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mgmat a

by resilient » Sun Jun 08, 2008 7:55 pm
If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9


IN statement two,Cant x be 02 (two digit integer) + 7, which is divisible by 9?

for this reason, I disregarded the statement


qa is d
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by meet.anup » Sun Jun 08, 2008 11:20 pm
Use divisibility test for 9: sum of the digits should be divisible by 9
Thus, if a+b = 11 this is not divisible by 9 and would give the remainder as 2.

Similarly, if X+7 is divisible by 9 then X is obviously not divisible by nine.
As 7 is needed to be added to X to make it divisible by 9, thus the remainder would be 2 (same as in 1).
e.g: 2+7/9 so 2/9 remainder is 2
11+7/9 is integer so 11/9 gives remainder 2 and so on..
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yes

by resilient » Mon Jun 09, 2008 9:09 am
yes my question is not the actual answer or the approach. My specific question is cant x be 02. If so, statement two is insuffucient.
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Re: yes

by Ian Stewart » Mon Jun 09, 2008 10:19 am
resilient wrote:yes my question is not the actual answer or the approach. My specific question is cant x be 02. If so, statement two is insuffucient.
x cannot be '02', because '02' is not a two-digit number. It is the one-digit number 2. Still, even if x is equal to 2, the remainder when you divide 2 by 9 is indeed equal to 2, just as for any other example of x you might choose when considering Statement 2. So even if you allow x to be equal to 2, Statement 2) is still sufficient.
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by andy9923 » Mon Jul 28, 2008 5:00 am
i have a problem with the answer choice D.

Can't X be negative? If X=-16, then X+7=-9, which i believe is divisible by 9.

A negative two-digit number is still a two-digit number right?

I posted this on MGMAT forum earlier but haven't gotten a response yet.

thanks a bunch
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