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When the positive integer x divided by the positive integer
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17 Jun 2014, 01:58
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When the positive integer x divided by the positive integer
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17 Jun 2014, 01:59
SOLUTIONWhen 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. 32A. 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 \(x=3*10+2=32\). Answer: C. Try NEW remainders DS question.
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Re: When the positive integer x divided by the positive integer
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17 Jun 2014, 02:21
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
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17 Jun 2014, 03:11
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
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17 Jun 2014, 07:50
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. hence answer should be C



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Re: When the positive integer x divided by the positive integer
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17 Jun 2014, 10:27
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
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17 Jun 2014, 20:42
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 \(z= x3y\) When \(z\) is divided by \(y\), the remainder is 2 So we have \(z= by+2\) or by+2=x3y 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
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18 Jun 2014, 00:27
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
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18 Jun 2014, 01:49
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.



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Re: When the positive integer x divided by the positive integer
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18 Jun 2014, 06:38
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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Re: When the positive integer x divided by the positive integer
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22 Jun 2014, 03:14
SOLUTIONWhen 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. 32A. 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 \(x=3*10+2=32\). Answer: C. 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
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22 Jun 2014, 04:39
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
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16 Jul 2014, 11:31
Bunnel, Z can be 1 also, right ?



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Re: When the positive integer x divided by the positive integer
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16 Jul 2014, 11:38
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
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05 Nov 2015, 17:53
Bunuel wrote: SOLUTIONWhen 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. 32A. 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 \(x=3*10+2=32\). Answer: C. Try NEW remainders DS question. 1 question about this solution Why is 2<y for x = 3y+2 and not 2<3y ? This is the only part i cannot grasp.



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Re: When the positive integer x divided by the positive integer
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05 Nov 2015, 19:56
GMATDemiGod wrote: Bunuel wrote: SOLUTIONWhen 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. 32A. 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 \(x=3*10+2=32\). Answer: C. Try NEW remainders DS question. 1 question about this solution 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
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05 Nov 2015, 21:19
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. Answer (C) To understand the concept of grouping, check: http://www.veritasprep.com/blog/2011/04 ... unraveled/http://www.veritasprep.com/blog/2011/04 ... yapplied/http://www.veritasprep.com/blog/2011/05 ... emainders/
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Re: When the positive integer x divided by the positive integer
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06 Nov 2015, 11:09
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
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09 Nov 2016, 10:26
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. Answer: C
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When the positive integer x divided by the positive integer
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08 Jun 2018, 18:56
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, (xz)/y3=(z2)/y➡ x=3y+2z2 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.




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