime numbers of 80 ==>
$$2x2x2x2x5$$

If 80 is a factor of r, then
$$r = 2x2x2x2x5x(...)$$

The (...) means we don't know if there are any other integers in r.
Quote:
So two possibilities,
1) No other prime numbers in r
2) There are other prime numbers. It could be 3. It could be not.

Since we are not certain..

15 may or may not be a factor of r. CANNOT DETERMINE!
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Re: Prime Boxes [#permalink]  30 Apr 2011, 03:19
svagmat wrote:
Hi Friends

I just began my preparation by starting off with Manahattan GMAT "The Number Properties Book", in it he did explain the concept of a prime box.As it states, a box that contains all the prime factors of any number is a prime box. It helps us to find out whether a giver no is a factor of another no or not.Okie, I have a doubt here, I understood what a prime box was and how to make a prime box for a specific number. But I have this new, using primebox is the only way to solve problems related to this stuff? or can we do it by any general procedure?..I thereby request any one explain me about the alternative procedure, if available...

The question was
If 80 is a factor of r,is 15 a factor of r?

regards
SVA

Prime factorization is the best method I know for these types of problem and I strongly recommend everyone to get acquainted with it.

For 15 to be a factor of r; r should contain at least as many factors as 15 has.

15=3*5
r must contain at least one 3 and one 5 as its factor to be a qualified multiple of 15.

80=2*2*2*2*5
This tells us that r contains at least four 2's and one 5 as its factors. But, we need just one 2 and one 3. We have one 2 but we don't know whether r contains 3. It may or may not contain 3. Thus, we won't know whether r is a multiple of 15 OR 15 is a factor of r.

e.g.
r=80=2*2*2*2*5. Not divisible by 15(No 3 in 80)
r=160=2*2*2*2*5. Not divisible by 15(No 3 in 160)
r=240=2*2*2*2*5*3. Divisible by 15(as now we have one 2 and one 3 in 240)
r=320=2*2*2*2*5*2*2. Not Divisible by 15(No 3 in 320)
r=400=2*2*2*2*5*5. Not Divisible by 15(No 3 in 400)
r=480=2*2*2*2*5*2*3. Divisible by 15(as now we have one 2 and one 3 in 480)

Conclusion:
In order for r to be a multiple 15, r should be a multiple of 3 as well in addition to be a multiple of 80.
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Re: Prime Boxes [#permalink]  30 Apr 2011, 16:38
Thanks a lott fluke and lawschoolsearcher..That definitely cleared my doubt..

regards
SVA
Re: Prime Boxes   [#permalink] 30 Apr 2011, 16:38
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