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tough multiple/factors problem #2

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by [email protected] » Sat Apr 12, 2014 12:13 pm
Hi Tom,

This question ultimately comes down to "prime factorization" - the idea that all positive integers that are not prime numbers can be "broken down" into a series of prime numbers that are multiplied together.

For example: 10 = 2x5

Here, we're told that 375Y = X^2 and that X and Y are integers.

X^2 = (X)(X).....a number multiplied by itself.....

That X^2 has to account for the 375 and whatever Y is.

375 = 3x5x5x5

Right now, we have one 3 and three 5s, but there's no way to equally distribute those numbers between the two Xs...this tells us some things about the value of Y....

Since there are two Xs, we need an even number of EACH of the prime factors.

If Y = 5, then combined with 375, we'd have...

3x5x5x5x5, so we could put "two 5s" into each X.

We still need to deal with the one 3 though, so Y has to also "hide" a 3.

If Y = 15 = 3x5, then with the 375 we'd have....

3x3x5x5x5x5 and we'd have X = 3x5x5

X^2 = (3x5x5)(3x5x5)

All of this goes to show that Y = 15 AT THE MINIMUM (it could technically be some greater multiple of 15, but with this information we can now answer the question.

Which of the 3 statements is an integer when Y = 15....

I. Y/15 = 15/15 = 1 YES
II. Y/30 = 15/30 = 1/2 NO
III. Y^2/25 = 15^2/25 = 225/25 = 9 YES

Final Answer: D

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by GMATGuruNY » Sun Apr 13, 2014 5:04 am
For more practice, here's a similar problem, along with my solution:
If n and y are positive integers and 450y = n³, which of the following must be an integer?

I. y/(3 x 2² x 5)

II. y/(3² x 2 x 5)

III. y/(3 x 2 x 5²)

a. None
b. I only
c. II only
d. III only
e. I, II, and III
Try to prove that I, II and III DON'T have to be integers.
To this end, plug in the MINIMUM POSSIBLE VALUE for y.

450y = n³ implies that 450y is the cube of an integer.

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor:
8 is the cube of an integer because 8 = 2³ = 2*2*2.
27 is the cube of an integer because 27 = 3³ = 3*3*3.

Thus, when we prime-factorize 450y, we need to get AT LEAST 3 of every prime factor.
Here's the prime-factorization of 450y:
450y = 2 * 3² * 5² * y

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y.
Thus, y must provide at at least two more 2's, one more 3, and one more 5.
Thus, the MINIMUM possible value of y = 2² * 3 * 5.

Plug y = 2² * 3 * 5 into the answer choices:

I. y/(3 x 2² x 5)
(2² * 3 * 5)/(3 x 2² x 5) = 1.
The smallest possible value of y yields an integer.
Eliminate every answer choice that does not include I.
Eliminate A, C and D.

II. y/(3² x 2 x 5)
(2² * 3 * 5)/(3² x 2² x 5) = 1/3.
Not an integer.
Eliminate every remaining answer choice that includes II.
Eliminate E.

The correct answer is B.
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by Matt@VeritasPrep » Sun Apr 13, 2014 10:47 am
Here's another way:

If two integers are equal to each other, they have the same prime factorization. For instance, 2 * 15 = 3 * 10, because each factors as 2 * 3 * 5. This seems obvious, but it'll be useful in a second!

If 375y = x * x, and both are integers, then the same rule applies: each side must have the same prime factorization. The left-hand side is 3*5*5*5*y, so the right hand side must also have be 3*5*5*5*something.

Now notice that the right hand side is a perfect square: x * x. That means we should be able to group it into two identical roots. (For instance, 36 = (2*3) * (2*3).) But if we try to group 375y into two identical roots, here's what we get:

(3 * 5 * 5) * (5 * y)

Yuck! So clearly we're missing a few numbers, but which ones? Well, if he had an extra 3 and an extra 5, we'd have (3 * 5 * 5) * (3 * 5 * 5). (The numbers I just made up are in red.) So (3*5*5)*(3*5*5) could equal x * x.

Since 3*5*5*5*y = x * x = (3*5*5) (3*5*5), notice that y = 15. From there, just plug y = 15 into each of the three options and see which are true.

You might be asking (rightly) why x * x couldn't have been something else, such as (3*5*5*7*11)² or something. It certainly COULD be that (we don't know!) However, since the question asks which of these MUST be an integer, we want to find the minimum possible value of y. Since each of our roots must have a 3 and two 5s in it, the minimum value of x * x = (3*5*5)², so the minimum value of y is 15. Whichever of the three statements is true of the minimum value of y will also be true of every other value of y, since every other value of y will just be a multiple of 15.
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by manjory » Mon Apr 14, 2014 1:24 pm
Hey,

the question says 375y=x^2
well.. 375= 5X5X5X3
therefore.. y is anything which makes this number a square
lets start from the nearest options, say y= 3X5 which makes the eq.= (5X5X5X3)X(3X5)
therefore from the given options in the question y/30 cannot be an integer as you will need 2 2's to make it a LHS a square.

moving from this argument y/15 and y^2/25 seems likely to result an integer..
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