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# When the positive integer x divided by the positive integer

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Manager
Joined: 31 Jul 2017
Posts: 197
Location: Tajikistan
Re: When the positive integer x divided by the positive integer  [#permalink]

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11 Jun 2018, 07:14
My answer is C and here is why:
I solved this question by a method of backsolving.
1. 5. if x is 5 then y cannot be integer to produce quotient 3, thus cross out 5.
2. 8. if x is 8 then y cannot be integer to produce quotient 3, thus cross out 8
3. 32, if x is 32, then y can be either 10 or 9 to produce quotient 3, and remainder (z) of 2 and 5, respectively. Consequently, if z is 2, then z/y or 2/10 gives reminder of 2 (given in the question), so it works. Lets check another option too; if, z is 5, then 5/9 gives remainder of 5, while the question states it is 2, so this option is out, leaving us only with x 32, y 10, z 2. Hope it helps
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Re: When the positive integer x divided by the positive integer  [#permalink]

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13 Jun 2018, 16:25
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 can create the equations:

x/y = 3 + z/y

x = 3y + z

and

z/y = Q + 2/y

z = Qy + 2

Since the remainder must be less than the divisor in division, we see that z < y. Therefore, Q must be 0 and thus z = 2. Therefore,

x = 3y + 2

Let’s check the Roman numerals.

I. x = 5 ----> 5 = 3y + 2 ----> y = 1

II. x = 8 ----> 8 = 3y + 2 ----> y = 2

III. x = 32 ----> 32 = 3y + 2 ----> y = 10

However, since z < y and z = 2, so 2 < y. Therefore, only 32 is a possible value of x.

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

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16 Jan 2019, 03:07
$$x$$ $$mod$$ $$y = z$$
$$z$$ $$mod$$ $$y=2$$

$$x=3y+z$$ $$\implies$$ $$o≤z<y$$

$$o≤z<y$$ $$in$$ $$conjunction$$ $$with$$ $$z$$ $$mod$$ $$y=2$$ $$\implies$$ $$z=2$$

$$\implies$$ $$x=3y+2$$ $$\implies$$ $$2<y$$

$$\implies$$ $$min_{x}=3(3)+2=11$$
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Re: When the positive integer x divided by the positive integer  [#permalink]

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13 Mar 2019, 10:40
Hi,
here are my two cents for this question

we have X>0, Y>0.
we are told that $$\frac{X}{Y}$$= 3Y+Z--------(I), where Y>Z$$\geqslant{0}$$

Now we are told that $$\frac{Z}{Y}$$ , remainder is 2. this means Y>2

or Z= YK+2 -----(II) where K is integer

Here we have multiple ways to visualize this statement

So X-3Y=Z --------(III)

So dividing the (III) by Y
$$\frac{Z}{Y}$$=$$\frac{X-3Y}{Y}$$,

So $$\frac{X}{Y}- \frac{3Y}{Y}$$

we are told that when $$\frac{Z}{Y}$$we have remainder 2. and from above, the part that gives remainder 2 is $$\frac{X}{Y}$$

Since $$\frac{X}{Y}$$ gives us remainder as Z and from above we have that Z=2

Now
X= 3Y+2
X>0,Y>2,Z>0
We have if y=3 X=11

Since this leaves us with only one option which is D. Hence answer is D

or you could think of this as below

Logically we have , Y>Z$$\geqslant{0}$$ & Y>2.

so whenever we have Y>Z and$$\frac{Z}{Y}$$ the remainder is always Z

Eg. $$\frac{4}{9}$$ remainder will be 4 .

this tells us that Z=2

so Eq(I) becomes X= 3y+2
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Re: When the positive integer x divided by the positive integer   [#permalink] 13 Mar 2019, 10:40

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