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square of an integer

Expert replies
by nikhilagrawal » Fri Jul 25, 2008 2:20 am
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
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Source: — Problem Solving |

by amitansu » Fri Jul 25, 2008 2:44 am
It's simple ...14 would be ans.

factorize 3150 and try adding each of these choices to see if they form a square.

Amit
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by pepeprepa » Fri Jul 25, 2008 2:51 am
3150y=x^2
3150 is the core of the problem so analize it.
3150=(3^2)*(5^2)*2*7

So it is y=14
3150*14=x^2
(2*3*5*7)^2=x^2
and x=(2*3*5*7)
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by ben2pop » Fri Nov 13, 2009 4:47 am
hey ,
can you explain how did you get to 3150=(3^2)*(5^2)*2*7

I mean did you do it by head or do you have a technique ?
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by heshamelaziry » Fri Nov 13, 2009 7:08 am
IMO 14. When asked about the aquare of two numbers and one of them is x or y, prime factor the known number, then see what numbers don't have 2 of each. here, divide 3150 by 10, will give you 315 ----> 315 /5 ----> 63 ----> 63/3----> 7-----> 7/7 -----> 1. yu divide the number until you get 1. this is called "prime factorization" meaning dividing the nymber until yu get only 2,3,5,7,11,13,17,19. these are primes because they only divid by themselves and 1.

In this question, after prime factorization, yuo have two 5, two 3, one 7 and one 2. So, you need one more 7 and one more 2 and their product is Y. This is the only way to get a square of 3150 *y
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Re: square of an integer

by Stuart@KaplanGMAT » Fri Nov 13, 2009 5:11 pm
nikhilagrawal wrote: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
Some good solutions, let's review the general principle.

All numbers are made up of prime factors. All perfect squares are made up of pairs of primes.

So, in order for 3150y to be a perfect square, it must consist only of pairs of prime factors.

There are many different ways to find the prime factors of a number - my personal preference is a prime factorization tree (which is a bit hard to show on the computer).

3150 breaks up into 315 * 10 (no primes)

315 breaks up into 5 * 63 (5 is prime, put that aside)
63 breaks up into 7*9 (7 is prime, put that aside)
9 breaks up into 3*3 (both prime, put them aside).

10 breaks up into 2*5 (both prime, put them aside).

So, we've put aside: 5*7*3*3*2*5

writing in order:

2*3*3*5*5*7

Our 3s and 5s are paired off, so we don't need any more of those to create a perfect square.

We have a single 2, so we need a 2.
We have a single 7, so we need a 7.

So, the minimum possible value for y is 2*7 = 14... choose E.
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by rahul.s » Sat Jan 09, 2010 3:41 am
Thank you Stuart. Your explanation was simple and easy to understand :)
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