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

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If y is the smallest positive integer such that 3,150 [#permalink] New post 08 Dec 2005, 16:28
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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 [#permalink] New post 08 Dec 2005, 16:39
E.. 44100...
by brute force, but didnt take that long.. once i multiplied the given choices by 3150 it was pretty easy to identify if they were square of any integers or not..
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Re: Arithmetic [#permalink] New post 08 Dec 2005, 16:40
If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be


3150
= 5 * 630
= 5 * 5 * 126
= 5 * 5 * 3 * 42
= 5 * 5 * 3 * 3 * 14
= 5 * 5 * 3 * 3 * 2 * 7
= 5^2 * 3^2 * 2^1 * 7^1

3150 * y is a square of an integer, so y should be 2^1 * 7^1.

Thus, y = 14.
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 [#permalink] New post 08 Dec 2005, 22:08
Agree with gamjatang, answer should be 14
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Re: Arithmetic [#permalink] New post 08 Dec 2005, 22:16
gamjatang 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


3150
= 5 * 630
= 5 * 5 * 126
= 5 * 5 * 3 * 42
= 5 * 5 * 3 * 3 * 14
= 5 * 5 * 3 * 3 * 2 * 7
= 5^2 * 3^2 * 2^1 * 7^1

3150 * y is a square of an integer, so y should be 2^1 * 7^1.

Thus, y = 14.


Splendid diagramming! Convinced that the OA should be 14.

Laxie, can you explain your genius shortcut here?
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Re: Arithmetic [#permalink] New post 08 Dec 2005, 22:30
desiguy 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


Matt, if i were in the test process and saw this question, I would immediately solve as gamjatang did. However, upon your request, i try to look for some quick reasoning: that is to multiple/ divide 3150 by 2

NOTE: Any number(contains at least two digits) having 0 as unit digit have only one 2 as a factor. Why? coz after dividing it by 2, the unit digit is 5 ---> it's odd and thus can't be further divided by 2.
So we need an answer choice which is a multiple of 2 to evenize the factor 2 of 3150 ----> B and D out
. The blue part is not so important ...it gonna be useful incase most of the answer choices are odd.

Now multiple 3150 by 2 ---> 6300 = 63*10^2= 7*3^3*10^2 ---> we need one more 7 ---> it must be 14 ( = 2*7)
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 [#permalink] New post 08 Dec 2005, 22:34
I see. And you cancel out the last 2 because of your initial doubling.
Nice :P
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Nice reasoning [#permalink] New post 08 Dec 2005, 23:04
Laxy,
even i solved it using gajmat's principle and got 14 as answer. but ur concept looks nice..saves lot of time
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 [#permalink] New post 09 Dec 2005, 00:10
I understand gt's factoring

3150
= 5 * 630
= 5 * 5 * 126
= 5 * 5 * 3 * 42
= 5 * 5 * 3 * 3 * 14
= 5 * 5 * 3 * 3 * 2 * 7
= 5^2 * 3^2 * 2^1 * 7^1

but why do we cancel out the 5 and the 3, leaving only the 2 and 7?
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Re: Arithmetic [#permalink] New post 09 Dec 2005, 01:21
Laxieqv,
You said - ny number(contains at least two digits) having 0 as unit digit have only one 2 as a factor. Why? coz after dividing it by 2, the unit digit is 5 ---> it's odd and thus can't be further divided by 2.

Let's take 100, - divide by 2 -> you will get 50. and 100=10*10=2*2*5*5



laxieqv wrote:
desiguy 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


Matt, if i were in the test process and saw this question, I would immediately solve as gamjatang did. However, upon your request, i try to look for some quick reasoning: that is to multiple/ divide 3150 by 2

NOTE: Any number(contains at least two digits) having 0 as unit digit have only one 2 as a factor. Why? coz after dividing it by 2, the unit digit is 5 ---> it's odd and thus can't be further divided by 2.
So we need an answer choice which is a multiple of 2 to evenize the factor 2 of 3150 ----> B and D out
. The blue part is not so important ...it gonna be useful incase most of the answer choices are odd.

Now multiple 3150 by 2 ---> 6300 = 63*10^2= 7*3^3*10^2 ---> we need one more 7 ---> it must be 14 ( = 2*7)
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Re: Arithmetic [#permalink] New post 09 Dec 2005, 01:32
skif wrote:
Laxieqv,
You said - ny number(contains at least two digits) having 0 as unit digit have only one 2 as a factor. Why? coz after dividing it by 2, the unit digit is 5 ---> it's odd and thus can't be further divided by 2.

Let's take 100, - divide by 2 -> you will get 50. and 100=10*10=2*2*5*5



laxieqv wrote:
desiguy 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


Matt, if i were in the test process and saw this question, I would immediately solve as gamjatang did. However, upon your request, i try to look for some quick reasoning: that is to multiple/ divide 3150 by 2

NOTE: Any number(contains at least two digits) having 0 as unit digit have only one 2 as a factor. Why? coz after dividing it by 2, the unit digit is 5 ---> it's odd and thus can't be further divided by 2.
So we need an answer choice which is a multiple of 2 to evenize the factor 2 of 3150 ----> B and D out
. The blue part is not so important ...it gonna be useful incase most of the answer choices are odd.

Now multiple 3150 by 2 ---> 6300 = 63*10^2= 7*3^3*10^2 ---> we need one more 7 ---> it must be 14 ( = 2*7)


ah, sorry, i meant there's only 0 as the unit digit and the tens digit is odd ...as in here, 3150 ..thank you for reminding me.
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 [#permalink] New post 09 Dec 2005, 01:45
coz : for example : a5*2 ...2*5 = 10 , this 1 is add to a*2 ...a*2 is always even ---> 2*a+1 is odd ...so those numbers with odd tens digit and 0 as unit digit are fine.
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 [#permalink] New post 09 Dec 2005, 11:28
Answer should 14, desiguy this is a great problem, really hammers home the importance of factorization, thanks for posting it.
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 [#permalink] New post 09 Dec 2005, 11:46
Here is necessary (BUT NOT SUFFICIENT) condition of a perfect square:
- when divided by 4, it gives remainder 0(divisible by 4) or 1

B, D can be eliminated straight away, as when multiplied, they don't conform to above.

for rest..you have to some work :wink:
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  [#permalink] 09 Dec 2005, 11:46
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