When the positive integer x divided by the positive integer

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.

Answer; C
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\(x\) \(mod\) \(y = z\)
\(z\) \(mod\) \(y=2\)


\(x=3y+z\) \(\implies\) \(o≤z<y\)

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

\(\implies\) \(x=3y+2\) \(\implies\) \(2<y\)

\(\implies\) \(min_{x}=3(3)+2=11\)
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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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Here is an easy way.
If x = 5, then y can be 1, 2, 3, 4, or 5
But the condition is that y is multiplied 3 times (quotient is 3)
So, none of the 5 mentioned integer values can be equal to y because 1*5 = 5 but the quotient should be 3.
2*3 = 6 but that again isn't possible because the remainder must be non-negative.
Thus 2, 3, 4, and 5 will leave a negative remainder when the quotient of y is 3 and x = 5.
So, x = 5 doesn't work

Similarly, if x = 8 then y = 1, 2, 3, 4, 5, 6, 7, or 8
1*8 = 8 but quotient should be 3. Doesn't work
2*4 = 8 but quotient should be 3. Doesn't work
3*3 = 9 negative remainder.
for all other values of y, remainder will be negative so they don't work.
x cannot be 8

So, x has got to be 32 because there is no option saying "none of the above"

Still, let's see
x = 32
quotient = 3
y < 32/3 ~11
y = 10
x = 3y + 2
z = 2
z = 0y + 2
Everything matches.
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