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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:
3, 6, 9 and 18

Nice one.
I had left out 3. :oops:

Remainder 0 is a possibility too.
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Bunuel
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:
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.

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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Bunuel
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:
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.

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 re-read the highlighted part.

Hope it helps.
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15=ya+y-3
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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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:
3, 6, 9 and 18

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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