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# M17-22

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Joined: 02 Sep 2009
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16 Sep 2014, 00:01
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Difficulty:

5% (low)

Question Stats:

81% (01:11) correct 19% (01:07) wrong based on 279 sessions

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If $$x$$ and $$y$$ are positive integers, what is the remainder when $$xy$$ is divided by 4?

(1) When $$x$$ is divided by 4 the remainder is 3

(2) When $$y$$ is divided by 4 the remainder is 2

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16 Sep 2014, 00:01
Official Solution:

Note: Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0 \le r \lt d$$ (remainder is non-negative integer and always less than divisor).

(1) When $$x$$ is divided by 4 the remainder is 3. Not sufficient as no info about $$y$$.

(2) When $$y$$ is divided by 4 the remainder is 2. Not sufficient as no info about $$x$$.

(1)+(2) From (1) $$x=4q+3$$ and from (1) $$y=4p+2$$, so $$xy=(4q+3)(4p+2)=16qp+8q+12p+6$$: now, all terms but the last are divisible by 4 and the last term, 6, yields remainder of 2 when divided by 4. Sufficient.

Or another way: $$x$$ can be: 3, 7, 11, ... and $$y$$ can be: 2, 6, 10, 14, ... You can try several values to see that $$xy$$ always yields remainder of 2 when divided by 4.

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10 May 2017, 02:30
The answer must be option C.

Statement 1 - This gives us info about x not about y. Not sufficient
Statement 2 - This gives us info about y not about x. Not sufficient
Combine 1 & 2 - We get info about both x & y. Hence sufficient.
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11 Nov 2017, 07:38
Bunuel wrote:
Official Solution:

Note: Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0 \le r \lt d$$ (remainder is non-negative integer and always less than divisor).

(1) When $$x$$ is divided by 4 the remainder is 3. Not sufficient as no info about $$y$$.

(2) When $$y$$ is divided by 4 the remainder is 2. Not sufficient as no info about $$x$$.

(1)+(2) From (1) $$x=4q+3$$ and from (1) $$y=4p+2$$, so $$xy=(4q+3)(4p+2)=16qp+8q+12p+6$$: now, all terms but the last are divisible by 4 and the last term, 6, yields remainder of 2 when divided by 4. Sufficient.

Or another way: $$x$$ can be: 3, 7, 11, ... and $$y$$ can be: 2, 6, 10, 14, ... You can try several values to see that $$xy$$ always yields remainder of 2 when divided by 4.

Thanks Bunuel for your explanation. Where can I find similar questions like this?
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11 Nov 2017, 07:42
septwibowo wrote:
Bunuel wrote:
Official Solution:

Note: Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0 \le r \lt d$$ (remainder is non-negative integer and always less than divisor).

(1) When $$x$$ is divided by 4 the remainder is 3. Not sufficient as no info about $$y$$.

(2) When $$y$$ is divided by 4 the remainder is 2. Not sufficient as no info about $$x$$.

(1)+(2) From (1) $$x=4q+3$$ and from (1) $$y=4p+2$$, so $$xy=(4q+3)(4p+2)=16qp+8q+12p+6$$: now, all terms but the last are divisible by 4 and the last term, 6, yields remainder of 2 when divided by 4. Sufficient.

Or another way: $$x$$ can be: 3, 7, 11, ... and $$y$$ can be: 2, 6, 10, 14, ... You can try several values to see that $$xy$$ always yields remainder of 2 when divided by 4.

Thanks Bunuel for your explanation. Where can I find similar questions like this?

6. Remainders

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26 Apr 2018, 19:18
I was checking for possible values of x and y, so when x=7 and y=2, xy will be 72. In this case 72/4, the remainder will be zero. Not sure if option C will be the correct answer here.
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26 Apr 2018, 22:33
VishakhaSingh13 wrote:
I was checking for possible values of x and y, so when x=7 and y=2, xy will be 72. In this case 72/4, the remainder will be zero. Not sure if option C will be the correct answer here.

xy there means x*y, x multiplied by y not a two-digit integer xy, recall that multiplication sign is often omitted. If xy were a two-digit number it would have been mentioned explicitly.
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Re: M17-22 &nbs [#permalink] 26 Apr 2018, 22:33
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# M17-22

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