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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.

(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.

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 ?

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.
_________________

(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...

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?

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.
_________________

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

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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.

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

Show Tags

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.

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

Show Tags

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.

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.
_________________