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# M02 Q31 (DS)

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14 Feb 2009, 23:15
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If $$x$$ and $$y$$ represent digits of a two digit number divisible by 3, is the two digit number less than 50?

1. Sum of the digits is a multiple of 18
2. Product of the digits is a multiple of 9

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I answered E (Statements 1+2 together are not sufficient), thinking that the number could be either 99 or -99. The answer says A (statement 1 is sufficient), because the only solution can be 99.

Is the answer wrong or am I missing something? What about -99?
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15 Feb 2009, 04:30
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Hi river,

If you consider -99, then what are the values of x & y?

It can only be (-9 & 9) OR (9 & -9) respectively. In either case, you get their sum as 0, which is not a multiple of 18. So, it doesn't satisfy option 1.

Does it make sense? HTH.

Regards,
Technext
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15 Feb 2009, 08:03
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Technext wrote:
Hi river,

If you consider -99, then what are the values of x & y?

It can only be (-9 & 9) OR (9 & -9) respectively. In either case, you get their sum as 0, which is not a multiple of 18. So, it doesn't satisfy option 1.

Does it make sense? HTH.

Regards,
Technext

0 is a multiple of any numer or integer. Even if the integer is 99 or -99, the sum is a multiple of 18. However it is confusing about to how to assign the value of x and y if the integer is -99.

So A is insuff...

river wrote:
If x and y represent digits of a two digit number divisible by 3, is the two digit number less than 50?
1. Sum of the digits is a multiple of 18
2. Product of the digits is a multiple of 9

I answered E (Statements 1+2 together are not sufficient), thinking that the number could be either 99 or -99. The answer says A (statement 1 is sufficient), because the only solution can be 99.

Is the answer wrong or am I missing something? What about -99?

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15 Feb 2009, 09:24
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Hi GMAT TIGER,

Thanks for clarifying this; I actually came across this fact just few months back only but it just skipped. Thanks again.

Although you've quoted that the problem is confusing, at last you've mentioned in your reply that 'So A is suff...'. If 99 & -99 both satisfy option 1, then how can we conclude that A is sufficient? Do you agree with A as the right answer?

Regards,
Technext
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15 Feb 2009, 10:16
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Technext wrote:
Hi GMAT TIGER,

Thanks for clarifying this; I actually came across this fact just few months back only but it just skipped. Thanks again.

Although you've quoted that the problem is confusing, at last you've mentioned in your reply that 'So A is suff...'. If 99 & -99 both satisfy option 1, then how can we conclude that A is sufficient? Do you agree with A as the right answer?

Regards,
Technext

Thats typo. A is insuff.
Thanks.
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15 Feb 2009, 12:00
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GMAT TIGER,
I agree with you. First of all, the number -99 is a negative integer with 2 digits (9 and 9). A digit is an integer from 0 to 9. There is no "negative digit", so we can't have a digit equal to -9.

So, from what I understand, the GMATclub answer to that question is wrong, right? It should be E instead of A.

(Indeed, even for the case of -9 and 9, the sum is 0, which i a multiple of 18, as 18*0=0.. but that's not part of this solution).
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16 Feb 2009, 21:40
river wrote:
GMAT TIGER,
I agree with you. First of all, the number -99 is a negative integer with 2 digits (9 and 9). A digit is an integer from 0 to 9. There is no "negative digit", so we can't have a digit equal to -9.

So, from what I understand, the GMATclub answer to that question is wrong, right? It should be E instead of A.

(Indeed, even for the case of -9 and 9, the sum is 0, which i a multiple of 18, as 18*0=0.. but that's not part of this solution).

That should be unless there is nothing hidden issues related digit/integer. I am little skeptical about -99.
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19 Feb 2009, 08:31
I am not very sure if for -99 we can take the digits as (-9,9) or (9,-9) because the -ve sign is for the whole number and not for individual digits I feel... Can somebody throw some light on this please...
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27 May 2009, 14:37
I came across this problem. Can anyone throw some light on the choices? Is it A or E.
Things to consider: Is 0 a multiple of any integer? If true then 0 will be LCM of nay number.

Pl. help.........
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27 May 2009, 15:43
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Expert's post
gmatprep09 wrote:
I came across this problem. Can anyone throw some light on the choices? Is it A or E.
Things to consider: Is 0 a multiple of any integer? If true then 0 will be LCM of nay number.

Pl. help.........

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23 Aug 2009, 15:40
bb wrote:
gmatprep09 wrote:
I came across this problem. Can anyone throw some light on the choices? Is it A or E.
Things to consider: Is 0 a multiple of any integer? If true then 0 will be LCM of nay number.

Pl. help.........

so what's the verdict bb?

the explanation concludes that 99 IS indeed THE only two digit number that justifies the sufficiency of S1. It doesn't, however, address the possibility of having 0 as a multiple of 18.
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23 Aug 2009, 22:01
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river wrote:
If x and y represent digits of a two digit number divisible by 3, is the two digit number less than 50?
1. Sum of the digits is a multiple of 18
2. Product of the digits is a multiple of 9

I answered E (Statements 1+2 together are not sufficient), thinking that the number could be either 99 or -99. The answer says A (statement 1 is sufficient), because the only solution can be 99.

Is the answer wrong or am I missing something? What about -99?

for 1), "Sum of the digits is a multiple of 18" tells that x=y=9 only, and 99 is divisible by 3. But it >50
it is a "No, but sufficient to tell No" question.

for 2), "Product of the digits is a multiple of 9" and with "a two digit number divisible by 3" xy can be 63 or 36, insuf.

So the answer is A for sure.
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29 Aug 2009, 02:36
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there is no way we can take a single digit of a two digit no as a negative.....not possible..its clearly A
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27 Sep 2010, 09:53
IMO A.

Agree with rohansherry...
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27 Sep 2010, 14:49
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imo no higher than 600-700 level. Not a difficult question.

As discussed above a "digit" cannot be negative, i.e. -3541—who can tell which digit is negative? Is it the 3 or the 1? Illogical point. So the only way a two-digit number can have its digits add up to 18 is if they were 9 and 9, hence 99 > 50. For (2) 3*3 can total 9, a multiple of 3, or 6*6 can total 36, which is also a multiple of 3, but > 50. A it is.
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28 Sep 2010, 17:48
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A. 0 is not a universal multiple. The negative in a negative number cannot be assigned to a digit of the number; it's a property of the entire number, not of a digit i.e. -(99) != (-9)(9) or (9)(-9).
Thanks for the post.

Last edited by rkurra on 28 Sep 2010, 18:40, edited 1 time in total.
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28 Sep 2010, 18:35
I opted A but, since x = y = 9 from first statement, the two digit number can be -99 or 99 that is divisible by 3. One is less than 50 and other is greater than 50.

Can someone clear this contention?
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29 Sep 2010, 09:50
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supoose the two digit number is xy

so the value of the number will be 10x + y which is divisible by 3 ( x=1 to 9 and y = 1to 9)

Now with the 1st option x+y=18 and the only case is 99 which is more than 50.

For the option 2, for example we can take 63 and 36 both are satisfying so we cant deduce the number.

Hence the Answer should be A.
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29 Sep 2010, 22:08
This is a Yes/No Data Sufficiency question, if you can answer yes or no to the options then the statement would be sufficient. If you can't answer definitively, then the statement is not sufficient.

If x and y represent digits of a two digit number divisible by 3, is the two digit number less than 50?

1. Sum of the digits is a multiple of 18
2. Product of the digits is a multiple of 9

1) Only option that answers this is 99. Answer to statement 1 is no, therefore it is sufficient.

2) 36 and 63 both meet the criteria for statement 2, therefore it is not sufficient.

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06 Sep 2011, 23:11
mbasaikiran wrote:
I opted A but, since x = y = 9 from first statement, the two digit number can be -99 or 99 that is divisible by 3. One is less than 50 and other is greater than 50.

Can someone clear this contention?

E for same reason
digits 9;9 - everyone agree
what nubmers can be made of these two?
clearly +(99) and -(99)
so E

about "every number multiple of 0"
so sum of the digits should be 0
the number that has the sum 0 is "00"
but there is no such a number
Re: M02 Q31 (DS)   [#permalink] 06 Sep 2011, 23:11

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