When the positive integer x divided by the positive integer

Question

When a positive integer  x  is divided by a 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

Bunuel · Accepted Answer

Official Solution:

When a positive integer  x  is divided by a 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 a positive integer  x  divided by a positive integer  y  gives a quotient of 3 and a remainder of  z . This can be expressed as  x = 3y + z , where  0 \leq z < y  (since the remainder is always less than the divisor).

The problem also states that when  z  is divided by  y , the remainder is 2. Since the remainder is always less than the divisor, it follows that  2 < y . Furthermore, we know that when the divisor ( y  in this case) is greater than the dividend ( z  in this case), the remainder must equal the dividend. Therefore, we can deduce that  z = 2 .

Substituting  z = 2  in  x = 3y + z , we get  x = 3y + 2 . As  y  is a positive integer and  2 < y , the smallest possible value for  y  is 3. Thus, the minimum value of  x  is  x = 3*3 + 2 = 11 . Therefore,  x  cannot be 5 or 8.

To check if  x  could be 32, we can set  y = 10 , which satisfies the condition that  2 < y . Substituting  y = 10  in  x = 3y + 2 , we get  x = 3*10 + 2 = 32 . Therefore,  x  could be 32, and option III is a possible value for  x .

Answer: C

Try NEW remainders  DS question .

Fanja91 · Answer

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.

manpreetsingh86 · Answer

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.

hence answer should be C

ankushbassi · Answer

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.

raul0002 · Answer

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.

justbequiet · Answer

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.

Answer is C.

dheeraj787 · Answer

Bunnel, Z can be 1 also, right ?

Bunuel · Answer

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.

KarishmaB · Answer

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. Answer (C) To understand the concept of grouping, check: https://youtu.be/A5abKfUBFSc https://anaprep.com/number-properties-b ... isibility/

JeffTargetTestPrep · Answer

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.

Answer: C

JeffTargetTestPrep · Answer

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

Probus · Answer

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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When the positive integer x divided by the positive integer

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When the positive integer x divided by the positive integer

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