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Number Properties Q

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Number Properties Q

by czechchamp » Mon Jul 14, 2008 5:50 am
I have exactly one week left before my GMAT. Took three GMATPrep exams and scored 640 (Q43, V35) - 640(Q42, V38) and 710(Q44, V42).
In math, my biggest weakness are number properties questions. Can someone please help me out with these two problems:
-------------------
PS:
If Y and N are positive integers and 450Y = N cubed, which of the following must be an integer?

I. Y / 3 x 2 squared x 5
II. Y / 3 squared x 2 x 5
III. Y / 3 x 2 x 5 squared

A) None
B) I only
C) II only
D) III only
E) I,II, III
-----------------
DS:
What is the tens digit of positive integer R?

1. The tens digit of R/10 is 3.
2) The hundreds digit of 10R is 6.


I studied the MGMAT number properties guide and it helped; however, I'm clueless as to how to approach the above problems.
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by asigheartau » Mon Jul 14, 2008 6:44 am
I think the answer to the second Q is 6.

Let's say R is a four digit number

if R/10 =_ _ 3 _ with 3 in the tens place then multiplying it by 10 we get R in the hundreds place => R= _ _ 3 _ 0

Example: 630 X10= 6300

Secondly,

if 10 R =_ 6 _ _ with 6 in the hundreds place, dividing by 10 gives us 6 in the tens place
therefore R= _ _ 6 _

Example:
1600:10= 160

The example make no reference to R whatsover. They are just illustrations of how digits change location if multiplied/ divided by 10.

Good Luck!
Alin Sigheartau
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Re: Number Properties Q

by Ian Stewart » Mon Jul 14, 2008 7:52 am
czechchamp wrote:
PS:
If Y and N are positive integers and 450Y = N cubed, which of the following must be an integer?

I. Y / 3 x 2 squared x 5
II. Y / 3 squared x 2 x 5
III. Y / 3 x 2 x 5 squared

A) None
B) I only
C) II only
D) III only
E) I,II, III
-----------------
For what it's worth, I think this is among the most difficult number theory questions you can find on the GMAT (along with others that test a similar concept). I'm sure you have a good command of prime factorizations. To illustrate the important concept in this question (because you might see it again, in different guises), let's take three different primes, p, q and r, and say (p^4)*(q^2)*(r^5) is the prime factorization of some number n. What will the prime factorization of n^3 look like?

n = (p^4)*(q^2)*(r^5)
n^3 = [(p^4)*(q^2)*(r^5)]^3
n^3 = (p^12)*(q^6)*(r^15)

That is, we just triple every exponent in the prime factorization of n. This is how you can recognize a perfect cube from a prime factorization: every exponent is divisible by 3. Similarly, the exponents are all even in prime factorizations of perfect squares, are all divisible by 4 in prime factorizations of perfect fourth powers, and so on.

Okay, back to the original question:

450Y = N^3

Prime factorize, of course!

(2^1)*(3^2)*(5^2)*Y = N^3

But the left side is equal to a perfect cube- the exponents on 2, 3 and 5 must be divisible by 3 after we multiply by Y. So Y must be divisible at least by (2^2)*(3)*(5). So only I) must be an integer (the others could be integers if Y is large enough, but don't need to be).
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by ildude02 » Fri Jul 18, 2008 1:05 pm
Ian,

We know that reducing 450 into it's prime factors gives, 2 * 3 ^2 * 5^2.
So, from equation 1, we can write y/ 2 ^2 * 3 * 5 as n ^3 / 2 ^3 * 3^3 * 5 ^3 => (n/2 * 3* 5) ^ 3. We know that n is an integer, so it could be that n can be an integer with prime factors of 2, 3 and 5 , meaning the result will always give us an integer. But if n were 4 or 8, the n/2 * 3 * 5 is NOT an integer. So how can we say it 's always an integer with my approach ? Appreciate your response.
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by Ian Stewart » Mon Jul 21, 2008 7:28 am
ildude02 wrote:
We know that reducing 450 into it's prime factors gives, 2 * 3 ^2 * 5^2.
So, from equation 1, we can write y/ 2 ^2 * 3 * 5 as n ^3 / 2 ^3 * 3^3 * 5 ^3 => (n/2 * 3* 5) ^ 3.
I think you're doing the following:

n^3 = 450y
n^3 = 2 * 3 ^2 * 5^2 * y
n^3/(2 * 3 ^2 * 5^2) = y
n^3/(2^3 * 3^3 * 5^3) = y/(2^2 * 3 * 5)
(n/(2*3*5))^3 = y/(2^2 * 3 * 5)

Now we know that n must be divisible by 2, 3 and 5 (from the equation n^3 = 450y), so the left side of the above must be an integer. That is, y must be divisible by (2^2 * 3 * 5). But that's the best we can say here.
ildude02 wrote: We know that n is an integer, so it could be that n can be an integer with prime factors of 2, 3 and 5 , meaning the result will always give us an integer. But if n were 4 or 8, the n/2 * 3 * 5 is NOT an integer. So how can we say it 's always an integer with my approach ? Appreciate your response.
From the original question, if n^3 = 450y, n *must* be divisible by 2, 3 and 5. The primes that divide the right side of the equation must also divide the left side (assuming we're dealing with integers).
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by aditikedia » Fri Aug 08, 2008 6:50 am
III. Y / 3 x 2 x 5 squared

Is this Y/ (3x2x5^2) OR is it (Y/3)X2X(5^2)

If it's the latter.. could someone explain why this can't be an integer?

Thanks
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by Ian Stewart » Fri Aug 08, 2008 12:12 pm
aditikedia wrote:III. Y / 3 x 2 x 5 squared

Is this Y/ (3x2x5^2) OR is it (Y/3)X2X(5^2)

If it's the latter.. could someone explain why this can't be an integer?

Thanks
It's the former- all of the numbers are in the denominator.
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by aditikedia » Fri Aug 08, 2008 5:48 pm
Thanks Ian!
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by Ancientmtk » Sat Nov 01, 2008 3:37 pm
you should put ( ) in the denominator... I thot it was ( Y/3 ) x 2 x 5 the while time..
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