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13 Nov 2010, 10:10
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
86% (01:13) correct
14%
(01:36)
wrong
based on 206
sessions
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Not Attempted Yet
If X and Y are positive integers, what is the remainder of \(\frac{XY}{4}\) ?
(1) The remainder of \(\frac{X}{4}\) is 3
(2) The remainder of \(\frac{Y}{4}\) is 2
(1) The remainder of \(\frac{X}{4}\) is 3
(2) The remainder of \(\frac{Y}{4}\) is 2
13 Nov 2010, 10:32
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vivaslluis
If \(X\) and \(Y\) are positive integers, what is the remainder of \(\frac{XY}{4}\) ?
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 \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).
(1) The remainder of \(\frac{X}{4}\) is 3 --> not sufficient as no info about \(y\).
(2) The remainder of \(\frac{Y}{4}\) is 2 --> not sufficient as no info about \(x\).
(1)+(2) From (1) \(x=4q+3\) and from (1) \(y=4p+2\) --> \(xy=(4q+3)(4p+2)=16qp+8q+12p+6\): 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.
Answer: C.
General Discussion
13 Nov 2010, 10:40
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Ohhhhhh! Ok... Now I see my mistake in the exercise.
Thank you!
Thank you!
