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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]

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

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New post 28 Jul 2010, 01:22
2
10
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.

Answer: B.
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Re: pretty hard one  [#permalink]

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New post 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.

Answer: B.

many thanks looks easy after the explanation :)

Do you have an idea about the level of this question ?
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Re: pretty hard one  [#permalink]

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New post 28 Jul 2010, 01:44
1
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.

Answer: B.

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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Re: pretty hard one  [#permalink]

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New post 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.
Answer: B.

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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Re: Even or Odd  [#permalink]

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New post 07 Sep 2011, 01:40
jamifahad 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.


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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Re: Even or Odd  [#permalink]

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New post 07 Sep 2011, 01:44
(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

Answer - B
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Re: pretty hard one  [#permalink]

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New post 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.

Answer: B.


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?

Please clarify

Thanks.
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Re: pretty hard one  [#permalink]

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New post 28 Apr 2014, 01:27
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.

Answer: B.


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?

Please clarify

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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New post 23 Jun 2017, 06:04
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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New post 24 Jun 2017, 05:41
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.

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

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New post 08 Aug 2017, 08:02
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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New post 26 Sep 2017, 23:44
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

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

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New post 18 Oct 2017, 07:39
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.

Answer: B.



What happens when t is odd?
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Re: The positive integers r, s, and t are such that r is  [#permalink]

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New post 18 Oct 2017, 07:43
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.

Answer: B.



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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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Re: The positive integers r, s, and t are such that r is &nbs [#permalink] 05 Nov 2018, 14:24
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