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# When 15 is divided by y, the remainder is y-3. If y must be

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When 15 is divided by y, the remainder is y-3. If y must be  [#permalink]

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02 Jun 2012, 23:28
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18
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64% (01:26) correct 36% (02:02) wrong based on 90 sessions

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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:
3, 6, 9 and 18
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Joined: 02 Sep 2009
Posts: 65062
Re: When 15 is divided by y, the remainder is y-3. If y must be  [#permalink]

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03 Jun 2012, 02:32
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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:
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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03 Jun 2012, 00:04
2
3
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)
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Re: When 15 is divided by y, the remainder is y-3. If y must be  [#permalink]

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03 Jun 2012, 03:02
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.

Remainder 0 is a possibility too.
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Re: When 15 is divided by y, the remainder is y-3. If y must be  [#permalink]

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04 May 2017, 09:59
Bunuel wrote:
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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Joined: 02 Sep 2009
Posts: 65062
Re: When 15 is divided by y, the remainder is y-3. If y must be  [#permalink]

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04 May 2017, 10:43
Shiv2016 wrote:
Bunuel wrote:
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 ?

Hope it helps.
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Re: When 15 is divided by y, the remainder is y-3. If y must be  [#permalink]

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04 May 2017, 13:52
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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Re: When 15 is divided by y, the remainder is y-3. If y must be  [#permalink]

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06 Mar 2020, 07:33
1
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).

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Re: When 15 is divided by y, the remainder is y-3. If y must be   [#permalink] 06 Mar 2020, 07:33