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When x is divided by 3 the remainder is 2, and when y is divided by 7

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When x is divided by 3 the remainder is 2, and when y is divided by 7  [#permalink]

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New post 30 Dec 2018, 22:56
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Question Stats:

43% (03:16) correct 57% (01:00) wrong based on 7 sessions

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When x is divided by 3 the remainder is 2, and when y is divided by \(7\), the remainder is \(4\). What is the remainder when \(x+y\) is divided by \(21\)?

I) \(x^2\) divided by \(7\) leaves a remainder of \(4\).
II) \(y-4\) is divisible by \(3\).

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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7  [#permalink]

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New post 30 Dec 2018, 23:05
So what I don't understand is that my algebraic solution would indicate C as the correct answer. Please let me know where I went wrong:

Stem: \(x=3q+2\) and \(y=7p+4\) \(–>\) \(x+y=3q+7p+6\)
Hence, our question becomes: is \(3q+7p=21k\). This is the case whenever \(3q\) is divisible by \(7\) and \(7p\) is divisible by \(3\).

Statement I: \(x^2=9q^2+12q+4=7L+4 –> q=7L\)
Still, Not sufficient as no information is given about y

Statement II: \(y-4=3m\). Now from the stem, we know that \(y=7p+4\) hence, this statement tells us that \(7p=3m\)
Still, Not sufficient as no information is given about x

Statement I&II combined:
From I) we know that \(3q\) is divisible by \(7\), hence, \(3q\) must be a multiple of \(21\)
From II) we know that \(7p\) is divisible by \(3\), hence, \(7p\) must be a multiple of \(21\)

Thus, \(x+y=(3q+7p)+6= 21b+6\)

Please let me know where I went wrong!
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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7  [#permalink]

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New post 30 Dec 2018, 23:29
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Y can have multiple values that is 25 and 46 , hence the option E is correct

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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7  [#permalink]

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New post 30 Dec 2018, 23:35
vishumangal wrote:
Y can have multiple values that is 25 and 46 , hence the option E is correct

Posted from my mobile device


thanks, I am aware that this is true, however, what is incorrect in my solution?
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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7  [#permalink]

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New post 31 Dec 2018, 00:45
ghnlrug wrote:
When x is divided by 3 the remainder is 2, and when y is divided by \(7\), the remainder is \(4\). What is the remainder when \(x+y\) is divided by \(21\)?

I) \(x^2\) divided by \(7\) leaves a remainder of \(4\).
II) \(y-4\) is divisible by \(3\).


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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7 &nbs [#permalink] 31 Dec 2018, 00:45
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