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When 15 is divided by y, the remainder is y3. If y must be
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02 Jun 2012, 23:28
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When 15 is divided by y, the remainder is y3. If y must be an integer, what are all the possible values of y? OA:




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Re: When 15 is divided by y, the remainder is y3. If y must be
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03 Jun 2012, 02:32
AceofSpades wrote: When 15 is divided by y, the remainder is y3. If y must be an integer, what are all the possible values of y? OA: 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 nonnegative integer and always less than divisor).So, we have that \(15=qy+(y3)\) and \(remainder=y3\geq{0}\) > \(y\geq{3}\). From \(15=qy+(y3)\) > \(y(q+1)=18\) > \(y\) 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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03 Jun 2012, 00:04
Think of it this way 15=ky+y3 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)
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Re: When 15 is divided by y, the remainder is y3. If y must be
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03 Jun 2012, 03:02
AceofSpades wrote: When 15 is divided by y, the remainder is y3. If y must be an integer, what are all the possible values of y? OA: Nice one. I had left out 3. Remainder 0 is a possibility too.



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Re: When 15 is divided by y, the remainder is y3. If y must be
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04 May 2017, 09:59
Bunuel wrote: AceofSpades wrote: When 15 is divided by y, the remainder is y3. If y must be an integer, what are all the possible values of y? OA: 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 nonnegative integer and always less than divisor).So, we have that \(15=qy+(y3)\) and \(remainder=y3\geq{0}\) > \(y\geq{3}\). From \(15=qy+(y3)\) > \(y(q+1)=18\) > \(y\) must be factor of 18 but greater than or equal to 3. Thus \(y\) can be: 3, 6, 9, or 18. I directly solved it. 18=y(q+1) As y is an integer, it should divide 18 i.e. it has to be a factor of 18. Total values y can take are 6 Why do we have to consider y>=3 ?



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Re: When 15 is divided by y, the remainder is y3. If y must be
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04 May 2017, 10:43
Shiv2016 wrote: Bunuel wrote: AceofSpades wrote: When 15 is divided by y, the remainder is y3. If y must be an integer, what are all the possible values of y? OA: 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 nonnegative integer and always less than divisor).So, we have that \(15=qy+(y3)\) and \(remainder=y3\geq{0}\) > \(y\geq{3}\). From \(15=qy+(y3)\) > \(y(q+1)=18\) > \(y\) must be factor of 18 but greater than or equal to 3. Thus \(y\) can be: 3, 6, 9, or 18. I directly solved it. 18=y(q+1) As y is an integer, it should divide 18 i.e. it has to be a factor of 18. Total values y can take are 6 Why do we have to consider y>=3 ? Please reread the highlighted part. Hope it helps.
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Re: When 15 is divided by y, the remainder is y3. If y must be
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04 May 2017, 13:52
15=ya+y3 18=y(a+1) factors of 18 are 1,2,3,6,9,18 y can take 3,6,9,18



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Re: When 15 is divided by y, the remainder is y3. If y must be
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06 Mar 2020, 07:33
AceofSpades wrote: When 15 is divided by y, the remainder is y3. If y must be an integer, what are all the possible values of y? OA: We can create the equation: 15/y = Q + (y  3)/y 15 = Qy + y  3 18 = y(Q + 1) 18/y = Q + 1 Thus, we see that y must be a factor of 18. So y could be 1, 2, 3, 6, 9, or 18. However, since y has to be at least 3 (notice that that the remainder y  3 is at least 0), y couldn’t be 1 or 2 (but it could be any of the other integers mentioned above). Answer: 3, 6, 9, 18
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Re: When 15 is divided by y, the remainder is y3. If y must be
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