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

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Intern
Joined: 22 Jun 2010
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The positive integers r, s, and t are such that r is  [#permalink]

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28 Jul 2010, 02:01
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Question Stats:

47% (01:28) correct 53% (01:31) wrong based on 546 sessions

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 49275
The positive integers r, s, and t are such that r is  [#permalink]

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28 Jul 2010, 02:22
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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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28 Jul 2010, 02:37
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: 49275

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28 Jul 2010, 02:44
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.
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07 Aug 2010, 05:47
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...
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Joined: 20 Dec 2010
Posts: 1868

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07 Sep 2011, 02:40
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.

r/s->Integer
s/t->Integer

(1) st is odd.
Both s and t are odd.

r=9; s=3; t=1
r=6; s=3; t=1
Not Sufficient.

(2) rt is even.
Either r or t or both are even.
r=even. Fantastic.
t=Even; s becomes even; t has to be even.

See this:
s/t=Integer; t=Even; s=Even*Integer=Even;
r/s=Integer; s=Even; r=Even*Integer=Even;

So, r is definitely even.
Sufficient.

Ans: "B"
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Location: United States (IN)
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07 Sep 2011, 02:44
(1)

st is odd means that s and t are odd, but r can be even or odd

e.g r = 10, s = 5, t = 1

r = 15, s = 5, t = 1

Insufficient

(2)

If rt is even then at least one of t or r is even

So r = k*s

s = m*t

=> r = k*m*t (where k and m are positive integers)

=> rt = r * r/(km) = r^2/km is an even integer (as per question)

=> r^2 is even

=> r is even

Sufficient

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Manager
Joined: 10 Mar 2014
Posts: 205

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27 Apr 2014, 03:01
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: 49275

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28 Apr 2014, 02:27
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: The positive integers r, s, and t are such that r is  [#permalink]

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23 Jun 2017, 07:04
This is how I solved it:

Is r even?

r is divisible by s, hence r = ns
s is divisible by t, hence s = mt
combining, r = nmt

1-> st = odd. This implies, both s and t are odd. no impact on r. insuff.
2 -> rt = even
Hence, r = even or t = even or both = even
If t = even, then r = nmt = even
If r = odd, then r is even for rt = even
Suff.
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Re: The positive integers r, s, and t are such that r is  [#permalink]

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24 Jun 2017, 06:41
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.

Solution:

The Tricky part in this question is to remember that r,s and t are integers.

Statement 1: No information about r is given. r can be either even or odd. Insufficient.

Statement 2: r has to be even as Odd number divided by odd will never yield a even number and Odd number divided by even will not yield a integer.
Therefore r must be even.

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Joined: 23 Jul 2015
Posts: 161
Re: The positive integers r, s, and t are such that r is  [#permalink]

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08 Aug 2017, 09:02
r = sx
s = ty

r = tyx

1. st is odd
O = O*y ==> y = odd
r = O*O*x.. we don't know x. Therefore, not sufficient

2. rt = even
r = tyx
O=E*y*x or
E = O*y*x

evaluate whether both outcomes are possible.
when r is odd
t, y, and x must be odd which is not possible since t is even in this case.

thus, r is even

Ans. B
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Re: The positive integers r, s, and t are such that r is  [#permalink]

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27 Sep 2017, 00:44
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.

The thing about these problems is that even though they appear to be very simple math they are designed to be very misleading

Statement 1

you could have 12 3 1 or 9 3 1

insuff

Statement 2

R cannot be odd because t has to be a multiple of 2- - because t has to be a factor of r

suff

B
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Posts: 87
Re: The positive integers r, s, and t are such that r is  [#permalink]

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18 Oct 2017, 08:39
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.

What happens when t is odd?
Math Expert
Joined: 02 Sep 2009
Posts: 49275
Re: The positive integers r, s, and t are such that r is  [#permalink]

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18 Oct 2017, 08:43
zanaik89 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.

What happens when t is odd?

For (2) if t is odd then r must be even right away, because rt=even.
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Re: The positive integers r, s, and t are such that r is &nbs [#permalink] 18 Oct 2017, 08:43
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