if Y is the smallest positive integer (no. properties)

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mattnyc15: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Fri Jul 17, 2015 1:09 pm; Thanked: 1 times

if Y is the smallest positive integer (no. properties)

by mattnyc15 » Sun Nov 22, 2015 1:02 pm

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Q: If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be:

A: 2
B: 5
C: 6
D: 7
E: 14

[spoiler]
OA: E[/spoiler]

I know we break this down by primes, but I don't understand why we need to get the single prime factors and multiply. Maybe I don't fully understand the question? Why couldn't we say 6 (2x3) or even 2?

I'd love an explanation here. Thanks!

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Source: Beat The GMAT — Problem Solving | Sun Nov 22, 2015 1:02 pm

theCEO: Master | Next Rank: 500 Posts; Posts: 363; Joined: Sun Oct 17, 2010 3:24 pm; Thanked: 115 times; Followed by:3 members

by theCEO » Sun Nov 22, 2015 2:07 pm

mattnyc15 wrote:Q: If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be:

A: 2
B: 5
C: 6
D: 7
E: 14

[spoiler]
OA: E[/spoiler]

I know we break this down by primes, but I don't understand why we need to get the single prime factors and multiply. Maybe I don't fully understand the question? Why couldn't we say 6 (2x3) or even 2?

I'd love an explanation here. Thanks!

Let's see if this this helps.

3,150 * y = a^2; in order for this to be true 3150*y must be a perfect square
50 * 21 * 3 * y = a^2
25 * 2 * 7 * 3 * 3 * y = a^2
5^2 * 3^2 * 2 * 7 * y = a^2
(5+3)^2 * 14y = a^2

For the smallest possible integer and to be a perfect square y=14

In other words when is the following equation true?
3150y = (5+3+y)^2 = a^2

It's true when y=14

ans = e

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by Brent@GMATPrepNow » Sun Nov 22, 2015 2:25 pm

mattnyc15 wrote:Q: If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be:

A: 2
B: 5
C: 6
D: 7
E: 14

[spoiler]
OA: E[/spoiler]

Key concept: The prime factorization of a perfect square (the square of an integer) will have an EVEN number of each prime.
For example, 36 = (2)(2)(3)(3)
And 400 = (2)(2)(2)(2)(5)(5)
Likewise, 3150y must have an EVEN number of each prime in its prime factorization.

Aside: for more on this concept, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/829

So, 3150y = (2)(3)(3)(5)(5)(7)y
We have an EVEN number of 3's and 7's, but we have a single 2 and a single 7.
If y = (2)(7), then we get a perfect square.

That is: 3150y = (2)(2)(3)(3)(5)(5)(7)(7)

So, if y = 14, then 3150y is a perfect square.

Answer: E

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pm; Location: East Bay all the way; Thanked: 625 times; Followed by:119 members; GMAT Score:780

by Matt@VeritasPrep » Fri Nov 27, 2015 2:41 am

I'd take 3150 and get it as close to a square as you can. If it's a square, it should have two identical roots, so let's shoot for that.

3150 = 315 * 10
= 63 * 5 * 10
= 7 * 3 * 3 * 5 * 5 * 2

Close! 3*3 and 5*5 are squares, so we're almost there: we're short a 7 and a 2. So if y gives us those factors, we're set, and y = 14.

To see why this works, take y = 14.

Then 3150*y = 3150 * 14 = (3 * 3 * 5 * 5 * 7 * 2) * 2 * 7, or (3*5*7*2)Â².

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by vishnujthattil » Thu Jan 14, 2016 9:10 am

mattnyc15 wrote:Q: If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be:

A: 2
B: 5
C: 6
D: 7
E: 14

Since 3150 * y should be a perfect square, the resulting product must have 0 in the ten's place too. hence options B and D become invalid. further for option A it is visual calculation, 3150* 2= 6300 is not a perfect square. for option C, its 3 times 3150 * 2, i.e 6300*3=18900 again a visual calculation and it is not a perfect square. thus option E is correct. This Approach helps to save valuable time I believe!!!

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by Jeff@TargetTestPrep » Sun Jun 24, 2018 5:07 pm

mattnyc15 wrote:Q: If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be:

A: 2
B: 5
C: 6
D: 7
E: 14

This problem is testing us on the rule that when we express a perfect square by its unique prime factors, every prime factor's exponent is an even number.

Let's start by prime factorizing 3,150.

3,150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2

3,150 = 2^1 x 3^2 x 5^2 x 7^1

Notice that the exponents of both 2 and 7 are not even numbers. This tells us that 3,150 itself is not a perfect square. However, we are given that the product of 3,150 and y is a perfect square. Thus, we can write this as:

2^1 x 3^2 x 5^2 x 7^1 x y = square of an integer

According to our rule, we need all unique prime factors' exponents to be even numbers. Remember also that y is the smallest number that makes 3150*y a perfect square. Thus, we need one more 2 and one more 7. Therefore, y = 7 x 2 = 14

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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Re: if Y is the smallest positive integer (no. properties)

by [email protected] » Sun Apr 25, 2021 9:12 pm

Hi All,

We’re told that Y is the SMALLEST positive integer such that 3150(Y) is the SQUARE of an INTEGER. We’re asked for the value of Y. This question is based on a specific Number Property rule involving squares, so while you could potentially ‘brute force’ your way to the correct answer by TESTing THE ANSWERS (since one of those answers, when multiplied by 3150, WILL create a Perfect Square), you can use Prime Factorization to get to the correct answer quicker.

By definition, a ‘perfect square’ has an EVEN number of every one of its PRIME FACTORS. For example:

9 is a perfect square because 9 = (3)(3) → it has two 3s.
16 is a perfect square because 16 = (4)(4) = (2)(2)(2)(2) → it has four 2s
36 is a perfect square because 36 = (6)(6) = (2)(3)(2)(3) → it has two 2s and two 3s.
Etc.

We’re told that 3150(Y) is a perfect square, so we have to first break 3150 down into its prime factors, then figure out what Y has to equal so that there will be an EVEN number of each of those prime factors.

3150 = (315)(10) = (63)(5)(2)(5) = (7)(9)(5)(2)(5) = (7)(3)(3)(5)(2)(5)

Notice that there are two 3s and two 5s… but only one 2 and one 7. This means that for Y to be as small as possible, it has to include one 2 and one 7 in its prime factorization. Thus, Y = (2)(7)

Final Answer: E

GMAT Assassins aren’t born, they’re made,
Rich

Contact Rich at [email protected]