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The positive integers r, s, and t are such that r is
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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
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28 Jul 2010, 01:22
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
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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
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28 Jul 2010, 01:44
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. 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 Ctrap 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
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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
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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
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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
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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
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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
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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
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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
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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
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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 B



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