\(3,150=2*3^2*5^2*7\), now \(3,150*y\) to be a perfect square \(y\) must complete the odd powers of 2 and 7 to even number (perfect square has even powers of its primes), so the least value of \(y\) is 2*7=14. In this case \(3,150y=(2*3^2*5^2*7)*(2*7)=(2*3*5*7)^2=perfect \ square\).

Answer: E.

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_________________

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.
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
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This question is based on the concept that when you prime factorize a perfect square, the power of every distinct prime factor is even. The reason for this is given in the following blog posts:
http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-writing-factors-of-an-ugly-number/
http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-factors-of-perfect-squares/

3150 is not a perfect square i.e. square of an integer because
\(3150 = 2*3^2*5^2*7\)

Note that 3 and 5 have even powers (2 each) while 2 and 7 have a power of 1 each. Hence to make 3150 a perfect square, we need to make all powers even. So if we multiply 3150 by 2*7, powers of all prime factors will become even.
\(3150*2*7 = 2^2*3^2*5^2*7^2\) - A perfect square

Answer (E).

Similarly, if you had to divide 3150 by the smallest number to get a perfect square, you would have divided it by 2*7 to get rid of 2 and 7 completely so that the remaining prime factors have even powers,
\(3150/(2*7) = 3^2*5^2\) - A perfect square
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3,150=2*3^2*5^2*7, now 3,150*y to be a perfect square y must complete the odd powers of 2 and 7 to even number (perfect square has even powers of its primes), so the least value of y is 2*7=14. In this case 3,150y=(2*3^2*5^2*7)*(2*7)=(2*3*5*7)^2=perfect square.

Answer: E.

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answer E. factor out the number and find any prime numbers that are not paired. 7 & 2.

E.

Create a prime factor tree for 3,150 and see what numbers do not have a pair.
3150 = 5x5x3x3x7x2.... the 7 and 2 do not have a second pair to make it a perfect square of a number. so y must be 7*2 = 14

In such questions we need to break the number into the smallest possible prime factors. So the smallest prime factors of 3150 are:
315*10=63*5*2*5=7*9*5*2*5=7*3*3*5*2*5. In order to get a square of an integer we have to have at least two identical primes. In our case we have 3*3 and 5*5 corresponding to this condition but not 2*7 so our smallest number should be 14.

Answer is E

Where can I find the logic of how to answer this question?

Is there a section in the math guide that covers this?

Thanks,
Hunter
_________________

No special section covers this, but I can recommend similar questions:
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Hope it helps.
_________________

3150 =\(2*3^2*5^2*7\)
For it to be perfect square all the prime number should be least raised to the power 2
in 3150 ...only 2 and 7 needs to be multiplied so that all prime will be raised power 2
hence least value of \(4y = 2*7 = 14\)

hence E
_________________

Hello,

can anyone help me with this type of question? I don't get it why the remaining numbers, 7 and 2, are the smallest positive integer y. Which chapter in the MGMAT books should i restudy to deal with this kind of problem? I don't understand the explanation in the OG which says: "To be a perfect square, 3,150y must have an even number of each of its prime factors."

Any help is appreciated!

Have you checked this: if-y-is-the-smallest-positive-integer-such-that-65323.html#p1035828

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_________________

Solution:

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.)

We also are given that 3,150 multiplied by y is the square of an integer. 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. Thus, we need one more 2 and one more 7. Therefore, y = 7 x 2 = 14

Answer is E.
_________________

I wanted to share my GMAT Timing Tips for this question (the links below include growing lists of questions that you can use to practice these tips):

Prime factors of a perfect square, perfect cube, etc.: As others in this thread have pointed out, we need to know that all prime factors of a perfect square have exponents that are even. If we find that any of the prime factors of 3,150 do not have even exponents, y will need to contain each of those prime factors, so that the prime factorization of 3,150*y will have an even exponent for each of those prime factors.

Prime factorization: In order to determine the prime factors of y, we need to do the prime factorization of 3,150, so let's try to do it as efficiently as possible. Because factors of 10 and 5 are easy to see, I recommend starting by factoring 3,150 into 315*10, then 63*5*2*5. We can also recognize that 63 = 9*7 = 3^2 *7. This means that the prime factorization of 3,150 is 2 * 3^2 * 5^2 * 7. Since there are odd powers of 2 and 7, y must contain factors of 2 and 7, and the smallest possible value of y is 2*7 = 14.

Please let me know if you have any questions, or if you would like me to post a video solution!
_________________

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.


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
_________________

CONCEPT: A number is a perfect square when all the exponents of its distinct prime factors are even integers

\(3150 = 315*10 = 5*63*2*5 =\) \(2^1\)\(*3^2*5^2*\)\(7^1\)

In order to make powers of all distinct prime factors even we need one 2 and one 7 because their power are 1 each in 3150

i.e. We need to multiply (2*7=14) in 3150 in order to make it a perfect square

hence, y = 14

Answer: Option E
_________________

Thank you for your explanation. I was struggling to understand this question because I didn't understand what the question is asking (the wording of this question is confusing to me). With your explanation, I now understand the question and the solution to it.

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

Prime Factorial of 3150 = 2*3*3*5*5*7 = \(2^1*3^2*5^2*7^1\)

3150 multiplied by y is the square of an integer = \(2^1*3^2*5^2*7^1\) * y = Square of an integer

We need 2 and 7 to match the even powers to be equal to square of integer , so y = \(2^1*7^1\) = 14

Ans: E
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