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# When the positive integer x divided by the positive integer

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When the positive integer x divided by the positive integer  [#permalink]

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17 Jun 2014, 02:58
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When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

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Posts: 47977
When the positive integer x divided by the positive integer  [#permalink]

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17 Jun 2014, 02:59
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SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and $$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Try NEW remainders DS question.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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17 Jun 2014, 03:21
1
My solution might be awful. But I did it the following way:

X/Y = 3+Z/Y.

Z/Y=2/Y.

Then I just plugged in 5 and tried adjusting Y.

There is no way I could get 5/Y= 3+2/Y as Y = 2 gives me 2+1/Y, and 1 would give me 5+0/Y.
Hence I is not possible.

Then 8 wouldn't work either.
Set Y = 2, gives X/Y=4+0/Y, no remainder.
Set Y = 3, gives X/Y=2+2/Y (The remainder is correct here, but we wanted 3+2/Y.
Hence II is not a possible value of X.

Lastly, X = 32.
I directly saw that setting Y = 10 would give me 3+2/Y. I didn't need to try any other options.

Therefore, the answer is C. Only III is a possible value.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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17 Jun 2014, 04:11
x=3y+z---(1)
z=ky+2----(2)
Combining 1 & 2, x=(3+k)y + 2
Already, equation 1 tells us that the remainder when x is divided by y is z.
So, applying the above result to equation 2 would give z=2
Thus x=3y+2
In other words, x when divided by 3 leaves remainder 2. Thus 5,8,32 all satisfy this criteria. Hence the answer is E.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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17 Jun 2014, 08:50
1
Bunuel wrote:

When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

when x is divided by y, remainder will always be less than the denominator y.

also, if numerator is less than denominator then remainder is equal to the numerator.

now coming to the question x= 3y+ z ----------------1)
using the above property we can conclude that x>y>z --------------------2)

also when z is divided by y remainder is 2. now since z is less than y therefore z must be equal to 2.

therefore by substituting the value of z=2 in the equation 1 we have

x=3y+2

also from 2 we have y>2

thus values x=5 and x= 8 are eliminated and thus x=32 is the only possible answer.

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Re: When the positive integer x divided by the positive integer  [#permalink]

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17 Jun 2014, 11:27
2
We have got 2 equations :-
Eqn1 - x=3y+z
Eqn2 - z=Qy+2(where Q is the quotient).

32 is the only option that can be written in accordance with equation 1 and leaves remainder of 2 when divided by 10.Hence,answer is C.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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17 Jun 2014, 21:42
Bunuel wrote:
When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$

Can be written as $$x= 3y+z$$ or $$z= x-3y$$

When $$z$$ is divided by $$y$$, the remainder is 2

So we have $$z= by+2$$

or by+2=x-3y or
x = by+3y+2 = 3y+z ---------> by+2=z=x

Now if by= even, then z is even and hence x is even.
but if by is odd then z is odd and thus x is odd

Since we need to find what could be the possible value...then any of three options could be the value...So Ans E
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Re: When the positive integer x divided by the positive integer  [#permalink]

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18 Jun 2014, 01:27
1
When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

x = 3y+z. (Now y can't take the value zero as z/y can't be defined)

when x = 5
y can only be 1 and thus z = 2 and the remainder is 0 for z/y

when x = 8
y = 1 and z = 5 and the remainder is 0 for z/y

y = 2 and z = 2 and the remainder is 0 for z/y

thus, only option C is applicable.

when x = 32
y can take 3 or 10 and z will be 23 or 2 respectively which gives the remainder as 2 for z/y.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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18 Jun 2014, 02:49
1
I did number picking on this one.

$$x/y=3y+z/y$$ ----(1)
$$z/y=by+2/y$$ ----(2)

Both $$x=5$$ & $$x=8$$ are too small to have a quotient of $$3$$ and still have $$y$$ as an integer.

Only $$x=32$$ is big enough to have $$y$$ as a positive integer with a quotient of $$3$$.

With this, only $$y=10$$ can complete the solution.

for $$x/y$$ --> $$32/10=3(10)+2/10$$
for $$z/y$$ --> $$2/10=2$$

Therefore, only $$x=32$$ can be the value of $$x$$ given the restrictions in the question.

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Re: When the positive integer x divided by the positive integer  [#permalink]

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18 Jun 2014, 07:38
For me its E
i.e. All 3 values are possible

Given :
1. x = 3y + z

From here we can interpret that z < y .

2. Now given that when z is divided by y the remainder is 2.

Since z < y , i.e. Numerator is smaller that Denominator
Therefore we can safely assume the smallest possible value of z to be 2. (As we have to find the values that could be possible )

So x=3y + 2

Given Y is a +ve Integer
So checking for given Values which give us a Integer value of y

For 5 , y=1
For 8 , y= 2
For 32 , y=10

All satisfy the Equation.
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Posts: 47977
Re: When the positive integer x divided by the positive integer  [#permalink]

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22 Jun 2014, 04:14
1
2
SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and $$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Kudos points given to correct solutions above.

Try NEW remainders DS question.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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22 Jun 2014, 05:39
niyantg wrote:
For me its E
i.e. All 3 values are possible

Given :
1. x = 3y + z

From here we can interpret that z < y .
2. Now given that when z is divided by y the remainder is 2.

Since z < y , i.e. Numerator is smaller that Denominator
Therefore we can safely assume the smallest possible value of z to be 2. (As we have to find the values that could be possible )

So x=3y + 2

Given Y is a +ve Integer
So checking for given Values which give us a Integer value of y

For 5 , y=1
For 8 , y= 2
For 32 , y=10

All satisfy the Equation.

As per above two highlighted statements, y can't take value of 1 or 2. y>2. Only 32 can satisfy the given conditions.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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16 Jul 2014, 12:31
Bunnel, Z can be 1 also, right ?
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Re: When the positive integer x divided by the positive integer  [#permalink]

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16 Jul 2014, 12:38
dheeraj787 wrote:
Bunnel, Z can be 1 also, right ?

No.

We are told that when z is divided by y, the remainder is 2 and we know also that y>z. Now, if z=1, then the reminder would be 1, not 2.

When z is divided by y, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, z=2.

Hope it's clear.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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05 Nov 2015, 18:53
Bunuel wrote:
SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and$$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Try NEW remainders DS question.

Why is 2<y for x = 3y+2 and not 2<3y ?

This is the only part i cannot grasp.
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Posts: 47977
Re: When the positive integer x divided by the positive integer  [#permalink]

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05 Nov 2015, 20:56
GMATDemiGod wrote:
Bunuel wrote:
SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and$$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Try NEW remainders DS question.

Why is 2<y for x = 3y+2 and not 2<3y ?

This is the only part i cannot grasp.

Check highlighted part.
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Re: When the positive integer x divided by the positive integer  [#permalink]

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05 Nov 2015, 22:19
1
Bunuel wrote:

When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

This question is a great example of how understanding the concept of grouping for division can help you solve tricky questions easily.

x divided by y gives quotient 3 and remainder z.
What I see is x split into 3 groups having y balls each and z balls lying separately (these form the remainder)

When z is divided by y, remainder is 2.
z was the remainder when x was divided by y. So z is less than y. So when z is divided by y, the quotient will be 0 and the remainder will be z only. This means that z is 2. So now my figure morphs into x divided into 3 groups of y balls each with 2 balls lying on the side. The 3 groups will have more than 2 balls each, of course.

x = 3y + 2 where y is greater than 2.
So x cannot be 5 or 8. It can be 32 because y = 10 in that case.

To understand the concept of grouping, check:
http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/
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Re: When the positive integer x divided by the positive integer  [#permalink]

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06 Nov 2015, 12:09
Bunuel wrote:
When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$

$$\frac{x}{y} = 3 + z$$

Bunuel wrote:
When $$z$$ is divided by $$y$$, the remainder is 2.

$$\frac{z}{y} = k + 2$$

Bunuel wrote:
Which of the following could be the value of $$x$$?

Now lets start plugging in the options -

I. 5 - No possible satisfying conditions can be found out.
II. 8 - No possible satisfying conditions can be found out.

III. 32 - Lets try

$$\frac{32}{y} = 3 + z$$
$$\frac{32}{9} = 3(Quotient) + 5(Remainder)$$

Lets try the second step now -

$$\frac{z}{y} = k + 2$$

$$\frac{5}{y} = k +2$$

$$\frac{5}{3} = 1 (Quotient) +2(Remainder)$$

Only III Satisfices th condition m hence aswer is (C)
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Re: When the positive integer x divided by the positive integer  [#permalink]

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09 Nov 2016, 11:26
1
Bunuel wrote:

When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

We are given that when the positive integer x is divided by the positive integer y, the quotient is 3 and the remainder is z. When z is divided by y, the remainder is 2. To determine which of the Roman numeral choices can be x, we can test each choice.

I. x = 5

If x = 5, no positive integer y exists that can produce a quotient of 3 when 5 is divided by y.

For example:

5/1 produces a quotient of 5.

5/2 produces a quotient of 2.

5/3 or 5/4 produces a quotient of 1.

Thus x cannot be 5.

II. x = 8

If x = 8, no positive integer y exists that can produce a quotient of 3 when 8 is divided by y.

For example:

8/1 produces a quotient of 8.

8/2 produces a quotient of 4.

8/3 or 8/4 produces a quotient of 2.

8/5, 8/6, 8/7, or 8/8 produces a quotient of 1.

Thus x cannot be 8.

III. x = 32

We see that x can be 32 since 32/10 produces a quotient of 3 and a remainder of 2, and when 2 is divided by 10, the quotient is 0 and the remainder is 2.

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When the positive integer x divided by the positive integer  [#permalink]

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08 Jun 2018, 19:56
Bunuel wrote:

When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

we know that z<y because the remainder must be less than the divisor
so z/y must give a quotient of 0
thus, (x-z)/y-3=(z-2)/y➡
x=3y+2z-2
I. if x=5, then 3y+2z=7; y=1; z=2; z>y no
II. if x=8, then 3y+2z=10; y=2; z=2; z=y no
III. if x=32, then 3y+2z=34; y=10; z=2; z<y yes
III only
C.
When the positive integer x divided by the positive integer &nbs [#permalink] 08 Jun 2018, 19:56

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