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

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mattnyc15 Junior | Next Rank: 30 Posts
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if Y is the smallest positive integer (no. properties)

Post Sun Nov 22, 2015 1:02 pm
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


OA: E


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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vishnujthattil Newbie | Next Rank: 10 Posts Default Avatar
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Post Thu Jan 14, 2016 9:10 am
[quote="mattnyc15"]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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Post 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)².

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vishnujthattil Newbie | Next Rank: 10 Posts Default Avatar
Joined
12 Jan 2016
Posted:
2 messages
Post Thu Jan 14, 2016 9:10 am
[quote="mattnyc15"]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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GMAT/MBA Expert

Matt@VeritasPrep GMAT Instructor
Joined
12 Sep 2012
Posted:
2640 messages
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113 members
Upvotes:
625
Target GMAT Score:
V51
GMAT Score:
780
Most Active Expert Most Responsive Expert Most Thanked Expert
Post 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)².

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