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by jjayhart » Wed Oct 09, 2013 5:41 pm
This came out of the OG. The explanation wasn't too great. I know to start by getting the prime factors of 3150 (although this is more memorizing the method than understanding it), but once I get all the prime factors of 3150, I can't make the connection to the next step that gets to the answer. Thanks.

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

(a) 2
(b) 5
(c) 6
(d) 7
(e) 14
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by theCodeToGMAT » Wed Oct 09, 2013 6:14 pm
3150 = 5 x 5 x 3 x 3 x 7 x 2

Now, to make 3150 a perfect square of an integer we need to have factors twice .. so, we need another "7" & "2"

So, 14

Answer [spoiler]{E}[/spoiler]
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by jjayhart » Wed Oct 09, 2013 7:42 pm
Ok, so I know that doing the calculation, 14 is the only value of Y that gives the perfect square of an integer. But when going through process and not knowing that, what is the explanation for picking 7 and 2 as the factors to multiply? What if the original number were 1215 (3*3*3*3*3*5) instead of 3150? What would Y be in that case?
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by ganeshrkamath » Wed Oct 09, 2013 7:57 pm
jjayhart wrote:This came out of the OG. The explanation wasn't too great. I know to start by getting the prime factors of 3150 (although this is more memorizing the method than understanding it), but once I get all the prime factors of 3150, I can't make the connection to the next step that gets to the answer. Thanks.

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

(a) 2
(b) 5
(c) 6
(d) 7
(e) 14
3150 = 5 * 630 = 5 * 2 * 5 * 7 * 9
3150 = 2 * 3^2 * 5^2 * 7

Now, to get a perfect square, we have to make the powers of each factor even.
To do so, we have to multiply by at least 2 * 7
Hence we have to multiply be at least 14.

Choose E

Cheers

PS: 1215 = 3^5 * 5
We have to multiply this by 3*5 to make the powers even.
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by theCodeToGMAT » Wed Oct 09, 2013 8:06 pm
Sorry jjayhart, I din't understand what exactly you meant to ask.
The reason i selected "7" & "2" is that both "7" and "2" appears only once in 3150.. To have a perfect square root of 3150 we need all the multiples of 3150 to be re-written in power of two; here, 5^2 & 3^2 satisfies .. but not "7" & "2"

For 1215

1215 = 3*3*3*3*3*5 = (3*3) x (3*3) x (3) x (5) --> we have pairs of 3 i.e. (3*3) & (3*3) but (3) and (5) are alone.. to make pair of them we need another 3 & 5 so that they become (3*3) & (5*5).. So, we need "15"
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by jjayhart » Wed Oct 09, 2013 8:11 pm
Ok that makes sense. Thanks!
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by theCodeToGMAT » Wed Oct 09, 2013 8:13 pm
jjayhart wrote:Ok that makes sense. Thanks!
You are welcome!
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by sanju09 » Thu Oct 10, 2013 12:47 am
jjayhart wrote:This came out of the OG. The explanation wasn't too great. I know to start by getting the prime factors of 3150 (although this is more memorizing the method than understanding it), but once I get all the prime factors of 3150, I can't make the connection to the next step that gets to the answer. Thanks.

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

(a) 2
(b) 5
(c) 6
(d) 7
(e) 14
Here, y is a positive integer. In other words, we need 3150y to be the smallest square of an integer. For this, we must break up 3150 into prime bases multiplied together to get an exponent look. We'd leave those which are already an even exponent, and would supply one more base where exponent(s) is(are) odd. Multiply all such supplies together to get the minimum y.

3150
2*1575
2*3*525
2*3*3*175
2*3*3*5*35
2*3*3*5*5*7=focus here, 3 and 5 are happy, only 2 and 7 need one more appearance each, in order to make 3150y be what we want it to become.

[spoiler]So, the required value of y is 2*7 = 14.

Pick E[/spoiler]
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by sanju09 » Thu Oct 10, 2013 12:53 am
jjayhart wrote:Ok, so I know that doing the calculation, 14 is the only value of Y that gives the perfect square of an integer. But when going through process and not knowing that, what is the explanation for picking 7 and 2 as the factors to multiply? What if the original number were 1215 (3*3*3*3*3*5) instead of 3150? What would Y be in that case?
In this case, both 3 and 5 have an odd exponent, both need one more appearance each, so we'll supply 3*5 = 15 to be y here.
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by GMATGuruNY » Thu Oct 10, 2013 2:29 am
The problem should read as follows:
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 test-takers might find the reasoning easier to understand if we factor out known perfect squares and then plug in the answer choices.

3150y

= 25 * 126 * y

= 25 * 2 * 63 * y

= 25 * 2 * 7 * 9 * y

= 9 * 25 * 14y.

9 and 25 are perfect squares.
For 3150y to be the square of an integer, 14y must also be a perfect square.
Plugging in the answer choices for y, we get:
14*2 = 28.
14*5 = 70.
14*6 = 84.
14*7 = 98.
14*14 = 196.

Only the result in red is a perfect square.

The correct answer is E.
Last edited by GMATGuruNY on Thu Oct 10, 2013 2:55 am, edited 1 time in total.
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by gloube » Thu Oct 10, 2013 2:50 am
Hi,

I don't understand the question

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

=> which integer are we talking about here?
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by theCodeToGMAT » Thu Oct 10, 2013 2:53 am
gloube wrote:Hi,

I don't understand the question

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

=> which integer are we talking about here?
Integer that can be formed by multiplying: the smallest value of "y" to 3150 to make a perfect square.
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by GMATGuruNY » Thu Oct 10, 2013 2:59 am
gloube wrote:Hi,

I don't understand the question

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

=> which integer are we talking about here?
Good catch. The problem was transcribed incorrectly. It should read as follows:
If y is the smallest positive integer such that 3,150 multiplied by y is the square of AN integer, then y must be
I've made the correction in my post above.
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by gloube » Thu Oct 10, 2013 3:00 am
GMATGuruNY wrote:
gloube wrote:Hi,

I don't understand the question

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

=> which integer are we talking about here?
Good catch. The problem was transcribed incorrectly. It should read as follows:
If y is the smallest positive integer such that 3,150 multiplied by y is the square of AN integer, then y must be
I've made the correction in my post above.
Thank you, it makes much more sense now.
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