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HCF /LCM

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by nehakhas1 » Fri Jan 30, 2009 3:44 am
1. The HCF of three numbers is 12. Their product is 17280. What are the three numbers?
I. One of the numbers is greater than the other two.
II. One of the numbers is lesser than the other two.

Please solve
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Source: — Data Sufficiency |
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by shulapa » Fri Jan 30, 2009 4:41 am
Lets first do some calculations:
17280 = N1*N2*N3 = (2^7)*(3^3)*5

As we know that the N1, N2, N3 have a Highest common number of 12 we know that the can be showed as:
N1 = (2^2)*3*X
N2 = (2^2)*3*Y
N3 = (2^2)*3*Z

X,Y,Z are created from the left out number (2^1, 5) and 1 and therefore can hold the following values: 10, 1, 2, 5

(1) If we assume that the highest number is N1 - X can be either 10 or 5 - therefore, insufficient.

(2) if we assume N3 to be the lower number - N1 and N2 must be multiplied by some factor, so one of them would be multiplied by 2 and the other by 5:
N1 = (2^2)*3*5
N2 = (2^2)*3*2
N3 = (2^2)*3

Therefore, sufficient.

IMO B
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by sanju09 » Fri Jan 30, 2009 5:57 am
shulapa wrote:Lets first do some calculations:
17280 = N1*N2*N3 = (2^7)*(3^3)*5

As we know that the N1, N2, N3 have a Highest common number of 12 we know that the can be showed as:
N1 = (2^2)*3*X
N2 = (2^2)*3*Y
N3 = (2^2)*3*Z

X,Y,Z are created from the left out number (2^1, 5) and 1 and therefore can hold the following values: 10, 1, 2, 5

(1) If we assume that the highest number is N1 - X can be either 10 or 5 - therefore, insufficient.

(2) if we assume N3 to be the lower number - N1 and N2 must be multiplied by some factor, so one of them would be multiplied by 2 and the other by 5:
N1 = (2^2)*3*5
N2 = (2^2)*3*2
N3 = (2^2)*3

Therefore, sufficient.

IMO B
:) Good Reasoning. IMO D

Let me contradict your this reasoning:

(1) If we assume that the highest number is N1 - X can be either 10 or 5 - therefore, insufficient.

No, we can't take X = 10 here. Otherwise we will have to take Y = Z = 1, or N2 = N3, which may or may not be true.

Each statement alone is sufficient.
The mind is everything. What you think you become. -Lord Buddha



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by aroon7 » Fri Jan 30, 2009 6:44 am
But the numbers need not be different so 10,1,1 or 5,2,1 both satisfies even if both the statements are taken together.
then why not E...
someone pl help me :shock:
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by shulapa » Fri Jan 30, 2009 6:49 am
Hi nehakhas,

At a first glance of the question I had the same line of reasoning as yours, but then I figured there is nothing that prevents N2 from being the same number as N3. Am I missing something?

Eyal
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Re: HCF /LCM

by coffee5251 » Fri Jan 30, 2009 7:15 am
nehakhas1 wrote:1. The HCF of three numbers is 12. Their product is 17280. What are the three numbers?
I. One of the numbers is greater than the other two.
II. One of the numbers is lesser than the other two.

Please solve
Out of curiosity does anyone know what level a problem like this is??
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by krisraam » Fri Jan 30, 2009 8:48 am
aroon7 wrote:But the numbers need not be different so 10,1,1 or 5,2,1 both satisfies even if both the statements are taken together.
then why not E...
someone pl help me :shock:

Stmt 1 says one of the numbers is greater than other two
10,1,1 and 5,2,1 both satisfies the condition so not sufficient.

Stmt 2 says one of the number is lesser than other two.

Only 5,2,1 satisfies the condition. So sufficient
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by sanju09 » Mon Feb 02, 2009 2:21 am
shulapa wrote:Hi nehakhas,

At a first glance of the question I had the same line of reasoning as yours, but then I figured there is nothing that prevents N2 from being the same number as N3. Am I missing something?

Eyal
IMO B now, I always want a discussion on points that could have room for errors if pondered carelessly. I did "IMO D" for the same reasons. Thanks
The mind is everything. What you think you become. -Lord Buddha



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