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M02-31

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M02-31  [#permalink]

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New post 16 Sep 2014, 00:18
11
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (01:53) correct 35% (02:06) wrong based on 256 sessions

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Math Expert
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Re M02-31  [#permalink]

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New post 16 Sep 2014, 00:18
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Re: M02-31  [#permalink]

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New post 03 Dec 2014, 21:25
Hi, Bunuel

In exam how will I quickly get to the number 99 in statement 1? Please show me the process
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Re: M02-31  [#permalink]

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New post 04 Dec 2014, 04:50
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Re: M02-31  [#permalink]

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New post 04 Dec 2014, 09:37
Ok got it thanks
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Re: M02-31  [#permalink]

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New post 13 Jan 2016, 14:30
i am not sure i understand the question properly. does it mean the # will be greater than 50 or not or sum of two digits will be greater than 50 or not. If it is by # it self, 99 is greater than 50, but I think 42 works as well. 4+2=6 is multiple of 18 and 42 less than 50.
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Re: M02-31  [#permalink]

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New post 14 Jan 2016, 00:32
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Re: M02-31  [#permalink]

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New post 15 Aug 2018, 10:39
Bunuel wrote:
rtommy wrote:
i am not sure i understand the question properly. does it mean the # will be greater than 50 or not or sum of two digits will be greater than 50 or not. If it is by # it self, 99 is greater than 50, but I think 42 works as well. 4+2=6 is multiple of 18 and 42 less than 50.


6 is NOT a multiple of 18, it's a factor of 18.


Hi Bunuel

As always super helpful on your explanations

MY question is the following:

Assuming that 1 is a multiple of all integers, couldn't statement 1 also be interpreted as:
x+Y=1 X=1 Y=0 ->number is 10 (which is a 2 digit positive number and respects the premises)
Since there is no restriction of XorY being non zero, other than together must for a two digit integer, this option should be contemplated and therefore statement becomes invalid (an actually correct answer would be E)

What do you think?

Thank you very much,
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Re: M02-31  [#permalink]

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New post 15 Aug 2018, 13:48
bpegenaute wrote:
Bunuel wrote:
rtommy wrote:
i am not sure i understand the question properly. does it mean the # will be greater than 50 or not or sum of two digits will be greater than 50 or not. If it is by # it self, 99 is greater than 50, but I think 42 works as well. 4+2=6 is multiple of 18 and 42 less than 50.


6 is NOT a multiple of 18, it's a factor of 18.


Hi Bunuel

As always super helpful on your explanations

MY question is the following:

Assuming that 1 is a multiple of all integers, couldn't statement 1 also be interpreted as:
x+Y=1 X=1 Y=0 ->number is 10 (which is a 2 digit positive number and respects the premises)
Since there is no restriction of XorY being non zero, other than together must for a two digit integer, this option should be contemplated and therefore statement becomes invalid (an actually correct answer would be E)

What do you think?

Thank you very much,


1 is not a multiple of every integer, it's a FACTOR (a divisor) of every integer.
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Re: M02-31  [#permalink]

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New post 22 May 2019, 23:58
Bunuel wrote:
If \(x\) and \(y\) represent digits of a positive 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


Frankly, I dont see why "divisible by 3" is necessary here? :(.
We can make the question shorter and less confuse with new version of "If \(x\) and \(y\) represent digits of a positive two-digit, is the two-digit number less than 50?"
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Re: M02-31  [#permalink]

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New post 23 May 2019, 00:07
Bunuel wrote:
If \(x\) and \(y\) represent digits of a positive 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


Statement 2 is easy, can be quickly negated.
Multiples of 9: 9, 18, 27

Case 1. 33 <50, 3*3=9
case 2. But 93 >50, 9*3=27

Insufficient.

Statement 1.
Multiples of 18: 18, 36...

two digits x and y can add up to form maximum 18, so it has to be 99 (Also divisible by 3)
So, number is >50. Sufficient.

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Re: M02-31   [#permalink] 23 May 2019, 00:07
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