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Number System

by imhimanshu » Tue Sep 13, 2011 5:15 am
Hi,
Please provide approach to solve such questions-


Three is the largest number that an be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100
1 - 30
2- 70
3- 210
4 - 300
E - 700

Thanks
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by sl750 » Tue Sep 13, 2011 5:22 am
We have to pick a number for x that is divisible by both 3 and 10. We have 30, 60, 90, and so on.
Therefore the largest integer that divides both x (x=300) and 2100 will be 300
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by Brent@GMATPrepNow » Tue Sep 13, 2011 6:05 am
imhimanshu wrote:Hi,
Please provide approach to solve such questions-


Three is the largest number that an be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100
1 - 30
2- 70
3- 210
4 - 300
E - 700

Thanks
The fastest approach here might be to use the process of elimination (POE).

E) Can x=700? No - otherwise the greatest common divisor of x and 100 would be 100 (where the question tells us that the GCD of x and 100 is 10)

D) Can x=300? No - otherwise the greatest common divisor of x and 100 would be 100 (where the question tells us that the GCD of x and 100 is 10)

C) Can x=210? Sure, the GCD of 210 and 27 is 3, and the GCD of 210 and 100 is 10
SO, if x=210, then the GCD of x and 2100 will be 210
The answer is C

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Brent
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by knight247 » Tue Sep 13, 2011 6:05 am
SL that is not the right answer.

The largest number that is divisible by any two numbers is always the GCD
We write both 27 and x in the form of their prime factors

27=3^3
x=unknown

GCD=3 which means that x has exactly one 3 in its prime factorisation.(@himanshu...U may want to brush up on your LCM and GCD finding skills else u wouldn't be able to understand this)

So x is of the form 3a

Now, considering 100 and x
100=2^2*5^2
x=3a

GCD=2*5 meaning that x has to have exactly one 2 and one 5 among its prime factors else the GCD would be higher.

So x is of the form 3*2*5*b=30b

Now writing x and 2100 in the form of their prime factors
2100=2*2*3*5*5*7
x=2*3*5*b

GCD=2*3*5 or GCD=2*3*5*7......(as b could have 7 in its prime factorisation)

So the answer is either 30 or 210 i.e. either A or C. What is the OA?
Last edited by knight247 on Tue Sep 13, 2011 6:11 am, edited 2 times in total.
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by knight247 » Tue Sep 13, 2011 6:08 am
Hey Brent,
Could you if possible go thru my post and help me filter out the incorrect option. I'd appreciate it.

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by Brent@GMATPrepNow » Tue Sep 13, 2011 6:11 am
knight247 wrote:SL that is not the right answer.

The largest number that is divisible by any two numbers is always the GCD
We write both 27 and x in the form of their prime factors

27=3^3
x=unknown

GCD=3 which means that x has exactly one 3 in its prime factorisation.(@himanshu...U may want to brush up on your LCM and GCD finding skills else u wouldn't be able to understand this)

So x is of the form 3a

Now, considering 100 and x
100=2^2*5^2
x=3a

GCD=2*5 meaning that x has to has exactly one 2 and one 5 among its prime factors else the GCD would be higher.

So x is of the form 3*2*5*b=30b

Now writing x and 2700 in the form of their prime factors
2100=2*2*3*5*5*7
x=2*3*5*b

GCD=2*3*5 or GCD=2*3*5*7......(as b could have 7 in its prime factorisation)

So the answer is either 30 or 210 i.e. either A or C. What is the OA?
Good point, knight247
The answer could be A or C. With the process of elimination strategy, one has to assume that there is only one correct answer here. Of course, on this sort of question on the GMAT, there will be only 1 correct answer.

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by Brent@GMATPrepNow » Tue Sep 13, 2011 6:15 am
imhimanshu wrote: Three is the largest number that an be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100
1 - 30
2- 70
3- 210
4 - 300
E - 700
In addition to having more than 1 correct answer, this question is worded poorly. The GMAT would never write "x divides evenly into y" since this could give one the impression that y divided by x must be an even number.

GMAT wording would say something like "3 is greatest common divisor of x and 27"

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Brent
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by imhimanshu » Tue Sep 13, 2011 6:26 am
This question is from Kaplan and the OA is C
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by imhimanshu » Tue Sep 13, 2011 6:29 am
@ Knight 247, I did get this answer correct while giving my exam. However, I just wanted to see the different approaches in order to kill such questions. Thanks for the concern though ! :-) appreciated !
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by deepsun » Tue Sep 13, 2011 7:52 am
knight247 wrote:SL that is not the right answer.

The largest number that is divisible by any two numbers is always the GCD
We write both 27 and x in the form of their prime factors

27=3^3
x=unknown

GCD=3 which means that x has exactly one 3 in its prime factorisation.(@himanshu...U may want to brush up on your LCM and GCD finding skills else u wouldn't be able to understand this)

So x is of the form 3a

Now, considering 100 and x
100=2^2*5^2
x=3a

GCD=2*5 meaning that x has to have exactly one 2 and one 5 among its prime factors else the GCD would be higher.

So x is of the form 3*2*5*b=30b

Now writing x and 2100 in the form of their prime factors
2100=2*2*3*5*5*7
x=2*3*5*b

GCD=2*3*5 or GCD=2*3*5*7......(as b could have 7 in its prime factorisation)

So the answer is either 30 or 210 i.e. either A or C. What is the OA?

But cant B be another 2/3/5 instead of a 7? then the GCD would be 60/90/150 isnt it?

Please explain

thanks

update 9/14/2011 : never mind i got it
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by saketk » Wed Sep 14, 2011 9:33 pm
although we have 2 answers matching i.e. 30 and 210, we can safely eliminate 30 because the question is asking for the largest possible number.
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