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M02 Q31 (DS)

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Re: M02 Q31 (DS) [#permalink]

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New post 29 Sep 2011, 05:22

A) 9+9 = 18, since those are the largest single digits, it's our only option.
B) xy = z/9 supports more than one answer. 99, 72, 63...etc
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Re: M02 Q31 (DS) [#permalink]

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New post 29 Sep 2011, 07:50
Hey guys,

Restate the stem as follows:

Is the unknown number one of 12 15 18 21 24 27 30 33 36 39 42 45 48 ???

Lets translate
stmt 1 to : (X+Y)/18= Integer
stmt 2 to: XY/9= Integer

Consider stmt one alone: The digits in none of the listed numbers add up to a number divisible by 18. Sum of digits of 48 and 39 both yield 12, which is too low. So the answer to the stem is definitely NO. Suff

Consider stmt two alone: 36 is divisible by 9 (Yes to the stem) and also 90 is divisible by 9 (No to the stem) => Insuf

Guys, don't bother with 0. I don't think that in such questions regarding digits and digit place, we can consider one digit -ive and another +ve. Do we have a SINGLE number such as 1-2 or 4-5 (Don't read it minus: It's a negative two-digit number with the sign before the second digit)? :shock: If you consider 0, you can make, e.g., -99, but since the question doesn't determine the order of digits, then you have to also accept existence of 9-9

If you are still reluctant, ok, maybe we can consider -ive or +ive in such question, too, but even then you have to write the positive number like 10X+Y and -ive number like -(10X+Y). Even in this case your -99 and 99 are in fact the same number b/c the -ive sign is irrelevant to our discussion of what X and Y are. What counts is that X and y are both 9 in both numbers, so they are the same numbers actually.

Am I clear or just lost in my faulty reasoning? ;)

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Re: M02 Q31 (DS) [#permalink]

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New post 29 Sep 2011, 09:04
The answer is A.
The reasoning has been explained by others.

However, there are 2 doubts still floating around:
1. The '0' question
2. The negative number thing

As far as the point about zero is concerned, the question clearly states that it is a a two digit number. I don't think zero qualifies as a two digit number.

Regarding point 2 (the negative number thing), I think mathematicians will agree that negative numbers can be multiples, however, I think GMAT only considers positive numbers when problems related to multiples, factors etc. are designed.

Just to clarify the point regarding zero, zero is not the factor of any number, however, it can be the multiple of a number (if it is clearly stated that non-negative numbers are to be considered).

Hope this is helpful.

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Re: M02 Q31 (DS) [#permalink]

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New post 29 Sep 2011, 20:05
A is the answer.
Totally agree with the fact that in case of negative numbers one can't assign -ve to any one of the digits.

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Re: M02 Q31 (DS) [#permalink]

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New post 07 Nov 2011, 08:50
"A" is the answer
Sign of the number is something different from the digits. with 2 digits of "9" and "9", we can reach the numbers of "99" and "-99"

But, the question asks "is the two digit number less than 50?" ,and not "is the number less than 50"?
Plz notice the delicate difference between these 2 questions. the answer of the first question is just 99. but the answer of second question is 99 and -99


Last edited by shahideh on 07 Nov 2011, 08:51, edited 1 time in total.
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Re: M02 Q31 (DS) [#permalink]

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New post 14 Nov 2011, 21:45
Agree with the logic above. 99 is the only one that fits thereby making the statement sufficient to answer, no.
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Re: M02 Q31 (DS) [#permalink]

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New post 01 Oct 2012, 07:00
XY (one number, 2 digits) is divisble by 3.
Is XY < 50? This is a Y/N answer. If you can answer with a Y/N, sufficient.

1) sum of digits is a multiple of 18... No two digits can equal any multiple of 18 greater than 18. Therefore the sum of X and Y must equal 18. There are only 2 digits that do that.. 9 and 9. So XY=99. SUFFICIENT.

2) Products of digits is a multiple of 9...
1*9=9... i can stop here. one number is greater than 50, the other one is not. INSUFFICIENT.

Answer is A.
Re: M02 Q31 (DS)   [#permalink] 01 Oct 2012, 07:00

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