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# The positive integers r, s, and t are such that r is

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The positive integers r, s, and t are such that r is [#permalink]  28 Jul 2010, 01:01
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Question Stats:

43% (02:22) correct 57% (01:21) wrong based on 164 sessions
The positive integers r, s, and t are such that r is divisible by s and s is divisible by t. Is r even?

(1) st is odd.
(2) rt is even.
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 31286
Followers: 5348

Kudos [?]: 62176 [1] , given: 9444

Re: pretty hard one [#permalink]  28 Jul 2010, 01:22
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Expert's post
mehdiov wrote:
The positive integers r, s, and t are such that r is divisible by s
and s is divisible by t. Is r even?
(1) st is odd.
(2) rt is even.

(1) $$st=odd$$, clearly not sufficient as no info about $$r$$, for example if $$r=6$$, $$s=1$$ and $$t=1$$ then answer is YES but if $$r=3$$, $$s=1$$ and $$t=1$$ then the answer is NO.

(2) $$rt=even$$. For product of 2 integers to be even either one or both must be even. Can $$r$$ not to be even? The only chance would be if $$t$$ is even and $$r$$ is odd. Let's check if this scenario is possible: if $$t$$ is even, so must be $$s$$, as $$s$$ is divisible by $$t$$ (if an integer is divisible by even it's even too). Now, if $$s$$ is even so must be $$r$$ by the very same reasoning. So scenario when $$r$$ is not even is not possible --> $$r=even$$. Sufficient.

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Joined: 22 Jun 2010
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Kudos [?]: 59 [0], given: 1

Re: pretty hard one [#permalink]  28 Jul 2010, 01:37
Bunuel wrote:
mehdiov wrote:
The positive integers r, s, and t are such that r is divisible by s
and s is divisible by t. Is r even?
(1) st is odd.
(2) rt is even.

(1) $$st=odd$$, clearly not sufficient as no info about $$r$$, for example if $$r=6$$, $$s=1$$ and $$t=1$$ then answer is YES but if $$r=3$$, $$s=1$$ and $$t=1$$ then the answer is NO.

(2) $$rt=even$$. For product of 2 integers to be even either one or both must be even. Can $$r$$ not to be even? The only chance would be if $$t$$ is even and $$r$$ is odd. Let's check if this scenario is possible: if $$t$$ is even, so must be $$s$$, as $$s$$ is divisible by $$t$$ (if an integer is divisible by even it's even too). Now, if $$s$$ is even so must be $$r$$ by the very same reasoning. So scenario when $$r$$ is not even is not possible --> $$r=even$$. Sufficient.

many thanks looks easy after the explanation

Do you have an idea about the level of this question ?
Math Expert
Joined: 02 Sep 2009
Posts: 31286
Followers: 5348

Kudos [?]: 62176 [0], given: 9444

Re: pretty hard one [#permalink]  28 Jul 2010, 01:44
Expert's post
mehdiov wrote:
Bunuel wrote:
mehdiov wrote:
The positive integers r, s, and t are such that r is divisible by s
and s is divisible by t. Is r even?
(1) st is odd.
(2) rt is even.

(1) $$st=odd$$, clearly not sufficient as no info about $$r$$, for example if $$r=6$$, $$s=1$$ and $$t=1$$ then answer is YES but if $$r=3$$, $$s=1$$ and $$t=1$$ then the answer is NO.

(2) $$rt=even$$. For product of 2 integers to be even either one or both must be even. Can $$r$$ not to be even? The only chance would be if $$t$$ is even and $$r$$ is odd. Let's check if this scenario is possible: if $$t$$ is even, so must be $$s$$, as $$s$$ is divisible by $$t$$ (if an integer is divisible by even it's even too). Now, if $$s$$ is even so must be $$r$$ by the very same reasoning. So scenario when $$r$$ is not even is not possible --> $$r=even$$. Sufficient.

many thanks looks easy after the explanation

Do you have an idea about the level of this question ?

Not very hard (600+) but tricky, as it's C-trap question: the question which is obviously sufficient if we take statements together. When we see such questions we should become very suspicious.
_________________
Director
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Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
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Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
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Kudos [?]: 116 [0], given: 15

Re: pretty hard one [#permalink]  07 Aug 2010, 02:44
Not very hard (600+) but tricky, as it's C-trap question: the question which is obviously sufficient if we take statements together. When we see such questions we should become very suspicious.[/quote]

what is C-trap?
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Consider kudos, they are good for health

Manager
Joined: 24 Apr 2010
Posts: 62
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Kudos [?]: 7 [0], given: 0

Re: pretty hard one [#permalink]  07 Aug 2010, 04:47
Bunuel wrote:

(1) $$st=odd$$, clearly not sufficient as no info about $$r$$, for example if $$r=6$$, $$s=1$$ and $$t=1$$ then answer is YES but if $$r=3$$, $$s=1$$ and $$t=1$$ then the answer is NO.

thanks...i was able to get to B but may be in 3 minutes.....
i complicated the question thinking like 2 4 8 and not thinking infact one can be one number or 2 numbers can be same 8 2 2 and so on...
Manager
Joined: 10 Mar 2014
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Kudos [?]: 41 [0], given: 13

Re: pretty hard one [#permalink]  27 Apr 2014, 02:01
Bunuel wrote:
mehdiov wrote:
The positive integers r, s, and t are such that r is divisible by s
and s is divisible by t. Is r even?
(1) st is odd.
(2) rt is even.

(1) $$st=odd$$, clearly not sufficient as no info about $$r$$, for example if $$r=6$$, $$s=1$$ and $$t=1$$ then answer is YES but if $$r=3$$, $$s=1$$ and $$t=1$$ then the answer is NO.

(2) $$rt=even$$. For product of 2 integers to be even either one or both must be even. Can $$r$$ not to be even? The only chance would be if $$t$$ is even and $$r$$ is odd. Let's check if this scenario is possible: if $$t$$ is even, so must be $$s$$, as $$s$$ is divisible by $$t$$ (if an integer is divisible by even it's even too). Now, if $$s$$ is even so must be $$r$$ by the very same reasoning. So scenario when $$r$$ is not even is not possible --> $$r=even$$. Sufficient.

HI Bunnel,

I have a doubt on this.

Generally we treat both the statements as seprate statements. then why are you mixing them.

If I will go with st2 i can r can be even or odd because rt = even ( r and t both can be even or one of them is even) now if we refer even to r and t then st1 will contradict.

is this the reason you are not considering both r and t as even?

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 31286
Followers: 5348

Kudos [?]: 62176 [0], given: 9444

Re: pretty hard one [#permalink]  28 Apr 2014, 01:27
Expert's post
PathFinder007 wrote:
Bunuel wrote:
mehdiov wrote:
The positive integers r, s, and t are such that r is divisible by s
and s is divisible by t. Is r even?
(1) st is odd.
(2) rt is even.

(1) $$st=odd$$, clearly not sufficient as no info about $$r$$, for example if $$r=6$$, $$s=1$$ and $$t=1$$ then answer is YES but if $$r=3$$, $$s=1$$ and $$t=1$$ then the answer is NO.

(2) $$rt=even$$. For product of 2 integers to be even either one or both must be even. Can $$r$$ not to be even? The only chance would be if $$t$$ is even and $$r$$ is odd. Let's check if this scenario is possible: if $$t$$ is even, so must be $$s$$, as $$s$$ is divisible by $$t$$ (if an integer is divisible by even it's even too). Now, if $$s$$ is even so must be $$r$$ by the very same reasoning. So scenario when $$r$$ is not even is not possible --> $$r=even$$. Sufficient.

HI Bunnel,

I have a doubt on this.

Generally we treat both the statements as seprate statements. then why are you mixing them.

If I will go with st2 i can r can be even or odd because rt = even ( r and t both can be even or one of them is even) now if we refer even to r and t then st1 will contradict.

is this the reason you are not considering both r and t as even?

Thanks.

The statements do not contradict: st is odd and rt is even is possible when r is even and both s and t are odd.
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Re: pretty hard one   [#permalink] 28 Apr 2014, 01:27
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# The positive integers r, s, and t are such that r is

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