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p divisible by 9?

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by Anurag@Gurome » Sun Sep 16, 2012 7:30 pm
grandh01 wrote:If p and q are both positive integers, then is p divisible by 9?

1) p/10 + q/10 is an integer
2) p/9 + q/10 is an integer

OA IS B
(1) p/10 + q/10 is an integer.
If p = 10, q = 20, then p/10 + q/10 = 10/10 + 20/10 = 30/10 = 3, which is an integer. Here p is not divisible by 9.
If p = 90, q = 20, then p/10 + q/10 = 90/10 + 20/10 = 110/10 = 11, which is an integer. Here p is divisible by 9.
No definite answer; NOT sufficient.

(2) Since p/9 + q/10 is an integer, so p/9 will always have to be an integer and hence p will always be divisible by 9.

If p = 9, q = 20, then p/9 + q/10 = 9/9 + 20/10 = 3, which is an integer. Here p is divisible by 9; SUFFICIENT.

The correct answer is B.
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by everything's eventual » Sun Sep 16, 2012 8:03 pm
Statement A) - p/10 + q/10 is an integer.

Let p = 10 and q = 20

10/10 + 20/10 = 3 ( integer)

Let p = 90 and q = 20

90/10 + 20/10 = 11 ( integer)

Here p can be divisible by 9 or not divisible by 9. Therefore not sufficient.

Statement B) - p/9 + q/10 is an integer.

p/9 + q/10 = (10p + 9q)/90 is an integer.

10p is divisible by 90. Therefore, 10p is divisible by 9*10. 10 is divisible by 10 but not divisible by 9. Therefore p has to be divisible by 9.

Therefore, B is sufficient.

I hope my understanding is correct.
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by manihar.sidharth » Mon Sep 17, 2012 12:04 pm
Hi Anurag
I got a question in the reasoning for second option.
How can you say that "Since p/9 + q/10 is an integer, so p/9 will always have to be an integer"?
As we might have a case like this where two decimals are adding up to an integer
For ex : Imagine P as 12 and Q as 27 ..Then when you divide 12/9 you get 1.3 and similarly when you divide 27/10 you get 2.7 and there addition leads to 4.Hope I made my point
Please correct me If I am wrong...

Thanks
Sid
Anurag@Gurome wrote:
grandh01 wrote:If p and q are both positive integers, then is p divisible by 9?

1) p/10 + q/10 is an integer
2) p/9 + q/10 is an integer

OA IS B
(2) Since p/9 + q/10 is an integer, so p/9 will always have to be an integer and hence p will always be divisible by 9.

If p = 9, q = 20, then p/9 + q/10 = 9/9 + 20/10 = 3, which is an integer. Here p is divisible by 9; SUFFICIENT.

The correct answer is B.
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by RipMrThick » Tue Sep 18, 2012 6:36 am
Was manihar.sidharth ever answered? I have this very same question...
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by manihar.sidharth » Tue Sep 18, 2012 6:37 am
Nah I m still waiting!!!
RipMrThick wrote:Was manihar.sidharth ever answered? I have this very same question...
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by Shalini Suresh » Tue Sep 18, 2012 6:57 am

manihar.sidharth wrote

For ex : Imagine P as 12 and Q as 27 ..Then when you divide 12/9 you get 1.3 and similarly when you divide 27/10 you get 2.7 and there addition leads to 4.Hope I made my point
Please correct me If I am wrong...
12/9 is 1.3333 which is non terminating . So adding 12/9 and 27/10 gives you 4.0333 so thats not an integer !
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by manihar.sidharth » Tue Sep 18, 2012 7:13 am
I guess you are correct So Do we have any number that is not a multiple of 9 and give terminating decimal when divided by 9?

Shalini Suresh wrote:

manihar.sidharth wrote

For ex : Imagine P as 12 and Q as 27 ..Then when you divide 12/9 you get 1.3 and similarly when you divide 27/10 you get 2.7 and there addition leads to 4.Hope I made my point
Please correct me If I am wrong...
12/9 is 1.3333 which is non terminating . So adding 12/9 and 27/10 gives you 4.0333 so thats not an integer !
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by Shalini Suresh » Tue Sep 18, 2012 9:20 am

I guess you are correct So Do we have any number that is not a multiple of 9 and give terminating decimal when divided by 9?
no number n divided by 9 will give u a terminating decimal unless n is a multiple of 9.

For x/y to be A terminating decimal 'y'should contain only factors of 2 or 5 .

For ex. any number n divided by 4 , n/4 will always be a terminating decimal since 4 can be factorized to 2*2
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by eski » Tue Sep 18, 2012 10:41 am
I have a different way of attack.

9 is the only unique number , when it divided any number other than 9 , it will be a not finite.

eg 3/9 = 1.33333 .... thus

x/9 + anything will never be integer if x is not multiple of 9 .

and 1st option can be removed by pasting values only.

-eski
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