intergers and multiple - og

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by Brian@VeritasPrep » Wed Nov 03, 2010 8:57 am
Hey Clawhammer:

Keep in mind that the OG solutions are much more concerned with being precise than with being helpful (I think they'll even admit that), so I'm glad you brought this one to the board!

With divisibility questions, it's almost always a good decision to use PRIME FACTORIZATION to determine exactly what you need. In this question:

We're given: x is a multiple of 6 and y is a multiple of 14.

So we can derive: x brings prime factors 2 and 3 and y brings prime factors 2 and 7.

They ask: Is xy a multiple of 105?

So they're really asking: Does xy have prime factors 3, 5, and 7?

But the given information already tells us that x brings a 3 and y brings a 7, so all we really need to know is:

Is either x or y a multiple of 5?


Statement 1 gives us:

x is a multiple of 9, or therefore of 3 * 3

This is not sufficient...we need a 5, and this one doesn't give it to us.

Statement 2 gives us:

y is a multiple of 25, or therefore of 5 * 5

This gives us the 5 we need, and is therefore sufficient! The answer is B.



Prime factorization is a pretty efficient way to determine exactly what you need in a divisibility-related question, so when in doubt, factor it out!
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by clawhammer » Wed Nov 03, 2010 9:05 am
Thanks a lot Brian. It makes proper sense.

One thing that has really got me thinking is: "x can be = (2)(3)(3)(5)" then too is x a multiple of both 6 and 9. How can I make sure, I can find all possible primes? Like I wont miss out the 5.

For Example: IF the second option confirmed that there was no 5 as a prime factor for y, then I could think, by COMBINING both, I'm sure that the answer is NO, therefore 'C'. But that would be mistaken, as it would become insufficient, as there is also a possibility of 5 for x.

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by Testluv » Wed Nov 03, 2010 10:12 am
clawhammer wrote:
For Example: IF the second option confirmed that there was no 5 as a prime factor for y, then I could think, by COMBINING both, I'm sure that the answer is NO, therefore 'C'. But that would be mistaken, as it would become insufficient, as there is also a possibility of 5 for x.
Hi,

as Brian pointed out, we know that xy's prime factorization is minimally 2*2*3*7. But 105's prime factorization is 3*5*7. Because we already know that xy contains 3 and 7 as prime factors, in order for xy to be a multiple of 105, all we need to know is whether EITHER x or y is a multiple of 5 (i.e., has a 5 as a prime factor)...but we do NEED to KNOW whether at least one of them is a multiple of 5.

In your hypothetical, (2) states that 5 is not a prime factor of y. But if (2) stated that y was not a multiple of 5, it would still leave open the possibility that x is a multiple of 5.

We know from the original that x is a multiple of 6. From (1) we know that x is a multiple of 9 as well. But we wouldn't know whether or not x is a multiple of 5. If it is a multiple of 5 (for example, x could be 6*9*5) then the answer to the question would be "yes". But if it isn't a multiple of 5 (for example, x may just be 6*9) , then answer would be "no". So, in this hypo, the answer would be E.
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