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03 Jun 2012, 00:28
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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When 15 is divided by y, the remainder is y-3. If y must be an integer, what are all the possible values of y?
OA:
OA:
3, 6, 9 and 18
03 Jun 2012, 03:32
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AceofSpades
When 15 is divided by y, the remainder is y-3. If y must be an integer, what are all the possible values of y?
OA:
OA:
3, 6, 9 and 18
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).
So, we have that \(15=qy+(y-3)\) and \(remainder=y-3\geq{0}\) --> \(y\geq{3}\). From \(15=qy+(y-3)\) --> \(y(q+1)=18\) --> \(y\) must be factor of 18 but greater than or equal to 3. Thus \(y\) can be: 3, 6, 9, or 18.
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Expert reply
03 Jun 2012, 01:04
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Think of it this way
15=ky+y-3
Where k is an integer
I.e y=18/(k+1)
Now put values of k such that y is an integer
I.e k = 0,1,2,5
Y= 3,6,9,18
However, for k=8 the equation holds but the remainder in that case is 1 not -1(thats when y=2)
15=ky+y-3
Where k is an integer
I.e y=18/(k+1)
Now put values of k such that y is an integer
I.e k = 0,1,2,5
Y= 3,6,9,18
However, for k=8 the equation holds but the remainder in that case is 1 not -1(thats when y=2)
