If y is the smallest positive integer such that 3,150 multip

[imane81]

Please, need your help on this one. Thanks in advance

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

[ajith]

Please, need your help on this one. Thanks in advance

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

3150 = 2*3^2*5^2*7

to make it a square we must multiply by 2*7 =14

[linkinpark]

In questions like this always find factors of given number

3150 = 5 * 5 * 3 * 3 * 7 * 2, if 14 = 2 * 7 =y is multiplied to this number, we'll get a square of number (each factor is two times e.g. 5, 3, 7, 2 in our case)

so answer has to be E

[thephoenix]

factorizing 3150=(3^2 )* (5^2)*2*7

we are asked when y is multiplied with 3150 then the least value of y to make the result a perfect square

clearly if we multiply the no. 3150 with 2 and 7 it will be a perfact square

therefore ans is 14

[money9111]

Thank you... I think I would have gotten as far as breaking down the factors... but then wouldn't have realized to multiple 7*2

[blsbball]

I'm trying to understand this problem. I can get to the prime factors but don't understand why we multiply by 2 and 7. The OG says "at a minimum, y must have one factor of 2 and one factor of 7 so that 3150y has two factors of each of the primes 2, 3, 5, and 7. The smallest positive integer value of y is then (2)(7). = 14".
Could someone please explain this. Wouldn't (2)(3) be the smallest integer value of y given the prime factors?
Thanks

[Geva@EconomistGMAT]

I'm trying to understand this problem. I can get to the prime factors but don't understand why we multiply by 2 and 7. The OG says "at a minimum, y must have one factor of 2 and one factor of 7 so that 3150y has two factors of each of the primes 2, 3, 5, and 7. The smallest positive integer value of y is then (2)(7). = 14".
Could someone please explain this. Wouldn't (2)(3) be the smallest integer value of y given the prime factors?
Thanks

The demand for a 2 and a 7 stems from the fact that 3150y must be the square of an integer. This means that 3150 y must be a number that you can take the square root out of, and still get an integer result.

Taking the square root of a power is equivalent to multiplying the exponent by 1/2 (for example: sqrt(2) = 2^1/2). This is why odd powers do not have integer square roots: for example, 8 does not have an integer square root, because 8 is 2^3 - an odd power, and it is not possible to halve the exponent and result in an integer square root - 2^3/2 is not an integer.

in the breakdown of 3150 = (3^2 )* (5^2)*2*7 , the 3s and the 5s are squared, so you can take the square root from those and get an integer result. However, the 2 and the 7 are missing one power each to become even, and those powers need to be supplied by y. If y=14, then 3150y = *3^2 )* (5^2)*2*7 * (2*7), leaving all even exponents: 3^2 * 5^2 * 2^2 * 7^2, from which you can take the square root and come up with an integer 3*5*2*7. The same cannot be said for y=2*3, as it will not complete 3150 to even exponents.

[Scott@TargetTestPrep]

Please, need your help on this one. Thanks in advance

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

Solution:

This problem is testing us on the rule that when we express a perfect square (greater than 1) 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: E